Apex Calculus Ii

Improper Integral AP X Version 4.0 x E +∞ x dx x2 + 1 ∫-∞ 1 + x2 u = 1 + x2 +∞ x du = 2x dx CALCULUS II = 1 2 dx 0.4 2 ∫-∞ 1 +x2 +∞ 1 2 0.2 = 2 ln(1 + x ) ∞ -10 -5 - X = {∞ - ∞}, 5 10 indeterminate form -0.2 -0.4 Does Not Exist. Gregory Hartman, Ph.D. Department of Applied Mathematics Virginia Military Institute Arc Length What is the arc length, the length of the curve, of y = f(x), a ≤ x ≤ b? Y Contributing Authors Troy Siemers, Ph.D. y = f(x) ds Department of Applied dy dx Mathematics Virginia Military Institute Brian Heinold, Ph.D. Department of Mathematics and Computer Science X a x b Mount Saint Mary’s University Theorem Let y = f(x) be differentiable for a ≤ x ≤ b. Then its arc length on that interval is

b Dimplekumar Chalishajar, Ph.D. _dy 2 s = ∫ 1 +   dx. a dx Department of Applied Mathematics Derivation Virginia Military Institute f is differentiable on the interval. So f is differentiable at x and therefore locally linear or equivalently 'asymptotically straight' there. Therefore Editor 2 2 2 ds ≈ dx + dy Theorem of Pythagoras Jennifer Bowen, Ph.D. _dy 2 2 = [1 + ( dx   dx Department of Mathematics and Computing Science, + dy 2 ds ≈ 1 dx dx b The College of Wooster dy -2 ⟹ s = 1 +   dx. ∫a dx

Apex Infinitesimal Calculus, Volume II Second Preliminary Version William Freed Concordia University of Edmonton CLICK on desired section Volume II Review 1- Infinitesimals. Review 2 - Derivatives and Integrals 6 Other Elementary Functions. Inverse Functions. 6.1 Theory of Inverse Functions 6.2 Exponential Functions Review 6.3 The Natural Logarithmic Function. Other Bases 6.4 Inverse Trig Functions Algebra 6.5 Calculus of Inverse Trig Functions 6.6 Hyperbolic Functions 6.7 Inverse Hyperbolic Functions 6.8 L'Hôpital's Rule Exam #1 7 Techniques of Integration 7.1a Method of Substitution Review 7.1b Method of Substitution Advanced 7.2 Integration by Parts 7.3 Trig Integrals 7.4 Trig Substitutions

7.5 Partial Fractions 7.6 Improper Integrals 7.7 Numerical Integration Exam #2 8 Applications of Integration 8.1 Area between Curves 8.2 Volumes: Method of Slicing. The Disk Method 8.3 Volumes: the Shell Method 8.4 Arc Length: Surface Area 8.5 Work Problems 8.6 Present and Future Value Exam #3 9 Generalized Functions 9.1 Generalized Functions 9.2 Geometric Calculus of Generalized Functions 9.3 Analytic Calculus of Generalized Functions 10 Differential Equations 10.1 First order Differential Equations. Variables Separable 10.2 Homogeneous First Order Linear Equations 10.3 Nonhomogeneous First Order Linear Equations Exam #4 Preface

Infinitesimal Calculus, Volume II

In Volume I we set up, in detail, the basics of hyperreal calculus. The payoff was being able to do a thorough, elementary and intuitive development of calculus. Proofs of the all basic operational formulas and theorems were accomplished as well as more difficult foundational theorems such as the Extreme Value Theorem and the Riemann Integrability of Continuous Functions over a Closed Interval. In this Volume II, early/late transcendentals, we continue the theoretical development - especially applications of integration - using hyperreal analysis. Historically, the fundamental starting point of calculus was the differential . This is because the basic laws of science and geometric are often simpler to discover over an infinitely short interval of space or time. Now, with the basics of elementary applications better understood, it makes sense to start where students have better mathematical strengths. To find the rate of change of Q with respect to t, the derivative , you divide the differential by dt. To find the change in Q from time t1 to t2 , thea integral of f(t) over the interval, you sum the infinitesimal f(t) dt’s from t1 to t2 and find the closest real number to this sum.

dQ = f (t ) dt ↑ dQ = f (t) dt the differential ↓ Δ = t 2 ( ) Q ∫t 1 f t dt

The greatest benefit of infinitesimal methods is providing a reliable, intuitive and foolproof guide to setting up applications of integration.

The differential in modern textbooks is a real number and is relegated to a method of approximating a function near a point. It is also used somewhat dishonestly and without further explanation in order to be able to employ powerful infinitesimal techniques such as the method of change of variable in integration and separation of variables in differential equations. The symbol dy now is a fraction dx and needs no awkward discussion and the chain rule is obvious and proves itself (check it out in Volume I.

The great tools of infinitesimal calculus are infinitesimals, which facilitates the ultra-precise calculations used to do calculus and the equivalence relation ≈ which enables the rapid comparison and simplification of hyperreal expressions. Misuse of the key comparison symbols =, ≈ and ≈> is not a big problem because their outputs tend to differ only by an infinitesimal. ≐ is the approximate equality for real numbers. 4

Chapter 6 Advanced Transcendental Functions

6.1 Inverse Functions Theory - a Review The main purpose of the inverse function is to solve for the independent variable in y = f(x):

y = f(x) ⟺ x = f-1(y)

I. When does f have an inverse? Answer: if for each y in the range of f, there is exactly one x.

Y Y

y=g(x) y=f(x) y ⟶

↓ y ⟶ ↓ ↓ X X x x1 x2

Each y, one x. Some y's, more than one x. We say that f is one-to-one or invertible if for each y, there is exactly one x (or equivalently f passes the Horizontal Line Test.) Note that increasing functions are invertible; so are decreasing functions.

II. How to find the formula for f-1? y = f(x) Solve for x: x = f -1(y)

III. The inverse function y = f-1(x) We prefer as usual, in working with or studying f -1 to call the dependent variable y. So interchanging x and y: y = f -1(x) (In applications, f -1 is usually of interest only when it cannot be found exactly! f -1 is then found numerically by computer.) Note that y = f(x) and y = f -1(x) have different graphs and therefore are different functions.

-1 -1 IV. How to graph y = f (x)? Since we interchanged x and y to obtain the graph of y = f (x) from that of y = f(x): Reflect the graph of y = f(x) across the line y = x. Y y = x 4 (3, 4)

Y 3 (4, 3) x = f -1(y) y = x y = f(x) 2

y = f -1(x) (x, y) 1 x = f(y)

↘ X (y, x) 1 2 3 4 How it works The point (a, b) reflected across ithe line y = x is the point (b, a). X 5

V. Inverse Function Identities These identities are verified by following the f -1(f (x)) = x arrows in the graphs below. Start at x in the f f -1(y) = y left graph and at y in the right graph.

Y Y

y = f(x) y = f(x) ⟶ y ⟶ f(x) ⟵ ⟵ f( f -1(y)) ↑ ↓ ↑ ↓

X X x f -1(y) f -1(f(x))

VI. Application: Universal Equation Solver Solve: f(x) = c f -1(f (x)) = f -1(c) Taking f -1 of both sides The solution: x = f -1(c) . Inverse function identity

VII. Calculus of a inverse function y = f -1(x) ⟺ x = f (y) 1 = f '(y) dy Differentiating implicitly; Chain Rule dx dy = 1 dx f '(y) or

dy = 1 dx dx . Live math. dy

Easy Example y = f(x) = 2x + 1

Y 5

I. 3 2

X 1 2 Passes the Horizontal Line Test. So f -1(x) exists.

II. Find the formula for f-1 ? y = f(x) = 2x + 1 Solve for x : x = f -1 (y) 1 1 = f -1 ( ) = . x y 2y - 2

III. The inverse function. Interchange x and y. 1 1 f -1 ( ) y = x = 2 x - 2. 6 IV. Its graph. Y

5 y = f(x) 4 3 2 y = x 1 y = f -1(x) X -2 -1 1 2 3 4 5

V. Inverse function Identities. Let us check. 1 1 1 1 f -1(f(x)) = f(x) - = (2 x + 1)- = x. 2 2 2 2 1 1 f(f -1(x)) = 2f -1(x) + 1 = 2( x - ) + 1 = x. 2 2 Properties VI and VII are left as exercises. They are important for more difficult examples.

Harder Example y = g(x) = x - 2 ? If you solve for x and then interchange x and y, you get y = g -1(x) = x2 + 2. Lets see. Y 4 ? y = g -1(x) 3 2 y = g(x) 1

X -2 -1 1 2 3 4 5 -1 Recall that squaring an equation can give spurious solutions. Clearly, reflecting y = g(x) about the line y = x, the correct inverse is y = g-1(x) = x2 + 2, x ≥ 0.

Another Example with a problem y = h(x) = x2 + 1 ?? If you solve for x and then interchange x and y, you get y = h -1(x) = ± x - 1 . Let's see. Y 4 y = h(x) 3

X -2 -1 1 2 3 4 -1 The problem here is that the function y = h(x) is not one-to-one. So it does not have an inverse. Nevertheless, it will be useful at times to do the best we can in finding a related inverse. What we do is restrict the domain to x ≥ 0; what remains is one-to-one. (We could, of course, made the domain choice x ≤ 0. In many advanced applications, there are good reasons for one choice over another.) Conclusion: 2 y = h1(x) = x + 1, x ≥ 0 -1 y = h1 (x) = x - 1

Y 4 y = h1(x) 3

-1 2 y = h1 (x)

X -1 1 2 3 4 5 -1 7

A Very Difficult Problem y = k(x) = 2x Y 4 y = k(x) 3

X -2 -1 1 2 3 4 -1 y = k(x) is a one-to-one function. So its inverse exists. However, there is no elementary way to solve for x in terms of y. What one does in such cases is to give the inverse function a name and let a computer program figure out the value of y for each x. Inverse functions are most important when you can’t solve for x explicitly!

In this case, as you may remember from high school math: y = k(x) = 2x -1 y = k (x) = log2(x), “ log, base 2, of x.”

Every one-to-one function has an inverse which automatically satisfies properties I to VI and VII also if the function is suitably differentiable. This knowledge gives you much information about the inverse function which it inherits from the original function. This is very helpful in that beginners often find inverse functions particularly difficult. In addition, any properties peculiar to to the original function also translate into properties of the inverse function. Keep this in mind as we continue through this chapter.

Exercises 6.1 In each problem, do steps I to V.a Check by graphing or calculating. Semi-memorize the seven steps. a 1. y = f(x) = 1 - x2 , 0 ≤ x ≤ 1. 2. y = g(x)= 1 - x2 , -1 ≤ x ≤ 0. y = h(x) 6

4 -x 3.y = h(x) 2 .. Hint: = 2+x 2

X -6 -4 -2 2 4 6 -2

-4 y = k(x) 4 -6

2 x3 4. y = k(x) = - 3x. Find one partial inverse. Hint: X You can use a CAS or Wolfram Alpha to solve for x. -4 -2 2 4

-2

5. y = 3x -4 8

Solutions

1. 2. Y Y

1 1 y = 1 - x y = 1 - x2

1 X -1 y = 1-x2

y = - 1 - x2 X -1 1

Y 4 #4. y = f(x) = x3 - 3 x. Solve for x and interchange x and y. We will get the leftmost of the three possible inverses using a CAS: 1/3 2 2 1/3 -x+ -4+x -1 2 y = f (x) = - 1/3 - . 21/3 -x+ -4+x2

X -4 -2 2 4

Graphing -2 y = x3- 3x

Y 4 ■ ■■ -4 y = f(x) ■

X -4 4

y = f -1(x)

-4 9

6.2 Exponential Functions Review. The Natural Exponential Function

Review3 of Exponents bx ← exponent base ↗ Definition of Exponents For n a natural number: b1 = b b2 = b·b b3 = b·b · b ⋮

Properties of Exponents (m, n natural numbers) m n m+n 1. b b = b Example: b2 ·b3 = (b·b)(b·b·b) = b2 + 3 m b m n 5 b - b b·b·b·b·b b5 -3 2. bn = Example: b3 = b·b·b = , cancelling 3 common factors m n m·n 3. (b ) = b Example: (b2)3 = (b·b)(b·b)(b·b) = b3·2 = b2·3, two groups of three x We would like to define b for other real numbers x. We do this in such a way that the Properties of Exponents hold. b 3 b0 b0 b3-3 Definition = 1 because, for example, = = b3 = 1 (Property 2)

0 -n 1 -n 0 - n b 1 b n b b n = n Definition = b because = = |b -b - --1 - 1 1 1 1 What about fractional powers? Consider b / 2 -:- _b /2 ·b /2 = (b /2 ) 2 = b1 = b ⟹ b /2 = b Definitions: n b1/n = b m n n m b /n = ( b )m = b Recall that if n is an even integer, b must be non-negative for m odd.

Definition Exponential Function with base b bx y =

Y 5 x y = 3

4 x Domain: all x = 2 y Range: 0 < y < +∞ 3 0 < b < 1 ⇒ decreasing function 2 ib > 1 ⇒ increasing function

1 x 1 = y ( 2 ) X -2 -1 0 1 2 3 4 10

Note: Exponential functions are not defined for bases b < 0.

Choosing a base for calculus Which base b has the ‘nicest’ derivative? d bx+dx- bx bdx- (bx) = = 1 bx dx dx dx

bdx- 1 Answer: if dx ≈> 1. Exploring this using limit approximations:

bh- 1 if h ⟶ 1 as h ⟶ 0 h or b - 1 ⟶ h as h ⟶ 0 h or b ⟶ 1 + h as h ⟶ 0 h or b ⟶ (1 + h)1/ as h ⟶ 0

Let us calculate (note that these calculations are elementary but tedious): h h (1 + h)1 1 (1 + 1)1 = 2 0.1 (1.1)10 = 2.5937424601 Why didn't we do an exact derivation or proof? 0.01 (1.01)100 ≐ 2.748138294 Unfortunately for advanced functions there is often no elementary algebra that can do the ( )1000 ≐ 0.001 1.001 2.716923 job. So we resort to a numerical procedure. 0.0001 (1.0001)10 000 ≐ 2.7181458 ↓ ↓ 0 2.718 ⋯

So the base which has the nicest derivative is e = 2 .718 ⋯ called Euler's Constant. Leonard Euler, 1707 to 1783, was the world's most prolific calculus mathematician.

Euler is pronounced 'Oiler'

Definition The natural exponential function is x y = e , e = 2.718281828459045 ⋯ Memorize this. Easier than π.

d x x x x (e ) = e e dx = e + C dx ∫ 11

Y x 5 y = e

4 3 . 2

1. . -2 -1 1 2 X

sTables of values for graphing by hand x ex -2 0.14 -1 0.37 0 1 1 2.7 2 7.4

Financial Application of The Natural Exponential Function You learned in high school about compound interest. Compounding means that after a period of time, the interest earned is added to the initial amount and the new larger amount continues earning interest. The future value F of a present value P at a yearly interest rate r compounded n times yearly (a high school formula) is r nt F = P1 + n  . What if compounding is done continuously (n → +∞)? r nt F = limn→+∞ P1 + n 

r n rt r . = limn→+∞ P1 + n  1 rt r h h = limh→0 P (1 + )  Letting h = n rt 1 = P e Note By the definition of e: e = limh→0 (1 + h) h F = Per t

Example You take the amount of money, $200,000, your mom gave you at age 20 to go to university but instead invested it at a rate of 10% compounded continuously. Will you be able to retire from your McRonald’s job at age 65 and live well? F = 200000 e0.1(65-20) = 200000 e4.5

= $ 18,000,000

Yes! (but watch out for inflation) 12

Exercises 6.2

x x x 1. On the same graph, plot by hand y = e , y = e2 and y = e /2. x x 2. On the graph, plot by hand y = e- and y = e-2 . x2 x2 3. On the same graph, plot by hand y = e and y = e- . x x 4. On the same graph, plot by hand y = e and y = e( -2)

3 5. Prove that x1/3 = x . Hint: cube the left hand side. d d (e3 x) = ex2  = 6. dx dx d d esin x = ex2 cos x = dx dx x2 d 2 sin x d e x + e  =   = dx 7 dx x + 1 d d ( (e3 x) ) = (etan x) = dx tan dx t t t 7. ∫ e dt = ∫ e cos(e )dt = x2 x ex ∫ x e dx ∫ e e dx

x 8. Find the area under the curve y = e for 0 ≤ x ≤ 1. Illustrate with a graph. 9. You decided to go to university anyway. But you decided to save the amount you budgeted for lunches for the four years, $15,000, and invest it at 10% for 45 years? Should you get a job with retirement benefits? _r nt 10. Derive the formula F = P1 + n . Start with the simple interest formula F = P(1 + rt). Derive this last formula first. 11. A certain bacterium divides every hour. How many will there be after 24 hours. a. Use the exact formula. b. Compare with the continuous compounding formula. F = Pe rt .

Solutions 6.2 3 x 6. e · 3 x2 e · 2x sin x e · cos x ex2 x cos (2x cos x - x2 sin x) sin x 2 sin x ex2 x x ex2 2x e + (x + 7) e · cos x _2 ( +1) - (x+1)2 2 e3 x e3 x x sec ( ) · 3 e tan sec2 x

et t 7. + C sin e + C x2 ex 1/2 e + C e + C

Y x 5 y = e 4

3 8. - 1 e 2

X 1 2 11 a. 16,777,216 b. 2.649 x 1010 This is vastly greater than the correct answer 16,777,216. Here the bacteria start growing immediately, whatever this means; they do not wait an hour. 13

6.3 The Natural Logarithmic Function. Other Bases The inverse of the natural exponential function is the natural logarithmic function. We will follow the steps of section 6.1 to discover its inverse function properties.

x x I. Does y = f(x) = e have an inverse so that y = f(x) = e ⟺ x = f -1(y)? Y 5

4 x y = e 3

1 2 X -1

Yes, it satisfies the horizontal line test.

II. Find its inverse function by solving x y = f(x) = e for x. This cannot be done by elementary algebra. So we pretend we can and write x = f -1(y) = ln(y) Note “logarithms are exponents” where ln y is called the natural logarithm of y. ln is the abbreviation of its Latin name Logarithimus Naturalis. It is pronounced ‘ell-n x’ or ‘lon (rhymes with Ron) x’ or in advanced work is written ‘log x’.

III. The natural logarithm function. x ↔ y.

y = ln x

IV. The graph of y = ln x Y Note The values of named functions 5 x y = e such as the natural logarithmic function 4 typically are calculated numerically by 3 computer and which you can then freely use in any application. 2 y = ln x

-2 -1 1 2 3 4 5 X -1

-2

V. The Inverse Function Identities x ln(e ) = x ‘easy logs’ eln x = x ‘exponentiation’ Example 2 in base e form is Examples aaa2 = eln 2 ≐ e0.693 y = ln x 1 -1 e e ln = ln = -1 2 ln 1 = ln e0 = 0 1 ln e = ln e1 = 1 X ln e2 = 2 1 2 3 4 5 -1-

These values are useful for graphing y = ln x by hand. 14 Compare with the case of easy roots, y = x useful for the same reason. 2 0 = 02 = 0 1 = 12 = 1 1 4 = 22 = 2 ⋮ X 1 2 3 4 5 VI. Solving Exponential Equations Example Solve x e2 = 7 2 x ln e = ln 7 Take the log of both sides 2x = ln 7 Easy logs 1 x = 2 ln 7 ≐ 0.973 Calculator.

VII. Calculus y y = ln x ⟺ x = e dy 1 1 1 = = y = x . dx dx/dy e

d 1 (ln x) = du = ln u + C dx x ∫ u

Verify that the absolute value sign in the integral is appropriate. d x x -1 x d d ln(-x) x < 0 ln(- ) < 0 x < 0  dx - 1 ≠ dx (ln|x|) = dx x x = d = 1 x = x , x 0. ln( ) > 0 (x) x > x > 0 dx ln 0 Example d 1 · dx (ln(sin x)) = sinx cos x = cot x Chain Rule

Example The absolute values sign allows us to find the ‘area’ below from x = -2 to x = -1. 1 x 2

-2 -1 1 2 X

-1

-2

-1 dx -1 x = ln|x| ∫-2 -2 = ln|-1| - ln|-2|

= 0 - ln 2

= - ln 2 ≐ - 0.693 15

Further Properties of Logarithms The Properties of Exponents x y x y 1. e e = e + ex e x-y 2. e y = x y xy 3. (e ) = e translate directly into

Properties of Logarithms

1. ln(x y) = ln x + ln y x 2. ln y = ln x - ln y

y 3. ln x = y ln x

Proof of 1 ln(x y) = ln x + ln y ln(xy) x y exponentiation = lneln eln  x y = lneln + ln  property 1 of exponents = lnx + lny easy logs

Other Bases u d x x u b (b ) = b ln b ∫ b du = + C dx ln b d (log x) = _1 dx x ln b

Proof x y = b ⟺ ln y = x ln b 1 dy

y dx = ln b dy

dx = y ln b x = b ln b

Proof y y = logbx ⟺ x = b y dy b 1 = dx lnb dy dx = by ln b 1 = x ln b or by general property VII of inverse functions. 16

Logarithmic Differentiation There is one more general derivative formula we need. We do not yet know how to differentiate g x y = f(x) ( ).

Method: take ln of both sides. g x ln y = ln(f(x) ( )) = g(x) ln(f(x)) Property 3 of logarithms 1 dy f ' (x) y dx = g’(x) ln(f(x)) + g(x) f(x) Implicit differentiation or dy f ' (x) dx = y (g’(x) ln(f(x)) + g(x) f(x) ) ( ) g x f ■(x) NOTE Do not memorize this formula. = f(x) ( )(g’(x) ln(f(x)) + g(x) ' f(x) ) Go through this process for each example.

x Example Graph y = x The natural domain of this function is x > 0. x x x 11 1 11 = 1 2 22 = 4

x You will learn how to do this exactly in Section 6 or now limx→0 x = 1 numerically, say, by evaluating 0.00010.0001 = 0.999079.

x = x Local extreme values y by logarithmic differentiation. 4 x y = x x ln y = ln x = x ln x 1 dy 1 · x· 3 y dx = 1 x + x dy dx = y(ln x + 1) xx = (ln x + 1) = 0 2 ⇒ ln x = -1 x = e-1 ≐ 0.37 y = 0.37 0.37 ≐ 0.69 1

1 2 X 17

Another application of logarithmic differentiation is to the differentiation of some complicated terms.

Example x2+  (x + )3 Find the derivative of y = 1 2 . (x + 4)5 2 ln y = ln((x + 1) + 3 ln(x+2) - 5 ln(x + 4) Properties of logs x 1 dy = 2 + 3 - 5 y dx x2+1 x+2 x+4 Logarithmic Differentiation dy 2 x 3 5   dx = y x2+1 + x+2 - x+4 x2+  (x+ )3 x = 1 2  2 + 3 - 5  (x+4)5 x2+1 x+2 x+4 This method is quick and produces a good looking answer.

Exercises 6.3

d e2 x+3 d ex2 d e(2 x+3)5 d esin x 1. dx ( ) = dx ( ) = dx ( ) = dx ( ) = d d e2 x d d (ex x   ( (e2 x+3 (x3 e2 x+3 dx tan ) = dx x2-3 = dx cos )) = dx ) =

x 1 x cos x cos x 2. ∫ 2 dx = x2 dx = ∫ 2 dx = ∫ dx = x + 1 ∫ 0 + 1 sin x+ 1 sin t π x x x2 3. e dt = cosx esin dx = 7 dx = x 10 dx = ∫ ∫ 0 ∫ ∫ x 4. Prove Property 2 of Logarithms ln y = ln x - ln y. y 5. Prove Property 3 of Logarithms ln x = y ln x.

d 1 d x2 d x3 d xn 6. dx (ln x ) = dx (ln ) = dx (ln ) = dx (ln ) =

d 3 x d x2 d (2 x+3)5 d sin x 7. dx (2 ) = dx (10 ) = dx (7 ) = dx (2 ) = 8. Work by logarithmic differentiation. d x3 x d xcos x dx ( ) = dx (sin ) = 9. Work by logarithmic differentiation. Compare with ordinary differentiation. xx2+ 3 x3- 5 a. y = x(x+3)(x+5) b. y = x4+ 2 x2+ 1

10. Write y = 10 x in base e form.

Solutions

x 8 b. y = sin x cos

ln y = cos x ln(sin x) 1 dy cos x

y dx = -sinx ln(sinx) + cosx sin x dy  cos2 x dx = y sin x - sin x ln(sin x))

2 x x cos x cos = sin sin x - sin x ln(sin x))

x x ln 10 x ln 10 10. 10 = e  = e 18

6.4 Inverse Trig Functions: Algebra

The Inverse Sine Function -1 If you want to solve sin θ = 0.347, you need the inverse sine function: θ = sin 0.347 ≐ 20.304°. The inverse sine function has many other applications in advanced calculus based applications.

Note: from the graph below, the function y = sin x is not one-to-one. The best we can do is observe π π ≤ ≤ . that it is one-to-one, for example, on the interval - -2 x - 2 We call the inverse function for this -1 interval sin x (if you are a mathematician) or for everyone else arcsin x (who might forget that -1 -1 sin x does not mean 1/sin x.) If you chose the magenta colored function your answers would be between 90° and 270° (OK, but nobody does that!) Y 2 π

3 π 2 ⟵ alternate arcsin x

π 2 y = arcsin x ⟶

y = sin x

π π 3 π X - -1 π 2 π 2 2 2

-1

- π 2

Inverse Trig Identities ≤ ≤ arcsin(sin x) = x, - π ≤ x ≤ π sin(arcsin x) = x, -1 x 1 2 2

Examples Be careful : see below

sin(arcsin 2) DNE arcsin(sin π) = 0

arcsin(sin x) sin(arcsin x) π 2 1

X π π -1 1 - X 2 2 -1 - π 2 19

Origin of the arcsin notation Look at the unit circle below.

(x, y)

1 y θ

θ x

y = sin θ means θ is an angle whose sine is y which means θ is an arc whose sine is y which said quickly is

θ = arcsin y

Other Inverse Trig Identities arcsin x means an ‘angle whose sine is x’. It is often useful to construct a triangle illustrating this.

1 x Picture of arcsin x

arcsinx 1 - x2

Full basic list of inverse sine identities (domains not shown) See picture above.

sin(arcsin x) = x cos(arcsin x) = 1 - x2

x 1 - x2 tan(arcsin x) = cot(arcsin x) = x 1 - x2

1 1 sec(arcsin x) = csc(arcsin x) = x 1 - x2 20

Basic Trig/ Inverse Trig Graphs , standard choices

sin x, arcsin x cos x, arccos x tan x, arctan x π π 2

π 2

X X π π 1 π π - - 2 2 2 2 - π 2 X -1 1 π

- π 2 -1

cot x, arccot x sec x, arcsec x csc x, arccsc x 3 3

2 2 π 2 1 1

X x x π -3 -2 -1 1 2 3 -3 -2 -1 1 2 3 2 -1 -1

-2 -2

-3 -3

Alternate Choices. Warning Some textbooks and computer algebra systems may also choose arccot, arcsec and arccsc differently! (You will see in evaluating integrals, an important use of inverse trig functions, any reasonable choice leads to the same answer.)

alternate arccot x our arccot x

π π 2

X -1 1

Note how both choices can be made.

arccot x selection graph cot x 4 4 2 2

-4 -2 2 4 -4 -2 2 4 -2 -2 -4 -4 21

π One would expect: ArcTan[x] + ArcCot[x] = 2 . Why? See diagram below.

arctan x + arccot x

π 2 arccot x

-3 -1 1 2 3 X

arctan x π 1 - 2

Is there a geometry problem when x < 0?

Exercises 6.4 1. Solve using inverse trig functions. Evaluate by calculator. π θ a. sin = .35 b. cos x = 13 c. tan y = 55 d. sec x = 2

2. Find the first three positive solutions of sin θ = 0.3 using inverse trig functions. Hint: Graph sin θ. 3. As in the first example, evaluate and check with a graph. a. cos(arccos 2) b. arccos(cos π)

4. Graph y = tan(arctan x) and y = arctan(tan x). Indicate domain and range of each. 5. State the full basic list of inverse cosine identities. Illustrate. State domains. 6. State the full basic list of inverse tangent identities. Illustrate. State domains. 7. Solve each. π c. tan y = 55 g. d. sec x = 2 8. a. sin θ = .358. b. cos x = 13 9. a. sin(arcsin 2x) b. tan(arcsin 3x) c. sin(2arcsin x) 10. Graph and compare both sides of cos(arcsin x)= 1- x2 . Solutions

Y 1

X ↑ ↑ π 2 π ↑

-1

arcsin(0.3), π - arcsin(0.3), 2π + arcsin(0.3)

3. a. D b. π

9. (with domain restrictions) sin(2arcsin x) a. a. . sin(arcsin(2x)) b. tan (arcsin(3x c. 3 x = 2 sin(arcsin x) cos(arcsin x) double angle formula = 2x = 1- x2 1- 9 x2 = 2 x 1 2 = 2 x 1 - x 22

6.5 Calculus of the Inverse Trig Functions

d Let us do dx (arcsin x) first.

d 1 dx (arcsin x) = 1-x2 Y 2 π

3 π 2 ⟵ alternate arcsin x

π 2 y = arcsin x ⟶

y = sin x

π π 3 π X - -1 π 2 π 2 2 2

-1

- π 2

Proof y = arcsin x ⟺ x = sin y dy

1 = cos y dx differentiating implicitly dy 1 dx = cos y = 1 Pythagorean Identity ± 1 - sin2 y 1 = Since our arcsin x has a positive slope 1 - sin2 y = 1 1 - x2 For the alternate arcsin x we would have chosen the negative square root.

d 1 dx (arctan x) = 1 + x2

Proof y = arctan x ⟺ x = tan y 2 dy

1 = sec y dx differentiating implicitly dy 1 dx = sec2 y 1 = 1 + tan2 y Pythagorean Identity 1 = 1 + x2 23

d 1 dx (arcsec x) = x x2- 1

Proof y = arcsec x ⟺ x = sec y dy

1 = sec y tan y dx differentiating implicitly dy 1 dx = sec y tan y = 1 Pythagorean Identity ±sec y tan2 y - 1 = 1 ±x x2 - 1 = 1 See graph below x x2 - 1

sec x, arcsec x 3

The slope of this arcsec x is alwayspositive. x -3 -2 -1 1 2 3 -1

-2

-3

d 1 d -1 (arcsin x) = (arccos x) = dx 1-x2 dx 1-x2 d 1 d -1 (arctan x) = (arccot x) = dx 1 + x2 dx 1 + x2 d 1 d -1 dx (arcsec x) = x dx (arccsc x) = x x2- 1 x2- 1

dx  = arcsin x + C 1 - x2 dx = arctan x + C ∫ 1 + x2 dx  = arcsec |x| + C x x2- 1 24

Exercises 6.5 1. Do a full derivation of the derivative of arccos x. Show a relevant graph.

d 2 d d ( x ) ( x x) x ax + b 2. dx arcsin = dx sin arcsin = dx arcsin = x 3. dt = dx = cos dx =  ∫ 4 + x2 ∫ 4 + sin2x 1 - t2 4. The inverse trigonometric integrals are often written du u  = arcsin a + C a2 - u2 du 1 u ∫ u2+ a2 = a arctan a + C du 1 u  = a arcsec a + C. u u2 - a2 Derive these. 5. A statue 5 high sits on a pedestal 4 meters high. Find the viewing angle θ as a function of the distance

5 .

θ 4

Solutions

9 4 5. Hint: θ = arctan x - arctan x Maximize θ Differentiate Simplify Solve x = 6 meters. 25

6.6 Hyperbolic Funcons y x2 + y2 = 1 1 The hyperbolic funcons are a set of funcons that have many applicaons to (cos θ,sin θ) mathemacs, physics, and engineering. Among many other applicaons, they are used to describe the formaon of satellite rings around planets, to describe θ the shape of a rope hanging from two points, and have applicaon to the theory 2 x of special relavity. This secon defines the hyperbolic funcons and describes −1 1 many of their properes, especially their usefulness to calculus. These funcons are somemes referred to as the “hyperbolic trigonometric funcons” as there are many, many connecons between them and the stan- −1 dard trigonometric funcons. Figure 6.6.1 demonstrates one such connecon...... 2 2 Just as cosine and sine are used to define points on the circle defined by x +y = y 1, the funcons hyperbolic cosine and hyperbolic sine are used to define points x2 − y2 = 1 on the hyperbola x2 − y2 = 1. (cosh θ,sinh θ) 2 We begin with their definions.

θ 2 x Definion 6.6.1 Hyperbolic Funcons −2 2 ex + e−x 1 1. cosh x = 4. sech x = − 2 cosh x 2 − .... ex − e x 1 2. sinh x = 5. csch x = . 2 sinh x Figure 6.6.1: Using trigonometric funcons to define points on a circle and hy- sinh x cosh x 3. tanh x = 6. coth x = perbolic funcons to define points on a cosh x sinh x hyperbola. The area of the shaded regions are included in them.

These hyperbolic funcons are graphed in Figure 6.6.2. In the graphs of cosh x and sinh x, graphs of ex/2 and e−x/2 are included with dashed lines. As x gets “large,” cosh x and sinh x each act like ex/2; when x is a large negave number, cosh x acts like e−x/2 whereas sinh x acts like −e−x/2. Noce the domains of tanh x and sech x are (−∞, ∞), whereas both coth x Pronunciaon Note: and csch x have vercal asymptotes at x = 0. Also note the ranges of these “cosh” rhymes with “gosh,” funcons, especially tanh x: as x → ∞, both sinh x and cosh x approach ex/2, “sinh” rhymes with “pinch,” and hence tanh x approaches 1. “tanh” rhymes with “ranch.” The following example explores some of the properes of these funcons that bear remarkable resemblance to the properes of their trigonometric coun- terparts.

The connection between the trigonometric and hyperbolic functions shown in Figure 6.6.1 is correct but not particularly enlightening. We would like to see the relationships between the ‘angles’ rather than the areas. In more advanced calculus that includes the imaginary number i = -1 , the relationships becomes clearer. That calculus is called Functions of a Complex Variable.

e i x+ e- i x x = cos 2 ↑ ex ↓ ex+ e- x cosh x = 2 26

y y 10 10

5 5

x x −2 2 −2 2

−5 −5 f(x) = cosh x f(x) = sinh x

..... −10 ..... −10

y y

2 f(x) = coth x 2 f(x) = csch x f(x) = sech x 1

x x −2 2 −2 2 −1 f(x) = tanh x −2 −2 − ...... 3

Figure 6.6.2: Graphs of the hyperbolic funcons.

Example 6.6.1 Exploring properes of hyperbolic funcons Use Deﬁnion 6.6.1 to rewrite the following expressions. ( ) 2 2 d 1. cosh x − sinh x 4. dx cosh x . (. ) 2 2 d 2. tanh x + sech x 5. dx sinh x ( ) d 3. 2 cosh x sinh x 6. dx tanh x

S ( ) ( ) ex + e−x 2 ex − e−x 2 1. cosh2 x − sinh2 x = − 2 2 e2x + 2exe−x + e−2x e2x − 2exe−x + e−2x = − 4 4 4 = = 1. 4 So cosh2 x − sinh2 x = 1. 27

sinh2 x 1 2. tanh2 x + sech2 x = + cosh2 x cosh2 x sinh2 x + 1 = Now use identy from #1. cosh2 x cosh2 x = = 1. cosh2 x So tanh2 x + sech2 x = 1. ( )( ) ex + e−x ex − e−x 3. 2 cosh x sinh x = 2 2 2 e2x − e−2x = 2 · 4 e2x − e−2x = = sinh(2x). 2 Thus 2 cosh x sinh x = sinh(2x). ( ) d ( ) d ex + e−x 4. cosh x = dx dx 2 ex − e−x = 2 = sinh x. ( ) d So dx cosh x = sinh x. ( ) d ( ) d ex − e−x 5. sinh x = dx dx 2 ex + e−x = 2 = cosh x. ( ) d So dx sinh x = cosh x. ( ) d ( ) d sinh x 6. tanh x = dx dx cosh x cosh x cosh x − sinh x sinh x = cosh2 x 1 = cosh2 x = sech2 x. ( ) d 2 So dx tanh x = sech x. 28

The following Key Idea summarizes many of the important idenes relat- ing to hyperbolic funcons. Each can be veriﬁed by referring back to Deﬁnion 6.6.1.

Key Idea 6.6.1 Useful Hyperbolic Funcon Properes. You actually should semi-memorize these! Basic Idenes Derivaves Integrals ( ) ∫ 2 − 2 1. d cosh x = sinh x 1. cosh x sinh x = 1 dx 1. cosh x dx = sinh x + C ( ) 2 2 d ∫ 2. tanh x + sech x = 1 2. dx sinh x = cosh x ( ) 2. sinh x dx = cosh x + C 2 − 2 = d 2 3. coth x csch x 1 3. dx tanh x = sech x ( ) ∫ 4. cosh 2x = cosh2 x + sinh2 x d − 4. dx sech x = sech x tanh x 3. tanh x dx = ln(cosh x) + C ( ) 5. sinh 2x = 2 sinh x cosh x d − ∫ 5. dx csch x = csch x coth x ( ) | | 2 cosh 2x + 1 4. coth x dx = ln sinh x + C 6. cosh x = 6. d coth x = − csch2 x 2 dx cosh 2x − 1 7. sinh2 x = 2

We pracce using Key Idea 6.6.1.

Example 6.6.2 Derivaves and integrals of hyperbolic funcons Evaluate the following derivaves and integrals. ∫ d ( ) ln 2 1. cosh 2x dx 3. cosh x dx ∫ 0 2. sech2(7t − 3) dt

S ( ) d 1. Using the Chain Rule directly, we have dx cosh 2x = 2 sinh 2x. Just to demonstrate that it works, let’s also use the Basic Identy found in Key Idea 6.6.1: cosh 2x = cosh2 x + sinh2 x. d ( ) d ( ) cosh 2x = cosh2 x + sinh2 x = 2 cosh x sinh x + 2 sinh x cosh x dx dx = 4 cosh x sinh x. 29

Using another Basic Identy, we can see that 4 cosh x sinh x = 2 sinh 2x. We get the same answer either way. 2. We employ substuon, with u = 7t − 3 and du = 7dt. Applying Key Ideas 6.1.1 and 6.6.1 we have: ∫ 1 sech2(7t − 3) dt = tanh(7t − 3) + C. 7

3. ∫ ln 2 ln 2 cosh x dx = sinh x = sinh(ln 2) − sinh 0 = sinh(ln 2). 0 0 We can simplify this last expression as sinh x is based on exponenals:

eln 2 − e− ln 2 2 − 1/2 3 sinh(ln 2) = = = . 2 2 4 30

Exercises 6.6 Terms and Concepts ∫ 1. In Key Idea 6.6.1, the equaon tanh x dx = ln(cosh x)+C In Exercises 23 – 28, find the equaon of the line tangent to the funcon at the given x-value. is given. Why is “ln | cosh x|” not used – i.e., why are absolute values not necessary? 23. f(x) = sinh x at x = 0 2. The hyperbolic funcons are used to define points on the 24. f(x) = cosh x at x = ln 2 right hand poron of the hyperbola x2 − y2 = 1, as shown in Figure 6.6.1. How can we use the hyperbolic funcons to 25. f(x) = tanh x at x = − ln 3 define points on the le hand poron of the hyperbola? 26. f(x) = sech2 x at x = ln 3 Problems In Exercises 29 – 44, evaluate the given indefinite integral. In Exercises 3 – 10, verify the given identy using Definion ∫ 6.6.1, as done in Example 6.6.1. 29. tanh(2x) dx

2 2 3. coth x − csch x = 1 ∫ 30. cosh(3x − 7) dx 4. cosh 2x = cosh2 x + sinh2 x ∫ cosh 2x + 1 5. cosh2 x = 31. sinh x cosh x dx 2

cosh 2x − 1 6. sinh2 x = 2 In Exercises 45 – 48, evaluate the given deﬁnite integral. ∫ d 1 7. [sech x] = − sech x tanh x dx 45. sinh x dx −1 d ∫ 8. [coth x] = − csch2 x ln 2 dx 46. cosh x dx ∫ − ln 2 9. tanh x dx = ln(cosh x) + C

∫ 10. coth x dx = ln | sinh x| + C

In Exercises 11 – 22, ﬁnd the derivave of the given funcon.

11. f(x) = sinh 2x

12. f(x) = cosh2 x

13. f(x) = tanh(x2)

14. f(x) = ln(sinh x)

15. f(x) = sinh x cosh x

16. f(x) = x sinh x − cosh x 31

[ ] d d ex + e−x 8. [coth x] = Solutions 6.6 dx dx ex − e−x (ex − e−x)(ex − e−x) − (ex + e−x)(ex + e−x) = (ex − e−x)2 e2x + e−2x − 2 − (e2x + e−2x + 2) = (ex − e−x)2 1. Because cosh x is always posive. 4 = − (ex − e−x)2 2. The points on the le hand side can be deﬁned as − 2 (− , ) = csch x cosh x sinh x . ∫ ∫ ( ) ( ) sinh x − 2 2 9. tanh x dx = dx ex + e x 2 cosh x 3. coth2 x − csch2 x = − ex − e−x ex − e−x Let u = cosh x; du = (sinh x∫)dx 1 (e2x + 2 + e−2x) − (4) = du = u e2x − 2 + e−2x = ln |u| + C e2x − 2 + e−2x = − = ln(cosh x) + C. e2x − 2 + e 2x ∫ ∫ cosh x = 1 10. coth x dx = dx sinh x ( ) ( ) ex + e−x 2 ex − e−x 2 Let u = sinh x; du = (cosh x)dx 4. cosh2 x + sinh2 x = + ∫ 1 2 2 = du e2x + 2 + e−2x e2x − 2 + e−2x u = + | | 4 4 = ln u + C 2e2x + 2e−2x = ln | sinh x| + C. = 4 11. 2 cosh 2x e2x + e−2x = 12. Taking the derivave of (cosh x)2 directly, one gets 2 cosh x sinh x; 2 2 = 1 ( + ) using the identy cosh x 2 cosh 2x 1 ﬁrst, one gets = cosh 2x. sinh 2x; by Key Idea 6.6.1, these are equal. ( ) 13. 2x sec2(x2) ex + e−x 2 5. cosh2 x = 2 14. coth x e2x + 2 + e−2x 15. sinh2 x + cosh2 x = 4 16. x cosh x 1 (e2x + e−2x) + 2 = 2 ( 2 ) 1 e2x + e−2x 23. 21. y = x = + 1 ■ 2 2 = ( − ) + 5 24. 22. y x ln 2 4 cosh 2x + 1 = . ■ 9 − 4 25. 23. y = (x + ln 3) 2 25 5 ■ 72 9 y = − (x − ln 3) + ( ) 26. 24. 125 25 ex − e−x 2 6. sinh2 x = ■ 2 29. 1/2 ln(cosh(2x)) + C e2x − 2 + e−2x = 30. 1/3 sinh(3x − 7) + C 4 − / 2 + / 2 + 1 (e2x + e 2x) − 2 31. 1 2 sinh x C or 1 2 cosh x C = 2 2 ( − ) 1 e2x + e 2x 45.43. 0 = − 1 ■ 2 2 46.44. 3/2 cosh 2x − 1 ■ = . 2 [ ] d d 2 7. [sech x] = dx dx ex + e−x −2(ex − e−x) = (ex + e−x)2 2(ex − e−x) = − (ex + e−x)(ex + e−x) 2 ex − e−x = − · ex + e−x ex + e−x = − sech x tanh x 32

6.7 Inverse Hyperbolic Funcons

Just as the inverse trigonometric funcons are useful in certain integraons, the inverse hyperbolic funcons are useful with others. Figure 6.6.3 shows the restricons on the domains to make each funcon one-to-one and the resulng domains and ranges of their inverse funcons. Their graphs are shown in Figure 6.6.4.

Because the hyperbolic funcons are deﬁned in terms of exponenal funcons, their inverses can be expressed in terms of logarithms as sh(o wn in√Key Ide)a − 6.6.2. It is oen more convenient to refer to sinh 1 x than to ln x + x2 + 1 , especially when one is working on theory and does not need to compute actual values. On the other hand, when computaons are needed, technology is oen − helpful but many hand-held calculators lack a convenient sinh 1 x buon. (Of- ten it can be accessed under a menu system, but not conveniently.) In such a situaon, the logarithmic representaon is useful. The reader is not encouraged to memorize these, but rather know they exist and know how to use them when needed.

Funcon Domain Range Funcon Domain Range − cosh x [0, ∞) [1, ∞) cosh 1 x [1, ∞) [0, ∞) − sinh x (−∞, ∞) (−∞, ∞) sinh 1 x (−∞, ∞) (−∞, ∞) − tanh x (−∞, ∞) (−1, 1) tanh 1 x (−1, 1) (−∞, ∞) − sech x [0, ∞) (0, 1] sech 1 x (0, 1] [0, ∞) − csch x (−∞, 0) ∪ (0, ∞)(−∞, 0) ∪ (0, ∞) csch 1 x (−∞, 0) ∪ (0, ∞)(−∞, 0) ∪ (0, ∞) − coth x (−∞, 0) ∪ (0, ∞)(−∞, −1) ∪ (1, ∞) coth 1 x (−∞, −1) ∪ (1, ∞)(−∞, 0) ∪ (0, ∞)

Figure 6.7.3: Domains and ranges of the hyperbolic and inverse hyperbolic funcons.

y y 10 y = cosh x 10

y = sinh x 5

5 x −1 y = cosh x −10 −5 5 10

− −5 y = sinh 1 x

x 5 10 −10

y y

3 2 − 2 y = coth 1 x 1

x x −2 2 −2 2 −1 − −1 −1 = 1 y = sech x y = tanh x y csch x − −2 2 −3

Figure 6.6.4: Graphs of the hyperbolic funcons and their inverses. 33

Note on Notation The ‘-1 power’ notation is widely used by mathematicians. This seems to make sense: y = f(x) ⇔ x = f-1(y)...... f-1 ? 1 It looks like to go from the left equation to the right one is take = f . But reciprocals are for -1(x) 1 variables not functions. Many beginners make the mistake of writing sinh = sinhx . Bad. Bad. Otherwise the notation sinh-1 is just fine. Some authors write arcsinh, but arcsinh has nothing much to do with arcs or angles. Others write argsinh because in y = sinh x, x is the argument of the sinh function. It is a good notation but not widely used. I think by Calculus II you should be able to handle sinh-1 ! ......

Key Idea 6.6.2 Logarithmic deﬁnions of Inverse Hyperbolic Funcons

( √ ) ( √ ) − − 1. cosh 1 x = ln x + x2 − 1 ; x ≥ 1 4. sinh 1 x = ln x + x2 + 1 ( ) ( ) − 1 1 + x − 1 x + 1 2. tanh 1 x = ln ; |x| < 1 5. coth 1 x = ln ; |x| > 1 2 1 − x 2 x − 1 ( √ ) ( √ ) 2 2 − 1 + 1 − x − 1 1 + x 3. sech 1 x = ln ; 0 < x ≤ 1 6. csch 1 x = ln + ; x ≠ 0 x x |x| 34

6.6 Hyperbolic Funcons

The following Key Ideas give the derivaves and integrals relang to the inverse hyperbolic funcons. In Key Idea 6.6.4, both the inverse hyperbolic and logarithmic funcon representaons of the anderivave are given, based on Key Idea 6.6.2. Again, these laer funcons are oen more useful than the former. Note how inverse hyperbolic funcons can be used to solve integrals we used Trigonometric Substuon to solve in Secon 6.4.

Key Idea 6.6.3 Derivaves Involving Inverse Hyperbolic Funcons

( ) ( ) d − 1 d − −1 1. cosh 1 x = √ ; x > 1 4. sech 1 x = √ ; 0 < x < 1 dx x2 − 1 dx x 1 − x2 ( ) ( ) d − 1 d − −1 2. sinh 1 x = √ 5. csch 1 x = √ ; x ≠ 0 dx x2 + 1 dx |x| 1 + x2 ( ) ( ) d − 1 d − 1 3. tanh 1 x = ; |x| < 1 6. coth 1 x = ; |x| > 1 dx 1 − x2 dx 1 − x2

Key Idea 6.6.4 Integrals Involving Inverse Hyperbolic Funcons ∫ ( ) √ 1 − x 1. √ dx = cosh 1 + C; 0 < a < x = ln x + x2 − a2 + C x2 − a2 a ∫ ( ) √ 1 − x 2. √ dx = sinh 1 + C; a > 0 = ln x + x2 + a2 + C x2 + a2 a  ( ) ∫ 1 −1 x 2 2  tanh + C x < a 1 a a 1 a + x 3. dx = = ln + C 2 − 2  ( ) − a x 1 −1 x 2 2 2a a x a coth a + C a < x ∫ ( ) ( ) 1 1 − x 1 x 4. √ dx = − sech 1 + C; 0 < x < a = ln √ + C x a2 − x2 a a a a + a2 − x2 ∫

1 1 − x 1 x 5. √ dx = − csch 1 + C; x ≠ 0, a > 0 = ln √ + C x x2 + a2 a a a a + a2 + x2

We pracce using the derivave and integral formulas in the following example. 35

Example 6.6.3 Derivaves and integrals involving inverse hyperbolic funcons Evaluate the following. [ ( )] ∫ d − 3x − 2 1 1. cosh 1 3. √ dx dx 5 9x2 + 10 ∫ 1 2. dx x2 − 1

S 1. Applying Key Idea 6.6.3 with the Chain Rule gives: [ ( )] − d −1 3x 2 1 3 cosh = √( ) · . dx 5 3x−2 2 − 5 5 1 ∫ 1 2. Mulplying the numerator and denominator by (−1) gives: dx = ∫ x2 − 1 −1 dx. The second integral can be solved with a direct applicaon 1 − x2 of item #3 from Key Idea 6.6.4, with a = 1. Thus ∫ ∫ 1 1 dx = − dx 2 − − 2 x 1  1 x −  − tanh 1 (x) + C x2 < 1 =  − − coth 1 (x) + C 1 < x2

−1 x + 1 = ln − + C 2 x 1

1 x − 1 = ln + C. (6.4) 2 x + 1

We should note that this exact problem was solved at the beginning of 1 | − |− 1 | Secon 6.5. In that example the answer was given as 2 ln x 1 2 ln x+ 1| + C. Note that this is equivalent to the answer given in Equaon 6.4, as ln(a/b) = ln a − ln b. 3. This requires a substuon, then item #2 of Key Idea 6.6.4 can be applied. Let u = 3x, hence du = 3dx. We have ∫ ∫ 1 1 1 √ dx = √ du. 9x2 + 10 3 u2 + 10 36

√ Note a2 = 10, hence a = 10. Now apply the integral rule.

( ) 1 − 3x = sinh 1 √ + C 3 10 √ 1 = ln 3x + 9x2 + 10 + C. 3

This secon covers a lot of ground. New funcons were introduced, along with some of their fundamental idenes, their derivaves and anderivaves, their inverses, and the derivaves and anderivaves of these inverses. Four Key Ideas were presented, each including quite a bit of informaon. Do not view this secon as containing a source of informaon to be memo- rized, but rather as a reference for future problem solving. Key Idea 6.6.4 contains perhaps the most useful informaon. Know the integraon forms it helps evaluate and understand how to use the inverse hyperbolic answer and the logarithmic answer. The next secon takes a brief break from demonstrang new integraon techniques. It instead demonstrates a technique of evaluang limits that return indeterminate forms. This technique will be useful in Secon 6.8, where limits will arise in the evaluaon of certain deﬁnite integrals. 37

Exercises 6.7 Solutions 6.7

In Exercises 11 – 22, ﬁnd the derivave of the given funcon. 16. x cosh x − 2x −1 2 17. √ 17. f(x) = sech (x ) (x2 ) 1−x4 18. √ 3 18. f(x) = sinh−1(3x) 9x2+1 19. √ 4x − 4x4−1 19. f(x) = cosh 1(2x2)

1 −1 20. 20. f(x) = tanh (x + 5) 1−(x+5)2

− 21. f(x) = tanh 1(cos x) 21. − csc x 22. sec x − 22. f(x) = cosh 1(sec x)

In Exercises 23 – 28, ﬁnd the equaon of the line tangent to 27. y = x the funcon at the given x-value. − 27. f(x) = sinh 1 x at x = 0 √ √ √ − − 28. f(x) = cosh 1 x at x = 2 28. y = (x − 2) + cosh 1( 2) ≈ (x − 1.414) + 0.881

In Exercises 29 – 44, evaluate the given indeﬁnite integral. ∫ 1 34. √ dx ( √ ) x2 + 1 34. sinh−1 x + C = ln x + x2 + 1 + C ( √ ) ∫ 35. cosh−1 x/3 + C = ln x + x2 − 9 + C 1 35. √ dx x2 − 9 ∫ 1  () 36. dx  1 tanh−1 x + C x2 < 9 9 − x2 3 3 36.  () = ∫ 1 −1 x + < 2 3 coth 3 C 9 x 2x 1 | + | − 1 | − | + 37. √ dx 2 ln x 1 2 ln x 1 C x4 − 4 √ 37. cosh−1(x2/2) + C = ln(x2 + x4 − 4) + C ∫ √ √ x 38. 2/3 sinh−1 x3/2 + C = 2/3 ln(x3/2 + x3 + 1) + C 38. √ dx 1 + x3 1 −1( / ) + 1 | − | + 1 | + | + 39. 16 tan x 2 32 ln x 2 32 ln x 2 C ∫ 40. ln x − ln |x + 1| + C 1 39. dx −1 x x4 − 16 41. tan (e ) + C ∫ 1 40. dx x2 + x ∫ ex 41. dx e2x + 1 ∫ − 44. sech x dx (Hint: mulply by cosh x ; set u = sinh x.) 44. 43. tan 1(sinh x) + C cosh x ■ In Exercises 45 – 48, evaluate the given deﬁnite integral. ∫ 2 1 ( √ ) 48. √ dx −1 . 2 48. sinh 2 = ln 2 + 5 1.444 0 x + 1 = 38

6.8 L’Hôpital’s Rule

While this chapter is devoted to learning techniques of integraon, this secon is not about integraon. Rather, it is concerned with a technique of evaluang certain limits that will be useful in the following secon, where integraon is once more discussed. Our treatment of limits exposed us to the noon of “0/0”, an indeterminate form. If lim f(x) = 0 and lim g(x) = 0, we do not conclude that lim f(x)/g(x) is x→c x→c x→c 0/0; rather, we use 0/0 as notaon to describe the fact that both the numerator and denominator approach 0. The expression 0/0 has no numeric value; other work must be done to evaluate the limit. Other indeterminate forms exist; they are: ∞/∞, 0·∞, ∞−∞, 00, 1∞ and ∞0. Just as “0/0” does not mean “divide 0 by 0,” the expression “∞/∞” does not mean “divide infinity by infinity.” Instead, it means “a quanty is growing without bound and is being divided by another quanty that is growing without bound.” We cannot determine from such a statement what value, if any, results in the limit. Likewise, “0 · ∞” does not mean “mulply zero by infinity.” Instead, it means “one quanty is shrinking to zero, and is being mulplied by a quanty that is growing without bound.” We cannot determine from such a descripon what the result of such a limit will be. This secon introduces l’Hôpital’s Rule, a method of resolving limits that produce the indeterminate forms 0/0 and ∞/∞. We’ll also show how algebraic manipulaon can be used to convert other indeterminate expressions into one of these two forms so that our new rule can be applied.

Theorem 6.8.1 L’Hôpital’s Rule, Part 1 Let lim f(x) = 0 and lim g(x) = 0, where f and g are diﬀerenable func- x→c x→c ons on an open interval I containing c, and g′(x) ≠ 0 on I except possi- bly at c. Then f(x) f ′(x) lim = lim ′ . x→c g(x) x→c g (x)

y = f(x) y = g(x)

y = f '(c)(x-c)

y = g'(c)(x-c)

X 39

We demonstrate the use of l’Hôpital’s Rule in the following examples; we will oen use “LHR” as an abbreviaon of “l’Hôpital’s Rule.”

Example 6.8.2 Using l’Hôpital’s Rule Evaluate the following limits, using l’Hôpital’s Rule as needed.

sin x x2 1. lim 3. lim x→0 x x→0 1 − cos x √ x + 3 − 2 x2 + x − 6 2. lim 4. lim x→1 1 − x x→2 x2 − 3x + 2

S 1. We proved this limit is 1 in Example 1.3.4 using the Squeeze Theorem. Here we use l’Hôpital’s Rule to show its power.

sin x by LHR cos x lim = lim = 1. x→0 x x→0 1 √ − x + 3 − 2 by LHR 1 (x + 3) 1/2 1 2. lim = lim 2 = − . x→1 1 − x x→1 −1 4 x2 by LHR 2x 3. lim = lim . x→0 1 − cos x x→0 sin x This laer limit also evaluates to the 0/0 indeterminate form. To evaluate it, we apply l’Hôpital’s Rule again. 2x by LHR 2 lim = = 2. x→0 sin x cos x x2 Thus lim = 2. x→0 1 − cos x 4. We already know how to evaluate this limit; ﬁrst factor the numerator and denominator. We then have: x2 + x − 6 (x − 2)(x + 3) x + 3 lim = lim = lim = 5. x→2 x2 − 3x + 2 x→2 (x − 2)(x − 1) x→2 x − 1 We now show how to solve this using l’Hôpital’s Rule.

x2 + x − 6 by LHR 2x + 1 lim = lim = 5. x→2 x2 − 3x + 2 x→2 2x − 3

Note that at each step where l’Hôpital’s Rule was applied, it was needed: the inial limit returned the indeterminate form of “0/0.” If the inial limit returns, for example, 1/2, then l’Hôpital’s Rule does not apply. 40

The following theorem extends our inial version of l’Hôpital’s Rule in two ways. It allows the technique to be applied to the indeterminate form ∞/∞ and to limits where x approaches ∞.

Theorem 6.8.2 L’Hôpital’s Rule, Part 2

1. Let lim f(x) = ∞ and lim g(x) = ∞, where f and g are diﬀer- x→a x→a enable on an open interval I containing a. Then

f(x) f ′(x) lim = lim ′ . x→a g(x) x→a g (x)

2. Let f and g be diﬀerenable funcons on the open interval (a, ∞) for some value a, where g′(x) ≠ 0 on (a, ∞) and lim f(x)/g(x) x→∞ returns either “0/0” or “∞/∞”. Then

f(x) f ′(x) lim = lim ′ . x→∞ g(x) x→∞ g (x)

A similar statement can be made for limits where x approaches −∞.

Example 6.8.2 Using l’Hôpital’s Rule with limits involving ∞ Evaluate the following limits.

3x2 − 100x + 2 ex 1. lim 2. lim . x→∞ 4x2 + 5x − 1000 x→∞ x3

1. We can evaluate this limit already using Theorem 1.6.1; the answer is 3/4. We apply l’Hôpital’s Rule to demonstrate its applicability.

3x2 − 100x + 2 by LHR 6x − 100 by LHR 6 3 lim = lim = lim = . x→∞ 4x2 + 5x − 1000 x→∞ 8x + 5 x→∞ 8 4

ex by LHR ex by LHR ex by LHR ex 2. lim = lim = lim = lim = ∞. x→∞ x3 x→∞ 3x2 x→∞ 6x x→∞ 6 Recall that this means that the limit does not exist; as x approaches ∞, the expression ex/x3 grows without bound. We can infer from this that ex grows “faster” than x3; as x gets large, ex is far larger than x3. (This 41

has important implicaons in compung when considering eﬃciency of algorithms.)

Indeterminate Forms 0 · ∞ and ∞ − ∞

L’Hôpital’s Rule can only be applied to raos of funcons. When faced with an indeterminate form such as 0·∞ or ∞−∞, we can somemes apply algebra to rewrite the limit so that l’Hôpital’s Rule can be applied. We demonstrate the general idea in the next example.

Example 6.8.3 Applying l’Hôpital’s Rule to other indeterminate forms Evaluate the following limits.

· 1/x 3. lim ln(x + 1) − ln x 1. lim x e →∞ x→0+ x 2 x · 1/x 4. lim x − e 2. lim x e x→∞ x→0−

S 1. As x → 0+, x → 0 and e1/x → ∞. Thus we have the indeterminate form e1/x 0 · ∞. We rewrite the expression x · e1/x as ; now, as x → 0+, we get 1/x the indeterminate form ∞/∞ to which l’Hôpital’s Rule can be applied. e1/x by LHR (−1/x2)e1/x lim x · e1/x = lim = lim = lim e1/x = ∞. x→0+ x→0+ 1/x x→0+ −1/x2 x→0+

Interpretaon: e1/x grows “faster” than x shrinks to zero, meaning their product grows without bound. 2. As x → 0−, x → 0 and e1/x → e−∞ → 0. The the limit evaluates to 0 · 0 which is not an indeterminate form. We conclude then that lim x · e1/x = 0. x→0− 3. This limit inially evaluates to the indeterminate form ∞−∞. By applying a logarithmic rule, we can rewrite the limit as ( ) x + 1 lim ln(x + 1) − ln x = lim ln . x→∞ x→∞ x

As x → ∞, the argument of the ln term approaches ∞/∞, to which we can apply l’Hôpital’s Rule. x + 1 by LHR 1 lim = = 1. x→∞ x 1 42

x + 1 Since x → ∞ implies → 1, it follows that x ( ) x + 1 x → ∞ implies ln → ln 1 = 0. x

Thus ( ) x + 1 lim ln(x + 1) − ln x = lim ln = 0. x→∞ x→∞ x Interpretaon: since this limit evaluates to 0, it means that for large x, there is essenally no diﬀerence between ln(x + 1) and ln x; their diﬀer- ence is essenally 0. 4. The limit lim x2 −ex inially returns the indeterminate form ∞−∞. We x→∞ ( ) ex can rewrite the expression by factoring out x2; x2 − ex = x2 1 − . x2 We need to evaluate how ex/x2 behaves as x → ∞:

ex by LHR ex by LHR ex lim = lim = lim = ∞. x→∞ x2 x→∞ 2x x→∞ 2

2 x 2 Thus limx→∞ x (1 − e /x ) evaluates to ∞ · (−∞), which is not an indeterminate form; rather, ∞ · (−∞) evaluates to −∞. We conclude that lim x2 − ex = −∞. x→∞ Interpretaon: as x gets large, the diﬀerence between x2 and ex grows very large.

Indeterminate Forms 00, 1∞ and ∞0

When faced with an indeterminate form that involves a power, it oen helps to employ the natural logarithmic funcon. The following Key Idea expresses the concept, which is followed by an example that demonstrates its use.

Key Idea 6.8.1 Evaluang Limits Involving Indeterminate Forms 00, 1∞ and ∞0 ( ) If lim ln f(x) = L, then lim f(x) = lim eln(f(x)) = e L. x→c x→c x→c 43

Example 6.8.4 Using l’Hôpital’s Rule with indeterminate forms involving exponents Evaluate the( following) limits. 1 x 1. lim 1 + 2. lim xx. x→∞ x x→0+

1. This is equivalent to a special limit given in Theorem 1.3.5; these limits have important applicaons within mathemacs and ﬁnance. Note that the exponent approaches ∞ while the base approaches 1, leading to the ∞ indeterminate form 1 . Let f(x) = (1 + 1(/x)x); the problem asks to evaluate lim f(x). Let’s ﬁrst evaluate lim ln f(x) . x→∞ x→∞ ( ) ( ) 1 x lim ln f(x) = lim ln 1 + →∞ →∞ x x ( x ) 1 = lim x ln 1 + →∞ x ( )x ln 1 + 1 = lim x x→∞ 1/x

This produces the indeterminate form 0/0, so we apply l’Hôpital’s Rule.

1 · − 2 + / ( 1/x ) = lim 1 1 x x→∞ (−1/x2) 1 = lim x→∞ 1 + 1/x = 1. ( ) Thus lim ln f(x) = 1. We return to the original limit and apply Key Idea x→∞ 6.8.1.

( ) 1 x lim 1 + = lim f(x) = lim eln(f(x)) = e1 = e. x→∞ x x→∞ x→∞

2. This limit leads to the indeterminate form 00. Let f(x) = xx and consider 44

( ) ﬁrst lim ln f(x) . y x→0+ ( ) 4 lim ln f(x) = lim ln (xx) x→0+ x→0+ 3 = lim x ln x x→0+ 2 ln x = lim . x→0+ 1/x 1 f(x) = xx −∞ ∞ x This produces the indeterminate form / so we apply l’Hôpital’s Rule...... 1 2 1/x Figure 6.8.1: A graph of f(x) = xx = lim x→0+ −1/x2 suppporng the fact that as x → 0+, f(x) → 1. = lim −x x→0+ = 0. ( ) Thus lim ln f(x) = 0. We return to the original limit and apply Key Idea x→0+ 6.8.1. lim xx = lim f(x) = lim eln(f(x)) = e0 = 1. x→0+ x→0+ x→0+ This result is supported by the graph of f(x) = xx given in Figure 6.8.1.

Our brief revisit of limits will be rewarded in the next secon where we consider improper integraon. So far, we have only∫ considered definite integrals 1 where the bounds are finite numbers, such as f(x) dx. Improper integraon 0 considers integrals where one, or both, of the bounds are “infinity.” Such integrals have many uses and applicaons, in addion to generang ideas that are enlightening.

Alternate one step approach if you are fluent with logs for exponential forms.

xx 0 limx → 0+ {0 } xx ln = limx → 0+ e x ln x = limx → 0+ e x {0·∞} x ln ∞ x + e 1/x = lim → 0  ∞  1/x LHR -1x2 = limx → 0+ e -x = limx → 0+ e = e0 = 1 45

Exercises 6.8 Terms and Concepts ex − x − 1 18. lim 2 x→0+ x 1. List the diﬀerent indeterminate forms described in this sec- x − sin x on. 19. lim 3 2 x→0+ x − x 2. T/F: l’Hôpital’s Rule provides a faster method of compung x4 derivaves. 20. lim x→∞ ex [ ] ′ d f(x) f (x) √ 3. T/F: l’Hôpital’s Rule states that = . x dx g(x) g′(x) 21. lim x→∞ ex ∞ 4. Explain what the indeterminate form “1 ” means. 1 22. lim ex x→∞ x2 5. Fill in the blanks: f(x) ex The Quoent Rule is applied to when taking 23. lim √ g(x) →∞ ; x x f(x) l’Hôpital’s Rule is applied to when taking certain ex g(x) 24. lim . x→∞ 2x

∞0 ex 6. Create (but do not evaluate!) a limit that returns “ ”. 25. lim x→∞ 3x 7. Create a funcon f(x) such that lim f(x) returns “00”. x→1 x3 − 5x2 + 3x + 9 26. lim x→3 x3 − 7x2 + 15x − 9 8. Create a funcon f(x) such that lim f(x) returns “0 · ∞”. x→∞ x3 + 4x2 + 4x 27. lim x→−2 x3 + 7x2 + 16x + 12

Problems 6.8 ln x 28. lim →∞ In Exercises 9 – 54, evaluate the given limit. x x ln(x2) x2 + x − 2 29. lim 9. lim x→∞ x x→1 x − 1 ( ) 2 x2 + x − 6 ln x 10. lim → 2 − 30. lim x 2 x 7x + 10 x→∞ x

sin x 31. lim x · ln x 11. lim → + x→π x − π x 0 √ sin x − cos x 32. lim x · ln x 12. lim x→0+ x→π/4 cos(2x) 33. lim xe1/x sin(5x) → + 13. lim x 0 x→0 x 34. lim x3 − x2 sin(2x) x→∞ 14. lim x→0 x + 2 √ 35. lim x − ln x x→∞ sin(2x) 15. lim x x→0 sin(3x) 36. lim xe x→−∞ sin(ax) 16. lim 1 −1/x → x 0 sin(bx) 37. lim 2 e x→0+ x ex − 1 ( + )1/x 17. lim 2 38. lim 1 x x→0+ x x→0+ 46

39. lim (2x)x 47. lim (1 + x2)1/x x→0+ x→∞

x 48. lim tan x cos x 40. lim (2/x) x→π/2 x→0+

x 49. lim tan x sin(2x) 41. lim (sin x) x→π/2 x→0+ 1 1 − 1−x 50. lim − 42. lim (1 x) x→1+ ln x x − 1 x→1+ 5 x 1/x 51. lim − 43. lim (x) → + 2 − − x→∞ x 3 x 9 x 3

x 52. lim x tan(1/x) 44. lim (1/x) x→∞ x→∞ (ln x)3 1−x 53. lim 45. lim (ln x) →∞ x→1+ x x x2 + x − 2 1/x 54. lim 46. lim (1 + x) x→1 x→∞ ln x x 55. lim 1 - x → x - Solutions 6.8 c x 0+

1. 0/0, ∞/∞, 0 · ∞, ∞ − ∞, 00, 1∞, ∞0 31. 0 2. F 32. 0 3. F 33. ∞ 4. The base of an expression is approaching 1 while its power is 34. ∞ growing without bound. 35. ∞ 5. derivaves; limits 36. 0 6. Answers will vary. 37. 0 7. Answers will vary. 38. e 8. Answers will vary. 39. 1 9. 3 40. 1 10. −5/3 41. 1 11. −1 42. 1 √ 12. − 2/2 43. 1 13. 5 44. 0 14. 0 45. 1 15. 2/3 46. 1 16. a/b 47. 1 17. ∞ 48. 1 18. 1/2 49. 2 19. 0 50. 1/2 20. 0 51. −∞ 21. 0 52. 1 22. ∞ 53. 0 23. ∞ 54. 3 ∞ 24. ∞ 55. + 25. 0 26. 2 27. −2 28. 0 29. 0 30. 0 47 Chapter 7 Techniques of Integration 7.1A Method of Substitution Review In applications, simple integrals like ∫cosx dx are rare. It is more likely you will encounter integrals x x like ∫cos(2 π ) dx or ∫cos(2.34 + 7 .49) dx. Fortunately these can often be worked with a slightly modified table of integrals. The idea of u is it can be any differentiable change of variable. Live math! Integral Table (Change of variable Form) ∫ du = u + C un+1 ∫ un du = + C u u n+1 ∫ e du = e + C au au du = + C ∫ ln a du  uu = ln|u| + C ∫ cosu du = sinu + C ∫ sinu du = -cosu + C 2 2 ∫ sec u du = tanu + C ∫ csc u du = -cotu + C

∫ secu tanu du = sec u + C ∫ cscucotu du = - csc u + C du du  = arcsin u + C  u2 = arctan u + C 1 - u2 1 +

Method of Substitution

u = g(x)  f(g(x)) g ' (x) dx = ∫ f(u) du du = g ' (x) dx Proof: the integral is live mathematics. 48

It is rare for an integral we are trying to evaluate to be exactly one on our Memory Integral List, really our only way so far of evaluating integrals other than making the list longer. The Method of Substitution is by far the most common and important way to transform an integral into one on the list. This method of evaluating integrals is so important we will spend two lectures on getting fluent at it.

The previous chapter we completed the basic calculus of most of the functions a well educated calculus user should know. Most combinations of these functions are easy to differentiate. However, many are difficult to integrate. This chapter is devoted to exploring techniques of andiﬀerenaon. While not every funcon has an anderivave in terms of elementary funcons we can sll ﬁnd anderivaves of a wide variety of funcons. Nevertheless, many remain impossible to integrate. For these we will learn approximate integration; however, realistically this job is best done by computer.

7.1 A Substuon

We movate this secon with an example. Let f(x) = (x2 + 3x − 5)10. ′ We can compute f (x) using the Chain Rule. It is: ′ f (x) = 10(x2 + 3x − 5)9 · (2x + 3) = (20x + 30)(x2 + 3x − 5)9.

∫ Now consider this: What is (20x + 30)(x2 + 3x − 5)9 dx? We have the answer in front of us; ∫ (20x + 30)(x2 + 3x − 5)9 dx = (x2 + 3x − 5)10 + C.

How would we have evaluated this indefinite integral without starng with f(x) as we did? This secon explores integraon by substuon. It allows us to “undo the Chain Rule.” Substuon allows us to evaluate the above integral without knowing the original funcon first. ∫ The underlying principle is to rewrite a∫ “complicated” integral of the form f(x) dx as a not–so–complicated integral h(u) du. We’ll formally establish la∫ ter how this is done. First, consider again our introductory indefinite integral, (20x + 30)(x2 + 3x − 5)9 dx. Arguably the most “complicated” part of the integrand is (x2 + 3x − 5)9. We wish to make this simpler; we do so through a substuon. Let u = x2 + 3x − 5. Thus

(x2 + 3x − 5)9 = u9. 49

We have established u as a funcon of x, so now consider the diﬀerenal of u:

du = (2x + 3)dx.

Keep in mind that (2x+3) and dx are mulplied; the dx is not “just sing there.” Return to the original integral and do some substuons through algebra: ∫ ∫ (20x + 30)(x2 + 3x − 5)9 dx = 10(2x + 3)(x2 + 3x − 5)9 dx ∫ 2 − 9 = 10(x| +{z3x 5}) |(2x +{z3) dx} ∫ u du = 10u9 du

10 = u + C (replace u with x2 + 3x − 5) = (x2 + 3x − 5)10 + C

One might well look at this and think “I (sort of) followed how that worked, but I could never come up with that on my own,” but the process is learnable. This secon contains numerous examples through which the reader will gain understanding and mathemacal maturity enabling them to regard substuon as a natural tool when evaluang integrals. We stated before that integraon by substuon “undoes” the Chain Rule. Speciﬁcally, let F(x) and g(x) be diﬀerenable funcons and consider the derivave of their composion: ( ( )) d ′ ′ F g(x) = F (g(x))g (x). dx Thus ∫ ′ ′ F (g(x))g (x) dx = F(g(x)) + C.

Integraon by substuon works by recognizing the “inside” funcon g(x) and replacing it with a variable. By seng u = g(x), we can rewrite the derivave as ( ( )) d ′ ′ F u = F (u)u . dx Since du = g′(x)dx, we can rewrite the above integral as ∫ ∫ ′ ′ ′ F (g(x))g (x) dx = F (u)du = F(u) + C = F(g(x)) + C.

This concept is important so we restate it in the context of a theorem. 50

Theorem 7.1.1 Integraon by Substuon Let F and g be diﬀerenable funcons, where the range of g is an interval I contained in the domain of F. Then ∫ ′ ′ F (g(x))g (x) dx = F(g(x)) + C.

If u = g(x), then du = g′(x)dx and ∫ ∫ ′ ′ ′ F (g(x))g (x) dx = F (u) du = F(u) + C = F(g(x)) + C.

The∫ point of substuon is to make the integraon step easy. Indeed, the step F′(u) du = F(u)+C looks easy, as the anderivave of the derivave of F is just F, plus a constant. The “work” involved is making the proper substuon. There is not a step–by–step process that one can memorize; rather, experience will be one’s guide. To gain experience, we now embark on many examples.

Example ∫7.1.1 Integrang by substuon Evaluate x sin(x2 + 5) dx.

S Knowing that substuon is related to the Chain Rule, we choose to let u be the “inside” funcon of sin(x2 + 5)*. (This is not always a good choice, but it is oen the best place to start.) Let u = x2 + 5, hence du = 2x dx. The integrand has an x dx term, but not a 2x dx term. (Recall that mulplicaon is commutave, so the x does not physically have to be next to dx for there to be an x dx term.) We can divide both sides of the du expression by 2:

1 du = 2x dx ⇒ du = x dx. 2

We can now substute. ∫ ∫ 2 2 x sin(x + 5) dx = sin(x| {z+ 5}) |{z}x dx u 1 du ∫ 2 1 = sin u du 2

*alternate way of thinking that is productive and insightful: The Method Choose a u for which there is (up to a constant) a du in the correct position with nothing left remaining. 51

1 = − cos u + C (now replace u with x2 + 5) 2 1 = − cos(x2 + 5) + C. 2 ∫ 2 − 1 2 Thus x sin(x + 5) dx = 2 cos(x + 5) + C. We can check our work by evaluang the derivave of the right hand side.

Example ∫7.1.2 Integrang by substuon Evaluate cos(5x) dx.

S Again let u replace the “inside” funcon. Leng u = 5x, we have du = 5dx. Since our integrand does not have a 5dx term, we can divide 1 the previous equaon by 5 to obtain 5 du = dx. We can now substute. ∫ ∫ cos(5x) dx = cos(|{z}5x ) |{z}dx u 1 du ∫ 5 1 = cos u du 5 1 = sin u + C 5 1 = sin(5x) + C. 5 We can again check our work through diﬀerenaon.

The previous example exhibited a common, and simple, type of substuon. The “inside” funcon was a linear funcon (in this case, y = 5x). When the inside funcon is linear, the resulng integraon is very predictable, outlined here.

Key Idea 7.1.1 Substuon With A Linear Funcon ∫ Consider F′(ax + b) dx, where a ≠ 0 and b are constants. Leng u = ax + b gives du = a · dx, leading to the result ∫ ′ 1 F (ax + b) dx = F(ax + b) + C. a

∫ − − 1 − Thus sin(7x 4) dx = 7 cos(7x 4) + C. Our next example can use Key Idea 6.1.1, but we will only employ it aer going through all of the steps. 52

Example ∫7.1.3 Integrang by substung a linear funcon 7 Evaluate dx. −3x + 1

S View the integrand as the composion of funcons f(g(x)), where f(x) = 7/x and g(x) = −3x + 1. Employing our understanding of substuon, we let u = −3x+1, the inside funcon. Thus du = −3dx. The integrand lacks a −3; hence divide the previous equaon by −3 to obtain −du/3 = dx. We can now evaluate the integral through substuon. ∫ ∫ 7 7 du − dx = − 3x + 1 u∫ 3 −7 du = 3 u −7 = ln |u| + C 3 7 = − ln | − 3x + 1| + C. 3 Using Key Idea 6.1.1 is faster, recognizing that u is linear and a = −3. One may want to connue wring out all the steps unl they are comfortable with this parcular shortcut.

Not all integrals that beneﬁt from substuon have a clear “inside” funcon. Several of the following examples will demonstrate ways in which this occurs.

Example ∫7.1.4 Integrang by substuon Evaluate sin x cos x dx.

S There is not a composion of funcon here to exploit; rather, just a product of funcons. Do not be afraid to experiment; when given an integral to evaluate, it is oen beneﬁcial to think “If I let u be this, then du must be that …” and see if this helps simplify the integral at all. In this example, let’s set u = sin x. Then du = cos x dx, which we have as part of the integrand! The substuon becomes very straighorward: ∫ ∫ sin x cos x dx = u du 1 = u2 + C 2 1 = sin2 x + C. 2 53

One would do well to ask “What would happen if we let u = cos x?” The result is just as easy to find, yet looks very different. The challenge to the reader is to evaluate the integral leng u = cos x and discover why the answer is the same, yet looks different.

Our examples so far have required “basic substuon.” The next example demonstrates how substuons can be made that oen strike the new learner as being “nonstandard.”

Example ∫ .1.5 Integrang by substuon 7 √ variation Evaluate x x + 3 dx.

S Recognizing the composion of funcons, set u = x + 3. Then du = dx, giving what seems inially to be a simple substuon. But at this stage, we have: ∫ ∫ √ √ x x + 3 dx = x u du.

We cannot evaluate an integral that has both an x and an u in it. We need to convert the x to an expression involving just u. Since we set u = x+3, we can also state that u−3 = x. Thus we can replace √ 1 x in the integrand with u − 3. It will also be helpful to rewrite u as u 2 . ∫ ∫ √ 1 x x + 3 dx = (u − 3)u 2 du ∫ ( ) 3 1 = u 2 − 3u 2 du

2 5 3 = u 2 − 2u 2 + C 5 2 5 3 = (x + 3) 2 − 2(x + 3) 2 + C. 5 Checking your work is always a good idea. In this parcular case, some algebra will be needed to make one’s answer match the integrand in the original problem.

Example ∫7.1.6 Integrang by substuon 1 Evaluate dx. x ln x

S This is another example where there does not seem to be an obvious composion of funcons. The line of thinking used in Example 6.1.5 is useful here: choose something for u and consider what this implies du must 54

be. If u can be chosen such that du also appears in the integrand, then we have chosen well. Choosing u = 1/x makes du = −1/x2 dx; that does not seem helpful. How- ever, seng u = ln x makes du = 1/x dx, which is part of the integrand. Thus:

∫ ∫ 1 1 1 dx = dx x ln x |{z}ln x |x{z} ∫ u du 1 = du u = ln |u| + C = ln | ln x| + C.

The ﬁnal answer is interesng; the natural log of the natural log. Take the derivave to conﬁrm this answer is indeed correct.

Integrals Involving Trigonometric Funcons Secon 7.3 delves deeper into integrals of a variety of trigonometric func-ons; here we use substuon to establish a foundaon that we will build upon.

The next three examples will help ﬁll in some missing pieces of our anderiva-ve knowledge. We know the anderivaves of the sine and cosine funcons; what about the other standard funcons tangent, cotangent, secant and cosecant? We discover these next.

Example ∫7.1.7 Integraon by substuon: anderivaves of tan x Evaluate tan x dx.

S The previous paragraph established that we did not know the anderivaves of tangent, hence we must assume that we have learned something in this secon that can help us evaluate this indeﬁnite integral. Rewrite tan x as sin x/ cos x. While the presence of a composion of funcons may not be immediately obvious, recognize that cos x is “inside” the 1/x funcon. Therefore, we see if seng u = cos x returns usable results. We have 55

that du = − sin x dx, hence −du = sin x dx. We can integrate: ∫ ∫ sin x tan x dx = dx ∫ cos x 1 = |sin{zx dx} |cos{z}x −du ∫ u −1 = du u = − ln |u| + C = − ln | cos x| + C.

Some texts prefer to bring the −1 inside the logarithm as a power of cos x, as in:

− − ln | cos x| + C = ln |(cos x) 1| + C

1 = ln + C cos x = ln | sec x| + C. ∫ Thus the result they give is tan x dx = ln | sec x| + C. These two answers are equivalent.

Example ∫7.1.8 Integrang by substuon: anderivaves of sec x Evaluate sec x dx.

S This example employs a wonderful trick: mulply the integrand by “1” so that we see how to integrate more clearly. In this case, we write “1” as sec x + tan x 1 = . sec x + tan x

This may seem like it came out of le* ﬁeld, but it works beaufully. Consider: ∫ ∫ sec x + tan x sec x dx = sec x · dx + ∫ sec x tan x sec2 x + sec x tan x = dx. sec x + tan x

*No disparaging of left handed persons intended. 56

Now let u = sec x + tan x; this means du = (sec x tan x + sec2 x) dx, which is our numerator. Thus: ∫ du = u = ln |u| + C = ln | sec x + tan x| + C.

We can use similar techniques to those used in Examples 7.1.7 and 7.1.8 to ﬁnd anderivaves of cot x and csc x (which the reader can explore in the exercises.) We summarize our results here.

We explore one more common trigonometric integral.

Example ∫7.1.9 Integraon by substuon: powers of cos x and sin x Evaluate cos2 x dx.

( ) 2 S We have a composion of funcons as cos2 x = cos x . However, seng u = cos x means du = − sin x dx, which we do not have in the integral. Another technique is needed. The process we’ll employ is to use a Power Reducing formula for cos2 x (perhaps consult the back of this text for this formula), which states 1 + cos(2x) cos2 x = . 2 The right hand side of this equaon is not diﬃcult to integrate. We have: ∫ ∫ 1 + cos(2x) cos2 x dx = dx ∫ ( 2 ) 1 1 = + cos(2x) dx. 2 2 57

Now use Key Idea 7.1.1:

1 1 sin(2x) = x + + C 2 2 2 1 sin(2x) = x + + C. 2 4

We’ll make signiﬁcant use of this power–reducing technique in future secons.

Simplifying the Integrand

It is common to be reluctant to manipulate the integrand of an integral; at first, our grasp of integraon is tenuous and one may think that working with the integrand will improperly change the results. Integraon by substuon works using a different logic: as long as equality is maintained, the integrand can be manipulated so that its form is easier to deal with. The next two examples demonstrate common ways in which using algebra first makes the integraon easier to perform.

Example ∫7.1.10 Integraon by substuon: simplifying ﬁrst

x 3 + 42 x + 8x + 5 Evaluate dx. x2 + 2x + 1

S One may try to start by seng u equal to either the numerator or denominator; in each instance, the result is not workable. When dealing with raonal funcons (i.e., quoents made up of polynomial funcons), it is an almost universal rule that everything works beer when the degree of the numerator is less than the degree of the denominator. Hence we use polynomial division. We skip the speciﬁcs of the steps, but note that when x2 + 2x + 1 is divided into x3 + 4x2 + 8x + 5, it goes in x + 2 mes with a remainder of 3x + 3. Thus

x3 + 4x2 + 8x + 5 3x + 3 = x + 2 + . x2 + 2x + 1 x2 + 2x + 1

Integrang x + 2 is simple. The fracon can be integrated by seng u = x2 + 2x + 1, giving du = (2x + 2) dx. This is very similar to the numerator. Note that du/2 = (x + 1) dx and then consider the following: ∫ ∫ ( ) x3 + 4x2 + 8x + 5 3x + 3 dx = x + 2 + dx x2 + 2x + 1 x2 + 2x + 1 ∫ ∫ 3(x + 1) = (x + 2) dx + dx x2 + 2x + 1 ∫ 1 3 du = x2 + 2x + C + 2 1 u 2 1 3 = x2 + 2x + C + ln |u| + C 2 1 2 2 1 3 = x2 + 2x + ln |x2 + 2x + 1| + C. 2 2 In some ways, we “lucked out” in that aer dividing, substuon was able to be done. In later secons we’ll develop techniques for handling raonal funcons where substuon is not directly feasible. 58

Example∫ 6 7.1.11 Integraon by alternate methods x2 + 2x + 3 Evaluate √ dx with, and without, substuon. x S We already know how to integrate this parcular example. √ 1 Rewrite x as x 2 and simplify the fracon:

2 x + 2x + 3 3 1 − 1 = x 2 + 2x 2 + 3x 2 . x1/2 We can now integrate using the Power Rule: ∫ ∫ 2 ( ) x + 2x + 3 3 1 − 1 dx = x 2 + 2x 2 + 3x 2 dx x1/2 2 5 4 3 1 = x 2 + x 2 + 6x 2 + C 5 3 This is a perfectly ﬁne approach. We demonstrate how this can also be solved using substuon as its implementaon is rather clever. √ 1 Let u = x = x 2 ; therefore

1 − 1 1 1 du = x 2 dx = √ dx ⇒ 2du = √ dx. 2 2 x x ∫ ∫ x2 + 2x + 3 This gives us √ dx = (x2 + 2x + 3) · 2 du. What are we to do x 1 2 2 with the other x terms? Since u = x 2 , u = x, etc. We can then replace x and x with appropriate powers of u. We thus have ∫ ∫ x2 + 2x + 3 √ dx = (x2 + 2x + 3) · 2 du x ∫ = 2(u4 + 2u2 + 3) du 2 4 = u5 + u3 + 6u + C 5 3 2 5 4 3 1 = x 2 + x 2 + 6x 2 + C, 5 3 which is obviously the same answer we obtained before. In this situaon, substuon is arguably more work than our other method. The fantasc thing is that it works. It demonstrates how ﬂexible integraon is. 59

Exercises 7.1 A Solutions 7.1 A Terms and Concepts

1. Substuon “undoes” what derivave rule? 1. Chain Rule. 2. T 2. T/F: One can use algebra to rewrite the integrand of an in- 1 (x3 − 5)8 + C tegral to make it easier to evaluate. 8

Problems

In Exercises 3 – 14, evaluate the indeﬁnite integral to develop an understanding of Substuon. ∫ ( ) 3. 2 3 − 7 3. 3x x 5 dx 1 ( 2 − + )4 + 4. 4 x 5x 7 C ( ) ∫ 1 2 + 9 + ( ) 5. 18 x 1 C − 2 − 3 4. (2x 5) x 5x + 7 dx 1 ( 2 + − )6 + 6. 3 3x 7x 1 C ∫ 7. 1 ln |2x + 7| + C ( ) 2 2 8 √ 5. x x + 1 dx 8. 2x + 3 + C √ ∫ 9. 2 (x + 3)3/2 − 6(x + 3)1/2 + C =2 (x − 6) x + 3 + C ( ) 3 3 5 ( ) ( + ) 2 + − 2 3/2 2 − + 6. 12x 14 3x 7x 1 dx 10. 21 x 3x 7 C √ ∫ 11. 2e x + C √ 1 5 7. dx 12. 2 x +1 + C 2x + 7 5 − 1 − 1 + ∫ 13. 2x2 x C 1 √ 2 8. dx ln (x) + 2x + 3 14. 2 C ∫ x 9. √ dx x + 3 ∫ x3 − x 10. √ dx x ∫ √ e x 11. √ dx x ∫ x4 12. √ dx x5 + 1 ∫ 1 + 1 13. x dx x2 ∫ ln(x) 14. dx x

In Exercises 15 – 24, use Substuon to evaluate the indeﬁ- nite integral involving trigonometric funcons. 3 ∫ sin (x) + 15. 3 C 2 15. sin (x) cos(x)dx 4 − cos (x) + 16. 4 C ∫ 16. cos3(x) sin(x)dx

∫ ( ) ln x3 35. 3 (ln x)2 + C 35. dx 2 x 1 ■2 60

7.1B Advanced Methods of Substitution. Definite Integrals Integral Table * ∫ du = u + C un+1 ∫ un du = + C u u n+1 ∫ e du = e + C au au du = + C ∫ ln a du  uu = ln|u| + C ∫ cosu du = sinu + C ∫ sinu du = -cosu + C 2 2 ∫ sec u du = tanu + C ∫ csc u du = -cotu + C

∫ secu tanu du = sec u + C ∫ cscucotu du = - csc u + C du du  = arcsin u + C  u2 = arctan u + C 1 - u2 1 +

u = g(x) Method of Integration bf g x g x g(b) f u ∫a ( ( )) ' ( ) dx = ∫g(a) ( ) du du = g' (x) dx Definite Integral form * This is the basic integral list everyone should memorize and remember forever! Substuon and Inverse Trigonometric Funcons When studying derivaves of inverse funcons, we learned that ( ) d − 1 tan 1 x = . dx 1 + x2 Applying the Chain Rule to this is not diﬃcult; for instance, ( ) d − 5 tan 1 5x = . dx 1 + 25x2 We now explore how Substuon can be used to “undo” certain derivaves that are the result of the Chain Rule applied to Inverse Trigonometric funcons. We begin with an example.

Example ∫ 6 7.1.12 Integrang by substuon: inverse trigonometric funcons ■ 1 Evaluate dx. 25 + x2 S gent funcon. Note: 1 1 = 25 + x2 x2 25(1 + 25 ) 1 = ( ) x 2 25(1 + 5 ) 1 1 = () . 25 x 2 1 + 5

1 + 5 61

Thus ∫ ∫ 1 1 1 dx = ( ) dx. 25 + x2 25 x 2 1 + 5 This can be integrated using Substuon. Set u = x/5, hence du = dx/5 or dx = 5du. Thus ∫ ∫ 1 1 1 ( ) dx = 2 dx 25 + x2 25 1 + x ∫ 5 1 1 = du 5 1 + u2 1 − = tan 1 u + C 5 ( ) 1 − x = tan 1 + C 5 5

Example 7.1.12 demonstrates a general technique that can be applied to other integrands that result in inverse trigonometric funcons. The results are summarized here.

Theorem 7.1.3 Integrals Involving Inverse Trigonometric Funcons Let a > 0. ∫ ( ) 1 1 − x 1. dx = tan 1 + C a2 + x2 a a ∫ ( ) 1 − x 2. √ dx = sin 1 + C a2 − x2 a ∫ ( ) 1 1 − |x| 3. √ dx = sec 1 + C x x2 − a2 a a

Let’s pracce using Theorem 7.1.3.

Example 7.1.13 Integrang by substuon: inverse trigonometric funcons Evaluate the given indeﬁnite integrals. ∫ ∫ ∫ 1 1 1 . , . √ . √ . 1 2 dx 2 dx 3 dx 9 + x 2 − 1 5 − x2 x x 100 62

S Each can be answered using a straighorward applicaon of Theorem 7.1.3. ∫ 1 1 − x 1. dx = tan 1 + C, as a = 3. 9 + x2 3 3 ∫ √ 1 −1 1 2. dx = 10 sec 10x + C, as a = 10 . 2 − 1 x x 100 ∫ √ 1 − x 3. √ = sin 1 √ + C, as a = 5. 5 − x2 5

Most applicaons of Theorem 7.1.3 are not as straighorward. The next examples show some common integrals that can sll be approached with this theorem.

Example ∫7.1.14 Integrang by substuon: compleng the square 1 Evaluate dx. x2 − 4x + 13

S Inially, this integral seems to have nothing in common with the integrals in Theorem 7.1.3. As it lacks a square root, it almost certainly is not related to arcsine or arcsecant. It is, however, related to the arctangent funcon. We see this by compleng the square in the denominator. We give a brief reminder of the process here. Start with a quadrac with a leading coeﬃcient of 1. It will have the form of x2 +bx+c. Take 1/2 of b, square it, and add/subtract it back into the expression. I.e., b2 b2 x2 + bx + c = x2 + bx + − + c | {z 4} 4 (x+b/2)2 ( ) b 2 b2 = x + + c − 2 4 In our example, we take half of −4 and square it, geng 4. We add/subtract it into the denominator as follows:

1 1 = 2 − 2 − − x 4x + 13 |x {z4x + 4} 4 + 13 (x−2)2 1 = (x − 2)2 + 9 63

We can now integrate this using the arctangent rule. Technically, we need to substute ﬁrst with u = x − 2, but we can employ Key Idea 7.1.1 instead. Thus we have ∫ ∫ 1 1 1 − x − 2 dx = dx = tan 1 + C. x2 − 4x + 13 (x − 2)2 + 9 3 3

Example ∫7.1.15 Integrals requiring mulple methods 4 − x Evaluate √ dx. 16 − x2

S This integral requires two diﬀerent methods to evaluate it. We get to those methods by spling up the integral: ∫ ∫ ∫ 4 − x 4 x √ dx = √ dx − √ dx. 16 − x2 16 − x2 16 − x2 The ﬁrst integral is handled using a straighorward applicaon of Theorem 7.1.3; the second integral is handled by substuon, with u = 16−x2. We handle each separately.∫ 4 − x √ dx = 4 sin 1 + C. 16 − x2 4 ∫ x √ dx: Set u = 16 − x2, so du = −2xdx and xdx = −du/2. We 16 − x2 have ∫ ∫ x −du/2 √ dx = √ − 2 16 x ∫ u 1 1 = − √ du √2 u = − u + C √ = − 16 − x2 + C. Combining these together, we have ∫ √ 4 − x − x √ dx = 4 sin 1 + 16 − x2 + C. 16 − x2 4 64

Substuon and Deﬁnite Integraon

This secon has focused on evaluang indefinite integrals as we are learning a new technique for finding anderivaves. However, much of the me integraon is used in the context of a definite integral. Definite integrals that require substuon can readily be calculated.

b u = g(x) g(b) f(g(x)) g (x) f(u) du The integral is live mathematics. ∫a ' dx = ∫g(a) x=a ⇒ u=g(a) x=b ⇒ u=g(b)

Theorem .1.4 Substuon with Deﬁnite Integrals Let F and g be diﬀerenable funcons, where the range of g is an interval I that is contained in the domain of F. Then ∫ ∫ b ( ) g(b) ′ ′ ′ F g(x) g (x) dx = F (u) du. a g(a)

In eﬀect, Theorem 7.1.4 states that once you convert to integrang with re-spect to u, you do not need to switch back to evaluang with respect to x. A few examples will help one understand.

Example ∫7.1.16 Deﬁnite integrals and substuon: changing the bounds y 2 − − Evaluate cos(3x 1) dx using Theorem 7.1.4. 1 y = cos(3x 1) 0

0.5 S Observing the composion of funcons, let u = 3x − 1, hence du = 3dx. As 3dx does not appear in the integrand, divide the laer x equaon by 3 to get du/3 = dx. −1 1 2 3 4 5 By seng u = 3x − 1, we are implicitly stang that g(x) = 3x − 1. Theorem −0.5 7.1.4 states that the new lower bound is g(0) = −1; the new upper bound is

−1 (a) y 1

1 0.5 y = 3 cos(u)

u −1 1 2 3 4 5

−0.5

−1 7.1.1: Graphing the areas deﬁned by the deﬁnite integrals of Example 7.1.16. 65 y − 1 y = cos(3x 1)

0.5 g(2) = 5. We now evaluate the deﬁnite integral: ∫ ∫ x 2 5 du −1 1 2 3 4 5 cos(3x − 1) dx = cos u − 3 −0.5 0 1 5 1 = sin u −1 3 −1 1( ) (a) = sin 5 − sin(−1) ≈ −0.039. 3 y 1 Noce how once we converted the integral to be in terms of u, we never went ..... back to using x. 1 0.5 y = 3 cos(u) The graphs in Figure 7.1.1 tell more of the story. In (a) the area deﬁned by

the original integrand is shaded, whereas in (b) the area deﬁned by the new in- u tegrand is shaded. In this parcular situaon, the areas look very similar; the −1 1 2 3 4 5

new region is “shorter” but “wider,” giving the same area. −0.5 Example 7.1.17 Definite integrals and substuon: changing the bounds ∫6.1.1 −1 _π/2 Evaluate sin x cos x dx using Theorem 7.1.4. (b) 0 Figure 7.1.1: Graphing the areas defined ..... S We saw the corresponding indefinite integral in Example 7.1.4. by the definite integrals of Example 6.1.16. In that example we set u = sin x but stated that we could have let u = cos x. For variety, we do the laer here. y Let u = g(x) = cos x, giving du = − sin x dx and hence sin x dx = −du. The 1 new upper bound is g(π/2) = 0; the new lower bound is g(0) = 1. Note how the lower bound is actually larger than the upper bound now. We have y = sin x cos x 0.5 ∫ ∫ π/2 0 sin x cos x dx = −u du (switch bounds & change sign) 0 1 x ∫ π 1 1 2 = u du 0 −0.5 1 1 = u2 = 1/2. (a) 2 0 y In Figure 7.1.2 we have again graphed the two regions defined by our definite 1 y = u integrals. Unlike the previous example, they bear no resemblance to each other...... However, Theorem 7.1.4 guarantees that they have the same area. 0.5 Integraon by substuon is a powerful and useful integraon technique. The next secon introduces another technique, called Integraon by Parts. As u 1 π substuon “undoes” the Chain Rule, integraon by parts “undoes” the Product 2 Rule. Together, these two techniques provide a strong foundaon on which most other integraon techniques are based. −0.5 (b)

Figure 7.1.2: Graphing the areas de-ﬁned Expert Method by the deﬁnite integrals of Example .....6.1.17. Example God’s Method of Substitution:

x2 x 2 ∫(1 + 2 x)2 e dx Let u = x e x2 x2 du = (1·e + x·e ·2x)dx 2 = (1 + 2 x2) ex dx = ∫du = u + C x2 = x e + C 66

Exercises ∫ 7.1 B Solutions 7.1 B 17. − 1 sin(3 − 6x) + C 17. cos(3 − 6x)dx 6 18. − tan(4 − x) + C ∫ 19. 1 ln | sec(2x) + tan(2x)| + C 18. sec2(4 − x)dx 2 3 20. tan (x) + C ∫ 3 sin(x2) + 19. sec(2x)dx 21. 2 C − ∫ 22. tan(x) x + C 20. tan2(x) sec2(x)dx 23. The key is to rewrite cot x as cos x/ sin x, and let u = sin x. 24. The key is to mulply csc x by 1 in the form ∫ ( ) (csc x + cot x)/(csc x + cot x). 21. x cos x2 dx ∫ 22. tan2(x)dx ∫ 23. cot x dx. Do not just refer to Theorem 6.1.2 for the answer; jusfy it through Substuon. ∫ 24. csc x dx. Do not just refer to Theorem 6.1.2 for the answer; jusfy it through Substuon. ∫ 14 45. √ dx 5 − x2 ∫ 2 46. √ dx x x2 − 9 ∫ 5 47. √ dx x4 − 16x2

In Exercises 79 – 86, evaluate the deﬁnite integral. ∫ 79. − ln 2 3 1 79. − dx 80. 352/15 1 x 5 81. 2/3 ∫ 6 √ 82. 1/5 80. x x − 2dx 2 83. (1 − e)/2 ∫ π/ π/2 84. 2 2 81. sin x cos x dx 85. π/2 −π/2 86. π/6 ∫ 1 82. 2x(1 − x2)4 dx 0 ∫ −1 2 83. (x + 1)ex +2x+1 dx −2 ∫ 1 1 84. 2 dx −1 1 + x ∫ 4 1 85. dx 2 − 2 x 6x + 10 ∫ √ 3 1 86. √ dx 2 1 4 − x 67

Integraon by Parts

Here’s a simple integral that we can’t yet evaluate:

x cos x dx. ∫ It’s a simple maer to take the derivave of the integrand using the Product Rule, but there is no Product Rule for integrals. However, this secon introduces Integraon by Parts, a method of integraon that is based on the Product Rule for derivaves. It will enable us to evaluate this integral. The Product Rule says that if u and v are funcons of x, then (uv)′ = u′v+uv′. For simplicity, we’ve wrien u for u(x) and v for v(x). Suppose we integrate both sides with respect to x. This gives

(uv)′ dx = (u′v + uv′) dx. ∫ ∫ By the Fundamental Theorem of Calculus, the le side integrates to uv. The right side can be broken up into two integrals, and we have

uv = u′v dx + uv′ dx. ∫ ∫ Solving for the second integral we have

uv′ dx = uv u′v dx. − ∫ ∫ Using diﬀerenal notaon, we can write du = u′(x)dx and dv = v′(x)dx and the expression above can be wrien as follows:

u dv = uv v du. − ∫ ∫ This is the Integraon by Parts formula. For reference purposes, we state this in a theorem.

Theorem 7.. Integraon by Parts Let u and v be diﬀerenable funcons of x on an interval I containing a and b. Then u dv = uv v du, − ∫ ∫ and x=b b x=b u dv = uv v du. x=a a − x=a ∫ ∫

Alternate Notation Derivaton

d du dv v + dx (uv) = dx u dx Product Rule d(u v) = v du + u dv Differential Form ∫d(uv) = ∫vdu + ∫udv Integrating u v = ∫vdu + ∫udv

∫udv = u v - ∫vdu Rearranging Useful for: 1. 'unlikely products' 2. inverse functions 3. all else fails. 68

Let’s try an example to understand our new technique.

Example 7.. Integrang using Integraon by Parts Evaluate x cos x dx. ∫ The following rule almost always works: 1. Let dv = hardest part you can integrate 2. Let u = remaining part of the integral cc (up to a constant).

dv = cos x dx v = sin x u = x ⇒ du = dx

Figure 7..: Seng up Integraon by Parts.

Now substute all of this into the Integraon by Parts formula, giving

x cos x dx = x sin x sin x dx. − ∫ ∫ We can then integrate sin x to get cos x + C and overall our answer is − x cos x dx = x sin x + cos x + C. ∫ Note how the anderivave contains a product, x sin x. This product is what makes Integraon by Parts necessary.

The example above demonstrates how Integraon by Parts works in general. We try to idenfy u and dv in the integral we are given, and the key is that we usually want to choose u and dv so that du is simpler than u and v is hopefully not too much more complicated than dv. This will mean that the integral on the

In the example above, we chose u = x and■dv = cos x dx. Then du = dx was simpler than u and v = sin x is no more complicated∫ than dv. Therefore, instead of integrang x cos x dx, we could integrate sin x dx, which we knew how to do. right side of the Integraon by Parts formula, v du will be simpler to integrate than the original integral u dv. ∫ 69

We now consider another example.

Example 7.. Integrang using Integraon by Parts Evaluate xex dx. ∫ S The integrand contains an Algebraic term (x) and an Exponenal term (ex). Our mnemonic suggests leng u be the algebraic term, so we choose u = x and dv = ex dx. Then du = dx and v = ex as indicated by the tables below.

dv = ex dx v = ex u = x ⇒ du = dx

Figure 7..: Seng up Integraon by Parts.

We see du is simpler than u, while there is no change in going from dv to v. This is good. The Integraon by Parts formula gives

xex dx = xex ex dx. − ∫ ∫ The integral on the right is simple; our ﬁnal answer is

xex dx = xex ex + C. − ∫ Note again how the anderivaves contain a product term.

Example 7.. Integrang using Integraon by Parts Evaluate x cos x dx. ∫ S The mnemonic suggests leng u = x instead of the trigonometric funcon, hence dv = cos x dx. Then du = x dx and v = sin x as shown below.

dv = cos x dx v = sin x u = x ⇒ du = 2x dx

Figure 7..: Seng up Integraon by Parts. 70

The Integraon by Parts formula gives

x cos x dx = x sin x x sin x dx. − ∫ ∫ At this point, the integral on the right is indeed simpler than the one we started with, but to evaluate it, we need to do Integraon by Parts again. Here we choose u = x and dv = sin x and ﬁll in the rest below.

dv = sin x dx v = cos x − u = x ⇒ du = dx

Figure 7..: Seng up Integraon by Parts (again).

x cos x dx = x sin x x cos x cos x dx . − − − − ∫ ( ∫ ) The integral all the way on the right is now something we can evaluate. It evaluates to sin x. Then going through and simplifying, being careful to keep all − the signs straight, our answer is

x cos x dx = x sin x + x cos x sin x + C. − ∫

Example 7.. Integrang using Integraon by Parts Evaluate ex cos x dx. ∫ S This is a classic problem. In this parcular example, one can let dv be either cos x dx or ex dx.

dv = cos x dx v = sin x u = ex ⇒ du = ex dx

Figure 7..: Seng up Integraon by Parts.

Noce that du is no simpler than u, going against our general rule (but bear with us). The Integraon by Parts formula yields

ex cos x dx = ex sin x ex sin x dx. − ∫ ∫ 71

The integral on the right is not much diﬀerent than the one we started with, so it seems like we have goen nowhere. Let’s keep working and apply Integraon by Parts to the new integral, using u = ex and dv = sin x dx. This leads us to the following:

dv = sin x dx v = cos x − u = ex ⇒ du = ex dx

Figure 7..: Seng up Integraon by Parts (again).

The Integraon by Parts formula then gives:

ex cos x dx = ex sin x ex cos x ex cos x dx − − − − ∫ ( ∫ ) = ex sin x + ex cos x ex cos x dx. 'Oh, shoot' − ∫ It seems we are back right where we started, as the right hand side contains ex cos x dx. But this is actually a good thing. ∫ Add ex cos x dx to both sides. This gives ∫

x x x e cos x dx = e sin x + e cos x 'Boot strapping!' In cowboy lore, boot ∫ strapping was the idea that you could Now divide both sides by : lift yourself up into the air by pulling up on the straps at the top back of your cowboy boots. ex cos x dx = ex sin x + ex cos x . ∫ ( ) Simplifying a lile and adding the constant of integraon, our answer is thus ex cos x dx = ex (sin x + cos x) + C. ∫ Example 7.. Integrang using Integraon by Parts: anderivave of ln x Evaluate ln x dx.

S One may have noced that we have rules for integrang the familiar trigonometric funcons and ex, but we have not yet given a rule for integrang ln x. That is because ln x can’t easily be integrated with any of the rules we have learned up to this point. But we can ﬁnd its anderivave by a 72

clever applicaon of Integraon by Parts. Set u = ln x and dv = dx. This is a good, sneaky trick to learn as it can help in other situaons. This determines du = (/x) dx and v = x as shown below.

dv = dx v = x u = ln x ⇒ du = /x dx

Figure 7..: Seng up Integraon by Parts.

Pung this all together in the Integraon by Parts formula, things work out very nicely: ln x dx = x ln x x dx. − x ∫ ∫ The new integral simpliﬁes to dx, which is about as simple as things get. Its integral is x + C and our answer is ∫ ln x dx = x ln x x + C. − ∫

Example 7.. Integrang using Int. by Parts: anderivave of arctan x Evaluate arctan x dx. ∫ S The same sneaky trick we used above works here. Let dv = dx and u = arctan x. Then v = x and du = /( + x) dx. The Integraon by Parts formula gives x arctan x dx = x arctan x dx. − + x ∫ ∫ The integral on the right can be solved by substuon. Taking u = + x, we get du = x dx. The integral then becomes arctan x dx = x arctan x du. − u ∫ ∫ The integral on the right evaluates to ln u + C, which becomes ln( + x) + C | | (we can drop the absolute values as + x is always posve). Therefore, the answer is arctan x dx = x arctan x ln( + x) + C. − ∫ 73

Substuon Before Integraon

When taking derivaves, it was common to employ mulple rules (such as using both the Quoent and the Chain Rules). It should then come as no surprise that some integrals are best evaluated by combining integraon techniques. In parcular, here we illustrate making an “unusual” substuon ﬁrst before using Integraon by Parts.

Example 7.. Integraon by Parts aer substuon Evaluate cos(ln x) dx. ∫ S The integrand contains a composion of funcons, leading us to think Substuon would be beneﬁcial. Leng u = ln x, we have du = /x dx. This seems problemac, as we do not have a /x in the integrand. But consider: du = dx x du = dx. x ⇒ · Since u = ln x, we can use inverse funcons and conclude that x = eu. Therefore we have that

dx = x du · = eu du.

We can thus replace ln x with u and dx with eu du. Thus we rewrite our integral as cos(ln x) dx = eu cos u du. ∫ ∫ We evaluated this integral in Example ... Using the result there, we have:

cos(ln x) dx = eu cos u du ∫ ∫ = eu sin u + cos u + C ( ) = eln x sin(ln x) + cos(ln x) + C ( ) = x sin(ln x) + cos(ln x) + C. ( )

Deﬁnite Integrals and Integraon By Parts

So far we have focused only on evaluang indeﬁnite integrals. Of course, we can use Integraon by Parts to evaluate deﬁnite integrals as well, as Theorem 74

7.. states. We do so in the next example.

Example 7.2. Deﬁnite integraon using Integraon by Parts . _ Evaluate x ln x dx. ∫ S Our mnemonic suggests leng u = ln x, hence dv = x dx. We then get du = (/x) dx and v = x/ as shown below.

dv = x dx v = x/ u = ln x ⇒ du = /x dx

Figure ..: Seng up Integraon by Parts.

The Integraon by Parts formula then gives x x x ln x dx = ln x dx − x ∫ ∫ x x = ln x dx − ∫ x x = ln x − x x = ln x − ( )

= ln ln − − − ( ) ( ) = ln − .. ≈

In general, Integraon by Parts is useful for integrang certain products of funcons, like xex dx or x sin x dx. It is also useful for integrals involving logarithms and inverse trigonometric funcons. ∫ ∫ As stated before, integraon is generally more diﬃcult than derivaon. We are developing tools for handling a large array of integrals, and experience will tell us when one tool is preferable/necessary over another. For instance, consider the three similar–looking integrals

xex dx, xex dx and xex dx. ∫ ∫ ∫ 75

While the ﬁrst is calculated easily with Integraon by Parts, the second is best approached with Substuon. Taking things one step further, the third integral has no answer in terms of elementary funcons, so none of the methods we learn in calculus will get us the exact answer. Integraon by Parts is a very useful method, second only to Substuon. In the following secons of this chapter, we connue to learn other integraon techniques. The next secon focuses on handling integrals containing trigonometric funcons.

Let dv be the hardest part you can integrate u = rest of the integrand (up to a constant)

This rule seems always to work. 76

Exercises Terms and Concepts . tan− (x) dx ∫ . T/F: Integraon by Parts is useful in evaluang integrands that contain products of funcons. . x tan− x dx ∫ . T/F: Integraon by Parts can be thought of as the “opposite of the Chain Rule.” . sin− x dx ∫

. x ln x dx . T/F: If the integral that results from Integraon by Parts ap- ∫ pears to also need Integraon by Parts, then a mistake was . (x ) ln x dx made in the orginal choice of “u”. − ∫

. x ln(x ) dx − Problems ∫

In Exercises – , evaluate the given indeﬁnite integral. . x ln(x) dx ∫ . x sin x dx x x dx ∫ . ln ∫ x . xe− dx . (ln x) dx ∫ ∫ . x sin x dx . (ln(x + )) dx ∫ ∫ . x sin x dx . x sec x dx ∫ ∫ x . xe dx . x csc x dx ∫ ∫ xex dx . . x√x dx − ∫ ∫

x . xe− dx . x√x dx − ∫ ∫

x . e sin x dx . sec x tan x dx ∫ ∫

x . e cos x dx . x sec x tan x dx ∫ ∫

x . e sin(x) dx . x csc x cot x dx ∫ ∫

. ex cos(x) dx In Exercises – , evaluate the indeﬁnite integral aer ﬁrst ∫ making a substuon.

. sin x cos x dx . sin(ln x) dx ∫ ∫

x x . sin− x dx . e cos e dx ∫ ∫ ( ) 77

π/ . sin(√x) dx . x sin x dx π/ ∫ ∫−

π/ √ . ln( x) dx . x sin x dx ∫ π/ ∫−

√x √ln . e dx . xex dx ∫ ∫ . eln x dx . xex dx ∫ ∫ In Exercises – , evaluate the deﬁnite integral. Note: the x corresponding indeﬁnite integrals appear in Exercises – . . xe− dx ∫ π π . x sin x dx . ex sin x dx ∫ ∫

π/ x x . xe− dx . e cos x dx π/ ∫− ∫− Solutions 7.2 10. x3ex − 3x2ex + 6xex − 6ex + C 32. sec x + C − −2x − | | − 1 2x − e + 33. x sec x ln sec x + tan x + C 11. 2 xe 4 C x − 34. −x csc x − ln | csc x + cot x| + C 12. 1/2e (sin x cos x) + C ( ) − 13. 1/5e2x(sin x + 2 cos x) + C 35. 1/2x sin(ln x) cos(ln x) + C ( ) ( ) 14. 1/13e2x(2 sin(3x) − 3 cos(3x)) + C 36. cos ex + ex sin ex + C (√ ) √ (√ ) 15. 1/10e5x(sin(5x) + cos(5x)) + C 37. 2 sin x − 2 x cos x + C 16. −1/2 cos2 x + C 38. 1 x ln |x| − x + C √ 2 2 −1 √ √ √ 17. 1 − x2 + x sin (x) + C x − x 39. 2 xe 2e + C 18. x tan−1(2x) − 1 ln 4x2 + 1 + C 4 40. 1/2x2 + C 1 2 −1( ) − x + 1 −1( ) + 19. 2 x tan x 2 2 tan x C 41. π √ 20. 1 − x2 + x sin−1 x + C 42. −2/e 2 1 2 | | − x + 43. 0 21. 2 x ln x 4 C 3π2 2 44. − 12 − x + 1 2 | | + − | | + 2 22. 24 x ln x 2x 2x ln x C / 2 45. 1 2 23. − x + 1 x2 ln |x − 1| −x − 1 ln |x − 1| + C 4 2 2 2 46. 6 − 2e ( ) 2 24. 1 x2 ln x2 − x + C 3 − 5 2 2 47. 4e2 4e4 3 1 3 | | − x + 1 eπ 25. 3 x ln x 9 C 48. + 2 ( 2 ) 26. 2x + x (ln x)2 − 2x ln x + C 49. 1/5 eπ + e−π 27. 2(x + 1) + (x + 1)(ln(x + 1))2 − 2(x + 1) ln(x + 1) + C 28. x tan(x) + ln | cos(x)| + C 29. ln | sin(x)| − x cot(x) + C 2 ( − )5/2 +4 ( − )3/2 + 30. 5 x 2 3 x 2 C 1 ( 2 − )3/2 + 31. 3 x 2 C 78

Trigonometric Integrals

Funcons involving trigonometric funcons are useful as they are good at describing periodic behavior. This secon describes several techniques for ﬁnding anderivaves of certain combinaons of trigonometric funcons.

Integrals of the form sinm x cosn x dx ∫ In learning the technique∫ of Substuon, we saw the integral sin x cos x dx in Example 6.1.4. The integraon was not diﬃcult, and one could easily evaluate the indeﬁnite integral by leng u = sin x or by leng u = cos x. This integral is easy since the power of both sine and cosine is 1. ∫ We generalize this integral and consider integrals of the form sinm x cosn x dx, where m, n are nonnegave integers. Our strategy for evaluang these integrals is to use the identy cos2 x + sin2 x = 1 to convert high powers of one trigonometric funcon into the other, leaving a single sine or cosine term in the integrand. We summarize the general technique in the following Key Idea.

Key Idea 7.. Integrals Involving Powers of Sine and Cosine

Consider sinm x cosn x dx, where m, n are nonnegave integers. ∫ . If m is odd, then m = k + for some integer k. Rewrite

Semi-memorize these* sinm x = sink+ x = sink x sin x = (sin x)k sin x = ( cos x)k sin x. − Then

sinm x cosn x dx = ( cos x)k sin x cosn x dx = ( u)kun du, − − − ∫ ∫ ∫ where u = cos x and du = sin x dx. − . If n is odd, then using substuons similar to that outlined above we have

sinm x cosn x dx = um( u)k du, − ∫ ∫ where u = sin x and du = cos x dx. . If both m and n are even, use the power–reducing idenes + cos(x) cos(x) cos x = and sin x = − to reduce the degree of the integrand. Expand the result and apply the principles of this Key Idea again.

** means you should be aware these methods exist here in case you need them later. 79

We pracce applying Key Idea 7.. in the next examples.

Example 7.. Integrang powers of sine and cosine Evaluate sin x cos x dx. ∫ S The power of the sine term is odd, so we rewrite sin x as

sin x = sin x sin x = (sin x) sin x = ( cos x) sin x. − Our integral is now ( cos x) cos x sin x dx. Let u = cos x, hence du = − sin x dx. Making the substuon∫ and expanding the integrand gives − ( cos) cos x sin x dx = ( u)u du = u+u u du = u u+u du. − − − − − − − ∫ ∫ ∫ ∫ ( ) ( ) This ﬁnal integral is not diﬃcult to evaluate, giving u u + u du = u + u u + C − − − − ∫ ( ) = cos x + cos x cos x + C. − −

Example 7.. Integrang powers of sine and cosine Evaluate sin x cos x dx. ∫ S The powers of both the sine and cosine terms are odd, therefore we can apply the techniques of Key Idea 7.. to either power. We choose to work with the power of the cosine term since the previous example used the sine term’s power. We rewrite cos x as

cos x = cos x cos x = (cos x) cos x = ( sin x) cos x − = ( sin x + sin x sin x + sin x) cos x. − − We rewrite the integral as

sin x cos x dx = sin x sin x + sin x sin x + sin x cos x dx. − − ∫ ∫ ( ) 80

Now substute and integrate, using u = sin x and du = cos x dx. sin x sin x + sin x sin x + sin x cos x dx = − − ∫ ( ) u ( u + u u + u ) du = u u + u u + u du − − − − ∫ ∫ ( ) = u u + u u + u + C − − = sin x sin x + sin x + ... − sin x + sin x + C. −

y Technology Note: The work we are doing here can be a bit tedious, but the . g(x) skills developed (problem solving, algebraic manipulaon, etc.) are important. Nowadays problems of this sort are oen solved using a computer algebra system. The powerful program Mathemaca® integrates sin x cos x dx as . cos(x) cos(x) cos(x) cos(x) cos(x) cos(x) cos(x) f(x) = + + ∫ , − − − − − x which clearly has a diﬀerent form than our answer in Example .., which is

. f(x) − g(x) = sin x sin x + sin x sin x + sin x...... − −

Figure 7..: A plot of f(x) and g(x) from Figure 7.. shows a graph of f and g; they are clearly not equal, but they diﬀer Example 7.. and the Technology Note. only by a constant. That is g(x) = f(x) + C for some constant C. So we have two diﬀerent anderivaves of the same funcon, meaning both answers are correct.

Example 7.. Integrang powers of sine and cosine Evaluate cos x sin x dx. ∫ S The powers of sine and cosine are both even, so we employ the power–reducing formulas and algebra as follows. + cos(x) cos(x) cos x sin x dx = − dx ∫ ∫ ( ) ( ) + cos(x) + cos(x) cos(x) = − dx · ∫ = + cos(x) cos(x) cos(x) dx − − ∫ ( ) The cos(x) term is easy to integrate, especially with Key Idea 7... The cos(x) term is another trigonometric integral with an even power, requiring the power– reducing formula again. The cos(x) term is a cosine funcon with an odd power, requiring a substuon as done before. We integrate each in turn below. 81

cos(x) dx = sin(x) + C. ∫ + cos(x) cos(x) dx = dx = x + sin(x) + C. ∫ ∫ Finally, we rewrite cos(x) as ( )

cos(x) = cos(x) cos(x) = sin(x) cos(x). − Leng u = sin(x), we have du = cos(x)(dx, hence )

cos(x) dx = sin(x) cos(x) dx − ∫ ∫ ( ) = ( u) du − ∫ = u u + C − ( ) = sin(x) sin(x) + C − ( ) Pung all the pieces together, we have cos x sin x dx = + cos(x) cos(x) cos(x) dx − − ∫ ∫ ( ) = x + sin(x) x + sin(x) sin(x) sin(x) + C − − − [ ( ) ( )] = x sin(x) + sin(x) + C − [ ] The process above was a bit long and tedious, but being able to work a problem such as this from start to ﬁnish is important.

Integrals of the form sin(mx) sin(nx) dx, cos(mx) cos(nx) dx, ∫ ∫ and sin(mx) cos(nx) dx. ∫ Funcons that contain products of sines and cosines of diﬀering periods are important in many applicaons including the analysis of sound waves. Integrals of the form

sin(mx) sin(nx) dx, cos(mx) cos(nx) dx and sin(mx) cos(nx) dx ∫ ∫ ∫ 82

are best approached by ﬁrst applying the Product to Sum Formulas found in the back cover of this text, namely

sin(mx) sin(nx) = cos (m n)x cos (m + n)x − − [ ( ) ( )] Be aware of these cos(mx) cos(nx) = cos (m n)x + cos (m + n)x − [ ( ) ( )] sin(mx) cos(nx) = sin (m n)x + sin (m + n)x − Example 7.. [ ( ) ( )] ■ Integrang products of sin(mx) and cos(nx) Evaluate sin(x) cos(x) dx. ∫ S The applicaon of the formula and subsequent integraon are straighorward:

sin(x) cos(x) dx = sin(x) + sin(x) dx ∫ ∫ [ ] = cos(x) cos(x) + C − −

Integrals of the form tanm x secn x dx. ∫ When evaluang integrals of the form sinm x cosn x dx, the Pythagorean Theorem allowed us to convert even powers of sine into even powers of cosine, ∫ and vise–versa. If, for instance, the power of sine was odd, we pulled out one sin x and converted the remaining even power of sin x into a funcon using powers of cos x, leading to an easy substuon. The same basic strategy applies to integrals of the form tanm x secn x dx, albeit a bit more nuanced. The following three facts will prove useful: ∫ d • dx (tan x) = sec x,

d • dx (sec x) = sec x tan x , and • + tan x = sec x (the Pythagorean Theorem).

If the integrand can be manipulated to separate a sec x term with the remaining secant power even, or if a sec x tan x term can be separated with the remaining tan x power even, the Pythagorean Theorem can be employed, leading to a simple substuon. This strategy is outlined in the following Key Idea. 83

Key Idea 7.. Integrals Involving Powers of Tangent and Secant

Consider tanm x secn x dx, where m, n are nonnegave integers. ∫ . If n is even, then n = k for some integer k. Rewrite secn x as

n k k k sec x = sec x = sec − x sec x = ( + tan x) − sec x.

Then

m n m k m k tan x sec x dx = tan x( + tan x) − sec x dx = u ( + u ) − du, ∫ ∫ ∫ where u = tan x and du = sec x dx. . If m is odd, then m = k + for some integer k.Rewrite tanm x secn x as

m n k+ n k n k n tan x sec x = tan x sec x = tan x sec − x sec x tan x = (sec x ) sec − x sec x tan x. − Then

m n k n k n tan x sec x dx = (sec x ) sec − x sec x tan x dx = (u ) u − du, − − ∫ ∫ ∫ where u = sec x and du = sec x tan x dx. . If n is odd and m is even, then m = k for some integer k. Convert tanm x to (sec x )k. Expand − the new integrand and use Integraon By Parts, with dv = sec x dx. . If m is even and n = , rewrite tanm x as

m m m m m tan x = tan − x tan x = tan − x(sec x ) = tan − sec x tan − x. − − So m m m tan x dx = tan − sec x dx tan − x dx . − ∫ ∫ ∫ apply rule # apply rule # again | {z } | {z } Semi-memorize these methods

The techniques described in items and of Key Idea 7.. are relavely straighorward, but the techniques in items and can be rather tedious. A few examples will help with these methods. 84

Example 7.. Integrang powers of tangent and secant Evaluate tan x sec x dx. ∫ S Since the power of secant is even, we use rule # from Key Idea 7.. and pull out a sec x in the integrand. We convert the remaining powers of secant into powers of tangent.

tan x sec x dx = tan x sec x sec x dx ∫ ∫ = tan x + tan x sec x dx ∫ ( ) Now substute, with u = tan x, with du = sec x dx.

= u + u du ∫ ( ) We leave the integraon and subsequent substuon to the reader. The ﬁnal answer is = tan x + tan x + tan x + C.

Example 7.. Integrang powers of tangent and secant Evaluate sec x dx. ∫ S We apply rule # from Key Idea 7.. as the power of secant is odd and the power of tangent is even ( is an even number). We use Integra- on by Parts; the rule suggests leng dv = sec x dx, meaning that u = sec x.

dv = sec x dx v = tan x u = sec x ⇒ du = sec x tan x dx

Figure 7..: Seng up Integraon by Parts.

Employing Integraon by Parts, we have

sec x dx = sec x sec x dx · ∫ ∫ u dv = sec|x{ztan} x| {z sec} x tan x dx. − ∫ 85

This new integral also requires applying rule # of Key Idea 7..:

= sec x tan x sec x sec x dx − − ∫ ( ) = sec x tan x sec x dx + sec x dx − ∫ ∫ = sec x tan x sec x dx + ln sec x + tan x − | | ∫ In previous applicaons of Integraon by Parts, we have seen where the original integral has reappeared in our work. We resolve this by adding sec x dx to both sides, giving: ∫

sec x dx = sec x tan x + ln sec x + tan x | | ∫ sec x dx = sec x tan x + ln sec x + tan x + C | | ∫ ( ) We give one more example.

Example 7.. Integrang powers of tangent and secant Evaluate tan x dx. ∫ S We employ rule # of Key Idea 7...

tan x dx = tan x tan x dx ∫ ∫ = tan x sec x dx − ∫ ( ) = tan x sec x dx tan x dx − ∫ ∫ Integrate the ﬁrst integral with substuon, u = tan x; integrate the second by employing rule # again.

= tan x tan x tan x dx − ∫ = tan x tan x sec x dx − − ∫ ( ) = tan x tan x sec x dx + tan x dx − ∫ ∫ 86

Again, use substuon for the ﬁrst integral and rule # for the second.

= tan x tan x + sec x dx − − ∫ ( ) = tan x tan x + tan x x + C. − −

These laer examples were admiedly long, with repeated applicaons of the same rule. Try to not be overwhelmed by the length of the problem, but rather admire how robust this soluon method is. A trigonometric funcon of a high power can be systemacally reduced to trigonometric funcons of lower powers unl all anderivaves can be computed. The next secon introduces an integraon technique known as Trigonomet- ric Substuon, a clever combinaon of Substuon and the Pythagorean The- orem.

Integral Table (Change of variable Form)

∫ du = u + C un+1 ∫ un du = + C u u n+1 ∫ e du = e + C au au du = + C ∫ ln a du  uu = ln|u| + C ∫ cosu du = sinu + C ∫ sinu du = -cosu + C 2 2u = - u + C ∫ sec u du = tanu + C ∫ csc du cot u u = - u + C ∫ secu tanu du = secu + C ∫ csc cot du csc ∫tan u du = ln(sec u) + C ∫cot u du = ln(cos u) + C

∫ sec u du = ln|sec u + tan u| + C ∫csc u du = ln|csc u - cot u| + C du du  = arc sin u + C u  = arctan + C 1 - u2 1 + u2

Method of Substitution u = g(x) f(g(x)) g ' (x) dx = ∫ f(u) du  du = g' (x) dx Proof: the integrals are live mathematics. u = g(x) bf g x g x g(b) f u ∫a ( ( )) ' ( ) dx = ∫g(a) ( ) du du = g' (x) dx 87

Exercises Terms and Concepts . cos(x) cos(x) dx ∫ π . cos x cos(πx) dx . T/F: sin x cos x dx cannot be evaluated using the tech- ∫ niques∫ described in this secon since both powers of sin x ( ) and cos x are even. . tan x sec x dx ∫ . T/F: sin x cos x dx cannot be evaluated using the techniques∫ described in this secon since both powers of sin x . tan x sec x dx and cos x are odd. ∫

. T/F: This secon addresses how to evaluate indefinite inte- . tan x sec x dx grals such as sin x tan x dx. ∫ ∫ . tan x sec x dx . T/F: Somemes computer programs evaluate integrals in- ∫ volving trigonometric funcons differently than one would using the techniques of this secon. When this is the case, . tan x sec x dx the techniques of this secon have failed and one should ∫ only trust the answer given by the computer. . tan x sec x dx Problems ∫ . tan x dx In Exercises – , evaluate the indefinite integral. ∫

. sin x cos x dx . sec x dx ∫ ∫

. sin x cos x dx . tan x sec x dx ∫ ∫

. sin x cos x dx In Exercises – , evaluate the deﬁnite integral. Note: the ∫ corresponding indeﬁnite integrals appear in the previous set.

. sin x cos x dx π . sin x cos x dx ∫ ∫ π . sin x cos x dx ∫ . sin x cos x dx π ∫− . sin x cos x dx π/ ∫ . sin x cos x dx π/ ∫− . sin x cos x dx ∫ π/ . sin(x) cos(x) dx . sin(x) cos(x) dx ∫ ∫ π/ . cos(x) cos(x) dx π/ . sin(x) cos(x) dx ∫− ∫ π/ . tan x sec x dx . sin(x) sin(x) dx ∫ ∫ π/ . sin(πx) sin(πx) dx . tan x sec x dx π/ ∫ ∫− 88

Solutions 7.3

1. F 2. F 3. F 4. F − 1 5( ) + 5. 5 cos x C 1 4( ) + 6.4 sin x C 1 5 − 1 3 + 7.5 cos x 3 cos x C 1 6 − 1 4 + 8. 6 cos x 4 cos x C 1 11 − 2 9 + 1 7 + 9. 11 sin x 9 sin x 7 sin x C 7 5 3 − 1 9( ) + 3 sin (x) − 3 sin (x) + sin (x) + 10. 9 sin x 7 5 3 C x − 1 ( ) + 11. 8 32 sin 4x C 1 2 + − 1 2 + 12.2 sin x C or 2 cos x C, depending on the choice of substuon ( ) 1 − 1 ( ) − 1 ( ) + 13. 2 8 cos 8x 2 cos 2x C ( ) 1 − 1 ( ) + (− ) + 14. 2 3 cos 3x cos x C ( ) 1 1 ( ) − 1 ( ) + 15. 2 4 sin 4x 10 sin 10x C ( ) 1 1 (π ) − 1 ( π ) + 16. 2 π sin x 3π sin 3 x C ( ) 1 ( ) + 1 ( ) + 17. 2 sin x 3 sin 3x C 1 ( π ) + 1 (π ) + 18. π sin 2 x 3π sin x C 5 tan (x) + 19. 5 C 1 5 + 1 3 + 20.5 tan x 3 tan x C 6 4 tan (x) + tan (x) + 21. 6 4 C 4 tan (x) + 22. 4 C 5 3 sec (x) − sec (x) + 23. 5 3 C 9 7 5 sec (x) − 2 sec (x) + sec (x) + 24. 9 7 5 C 1 3 − + + 25.3 tan x tan x x C 1 3 +3 ( + | + |) + 26. 4 tan x sec x 8 sec x tan x ln sec x tan x C 1 ( − | + |) + 27. 2 sec x tan x ln sec x tan x C 1 3 − 1 ( + | + |) + 28.4 tan x sec x 8 sec x tan x ln sec x tan x C 2 29. 5 30. 0 31. 32/315 32. 1/2 33. 2/3 34. 1/5 35. 16/15 89

Trigonometric Substuon

In Secon . we defined the definite integral as the “signed area under the curve.” In that secon we had not yet learned the Fundamental Theorem of Calculus, so we only evaluated special definite integrals which described nice, geometric shapes. For instance, we were able to evaluate π x dx = (7.) − ∫− √ as we recognized that f(x) = √ x described the upper half of a circle with radius . − We have since learned a number of integraon techniques, including Sub- stuon and Integraon by Parts, yet we are sll unable to evaluate the above integral without resorng to a geometric interpretaon. This secon introduces Trigonometric Substuon, a method of integraon that fills this gap in our integraon skill. This technique works on the same principle as Substuon as found in Secon ., though it can feel “backward.” In Secon ., we set u = f(x), for some funcon f, and replaced f(x) with u. In this secon, we will set x = f(θ), where f is a trigonometric funcon, then replace x with f(θ). We start by demonstrang this method in evaluang the integral in Equaon (7.). Aer the example, we will generalize the method and give more examples.

Example 7.. Using Trigonometric Substuon Evaluate x dx. − ∫ − √ S We begin by nong that sin θ + cos θ = , and hence cos θ = sin θ. If we let x = sin θ, then x = sin θ = cos θ. − − − Seng x = sin θ gives dx = cos θ dθ. We are almost ready to substute. We also wish to change our bounds of integraon. The bound x = corre- − sponds to θ = π/ (for when θ = π/, x = sin θ = ). Likewise, the − − − bound of x = is replaced by the bound θ = π/. Thus

π/ x dx = sin θ( cos θ) dθ − π/ − ∫− ∫− √ π/ √ = √ cos θ cos θ dθ π/ ∫− π/ = cos θ cos θ dθ. π/ | | ∫− On [ π/, π/], cos θ is always posive, so we can drop the absolute value bars, − then employ a power–reducing formula: 90

π/ = cos θ dθ π/ ∫− / π = + cos(θ) dθ π/ ∫− ( π/) = θ + sin(θ) = π. π/ ( ) −

This matches our answer from before.

We now describe in detail Trigonometric Substuon. This method excels when dealing with integrands that contain √a x, √x a and √x + a. − − The following Key Idea outlines the procedure for each case, followed by more examples. Each right triangle acts as a reference to help us understand the re- laonships between x and θ.

Key Idea 7.. Trigonometric Substuon

(a) For integrands containing √a x: − Let x = a sin θ, dx = a cos θ dθ a Thus θ = sin− (x/a), for π/ θ π/. x − ≤ ≤ On this interval, cos θ , so ≥ . θ √a x = a cos θ √a2 x2 − − (b) For integrands containing √x + a:

2 Let x = a tan θ, dx = a sec θ dθ a 2 + x Thus θ = tan− (x/a), for π/ < θ < π/. √ x − On this interval, sec θ > , so . θ √x + a = a sec θ a (c) For integrands containing √x a: − Let x = a sec θ, dx = a sec θ tan θ dθ x 2 2 Thus θ = sec− (x/a). If x/a , then θ < π/; √x a ≥ ≤ − if x/a , then π/ < θ π. ≤ − ≤ . θ We restrict our work to where x a, so x/a , and ≥ ≥ a θ < π/. On this interval, tan θ , so ≤ ≥ √x a = a tan θ −

Be fluent with these diagrams. Be able to construct them as needed. 91

Example 7.. Using Trigonometric Substuon Evaluate dx. √ + x ∫

S Using Key Idea ..(b), we recognize a = √ and set x = √ tan θ. This makes dx = √ sec θ dθ. We will use the fact that √ + x = √ + tan θ = √ sec θ = √ sec θ. Substung, we have: dx = √ sec θ dθ √ + x √ + θ ∫ ∫ tan √ sec θ = dθ √ sec θ ∫ = sec θ dθ ∫ = ln sec θ + tan θ + C. While the integraon steps are over, we are not yet done. The original problem was stated in terms of x, whereas our answer is given in terms of θ. We must convert back to x. The reference triangle given in Key Idea ..(b) helps. With x = √ tan θ, we have x √x + tan θ = and sec θ = . √ √ This gives dx = ln sec θ + tan θ + C √ + x ∫

√x + x = ln + + C. √ √

We can leave this answer as is, or we can use a logarithmic identy to simplify it. Note: √x + x ln + + C = ln x + + x + C √ √ √ (√ ) = ln + ln x + + x + C √ √ = ln x + + x + C,

√ √ where the ln / term is absorbed into the constant C. (In Secon . we will learn another way of approaching this problem.) ( ) 92

Example 7.. Using Trigonometric Substuon Evaluate x dx. − ∫ √ S We start by rewring the integrand so that it looks like √x a − for some value of a:

x = x − √ − √ ( ) = x . − √ ( ) So we have a = /, and following Key Idea 7..(c), we set x = sec θ, and = θ θ θ hence dx sec tan d . We now rewrite the integral with these substuons:

x dx = x dx − √ − ∫ √ ∫ ( ) = sec θ sec θ tan θ dθ − ∫ √ ( ) = (sec θ ) sec θ tan θ dθ − ∫ √ ( ) = tan θ sec θ tan θ dθ ∫ √ ( ) = tan θ sec θ dθ ∫ = sec θ sec θ dθ − ∫ ( ) = sec θ sec θ dθ. − ∫ ( ) We integrated sec θ in Example 7.., ﬁnding its anderivaves to be sec θ dθ = sec θ tan θ + ln sec θ + tan θ + C. | | ∫ ( ) Thus x dx = sec θ sec θ dθ − − ∫ ∫ √ ( ) = sec θ tan θ + ln sec θ + tan θ ln sec θ + tan θ + C | | − | | ( ) ( ) = (sec θ tan θ ln sec θ + tan θ ) + C. − | | 93

We are not yet done. Our original integral is given in terms of x, whereas our ﬁnal answer, as given, is in terms of θ. We need to rewrite our answer in terms of x. With a = /, and x = sec θ, the reference triangle in Key Idea ..(c) shows that tan θ = x / (/) = x / and sec θ = x. − − / Thus √ √ sec θ tan θ ln sec θ + tan θ + C = x x / ln x + x / + C − · − − − ( ) ( √ √ ) = x x / ln x + x / + C. − − − ( √ √ ) The ﬁnal answer is given in the last line above, repeated here: x dx = x x / ln x + x / + C. − − − − ∫ √ ( √ √ )

Example■7. .. Using Trigonometric Substuon √ x Evaluate − dx. x ∫ S We use Key Idea 7..(a) with a = , x = sin θ, dx = cos θ and hence √ x = cos θ. This gives − √ x cos θ − dx = ( cos θ) dθ x sin θ ∫ ∫ = cot θ dθ ∫ = (csc θ ) dθ − ∫ = cot θ θ + C. − − We need to rewrite our answer in terms of x. Using the reference triangle found in Key Idea ..(a), we have cot θ = √ x/x and θ = sin− (x/). Thus − √ √ x x x − dx = − sin− + C. x − x − ∫ ( )

Trigonometric Substuon can be applied in many situaons, even those not of the form √a x, √x a or √x + a. In the following example, we ap- − − ply it to an integral we already know how to handle. 94

Example 7.. Using Trigonometric Substuon Evaluate dx. x +

S We know the answer already as tan− x+C. We apply Trigono- metric Substuon here to show that we get the same answer without inher- ently relying on knowledge of the derivave of the arctangent funcon. Using Key Idea 7..(b), let x = tan θ, dx = sec θ dθ and note that x + = tan θ + = sec θ. Thus

dx = sec θ dθ x + sec θ ∫ ∫ = dθ ∫ = θ + C.

Since x = tan θ, θ = tan− x, and we conclude that dx = tan− x+C. x + ∫ The next example is similar to the previous one in that it does not involve a square–root. It shows how several techniques and idenes can be combined to obtain a soluon.

Example 7.. Using Trigonometric Substuon Evaluate dx. (x + x + )

S∫ We start by compleng the square, then make the substuon u = x + , followed by the trigonometric substuon of u = tan θ:

dx = dx = du. (x + x + ) (u + ) ∫ ∫ (x + ) + ∫ ( ) Now make the substuon u = tan θ, du = sec θ dθ:

= sec θ dθ (tan θ + ) ∫ = sec θ dθ (sec θ) ∫ = cos θ dθ. ∫ 95

Applying a power reducing formula, we have

= + cos(θ) dθ ∫ ( ) = θ + sin(θ) + C. (.) We need to return to the variable x. As u = tan θ, θ = tan− u. Using the identy sin(θ) = sin θ cos θ and using the reference triangle found in Key Idea ..(b), we have u u sin(θ) = = . √u + · √u + u + Finally, we return to x with the substuon u = x + . We start with the expression in Equaon (.):

u θ + sin(θ) + C = tan− u + + C u + x + = tan− (x + ) + + C. (x + x + ) Stang our ﬁnal result in one line,

x + dx = tan− (x + ) + + C. (x + x + ) (x + x + ) ∫

Our last example returns us to definite integrals, as seen in our first example. Given a definite integral that can be evaluated using Trigonometric Substuon, we could first evaluate the corresponding indefinite integral (by changing from an integral in terms of x to one in terms of θ, then converng back to x) and then evaluate using the original bounds. It is much more straighorward, though, to change the bounds as we substute.

Example■ ..7. Deﬁnite integraon and Trigonometric Substuon x Evaluate dx. √ ∫ x + S Using Key Idea 7..(b), we set x = tan θ, dx = sec θ dθ, and √ x + = sec θ. As we substute, we can also change the bounds. boundsnote that of. i aon.- ._ = = θ θ The lower bound of the original integral is x . As x tan , we solve for _ and ﬁnd θ = tan− (x/). Thus the new lower bound is θ = tan− () = . The 96

original upper bound is x = , thus the new upper bound is θ = tan− (/) = π/. Thus we have / x π tan θ dx = sec θ dθ √x + sec θ ∫ ∫ π/ = tan θ sec θ dθ. ∫ We encountered this indeﬁnite integral in Example .. where we found tan θ sec θ dθ = sec θ tan θ ln sec θ + tan θ . − | | ∫ ( ) So / / π π tan θ sec θ dθ = sec θ tan θ ln sec θ + tan θ − | | ∫ ( ) = √ ln(√ + ) − .( . ) ≈

The following equalies are very useful when evaluang integrals using Trigono- metric Substuon.

Key Idea 7.. Useful Equalies with Trigonometric Substuon

. sin(θ) = sin θ cos θ . cos(θ) = cos θ sin θ = cos θ = sin θ − − − . sec θ dθ = sec θ tan θ + ln sec θ + tan θ + C ∫ ( ) . cos θ dθ = + cos(θ) dθ = θ + sin θ cos θ + C. ∫ ∫ ( ) ( )

The next secon introduces Paral Fracon Decomposion, which is an algebraic technique that turns “complicated” fracons into sums of “simpler” fracons, making integraon easier. 97

Exercises Terms and Concepts In Exercises – , evaluate the indefinite integrals. Some may be evaluated without Trigonometric Substuon. . Trigonometric Substuon works on the same principles as Integraon by Substuon, though it can feel “ ”. √x . − dx x ∫ . If one uses Trigonometric Substuon on an integrand containing √ x, then one should set x = . − . dx (x + ) ∫ . Consider the Pythagorean Identy sin θ + cos θ = . x (a) What identy is obtained when both sides are di- . dx √x vided by cos θ? ∫ − (b) Use the new identy to simplify tan θ + . . x√ x dx − ∫ . Why does K ey Idea 7..(a) state that √a x = a cos θ, − x and not acos| θ ?| . dx (x + )/ ∫ Problems x . dx √x In Exercises – , apply Trigonometric Substuon to eval- ∫ − uate the indefinite integrals. ■ . dx (x + x + ) ∫ . √x + dx / ∫ . x ( x )− dx − ∫ . √x + dx ∫ √ x . − dx x ∫ . √ x dx − ∫ x . dx √x + . √ x dx ∫ − ∫ In Exercises – , evaluate the definite integrals by making the proper trigonometric substuon and changing the . √x dx − bounds of integraon. (Note: each of the corresponding ∫ indefinite integrals has appeared previously in this Exercise set.) . √x dx − ∫ . √ x dx − . √x + dx ∫−

∫ . √x dx . √ x dx − − ∫ ∫ . √x + dx . √x dx − ∫ ∫ . dx . dx (x + ) √x + ∫− ∫ √ . dx . x dx √ x − ∫ − ∫− . dx . x √ x dx √x − ∫ − ∫− 98

Solutions 7.4

1. backwards 2. 5 sin θ 3. (a) tan2 θ + 1 = sec2 θ (b) 9 sec2 θ. 4. Because we are considering a > 0 and x = a sin θ, which means θ = sin−1(x/a). The arcsine funcon has a domain of −π/2 ≤ θ ≤ π/2; on this domain, cos θ ≥ 0, so a cos θ is always non-negave, allowing us to drop the absolute value signs. ( √ √ ) 1 2 + + | 2 + + | + 5. 2 x x 1 ln x 1 x C ( √ ) √ 2 x 2 + + | x +4 + x | + 6. 2 4 x 4 ln 2 2 C ( √ ) 1 −1 + − 2 + 7. 2 sin x x 1 x C ( √ ) 1 −1( / ) + − 2 + 8. 2 9 sin x 3 x 9 x C √ √ 1 2 − − 1 | + 2 − | + 9. 2 x x 1 2 ln x x 1 C √ √ 2− 1 2 − − x + x 16 + 10. 2 x x 16 8 ln 4 4 C √ √ 2 + / + 1 | 2 + / + | + = 11. x √x 1 4 4 ln √2 x 1 4 2x C 1 2 + + 1 | 2 + + | + 2 x 4x 1 4 ln 4x 1 2x C √ √ 1 −1( )+ 3 / − 2 + = 1 −1( )+ 1 − 2 + 12. 6 sin 3x 2 1 9 x C 6 sin 3x 2 1 9x C ( √ √ ) 13. 4 1 x x2 − 1/16 − 1 ln |4x + 4 x2 − 1/16| + C = √2 32 √ 1 x 16x2 − 1 − 1 ln |4x + 16x2 − 1| + C 2 8 √ x2+2 14. 8 ln √ + √x + C; with Secon 6.6, we can state the 2 2 √ answer as 8 sinh−1(x/ 2) + C. ( ) 15. 3 sin−1 √x + C (Trig. Subst. is not needed) 7

√ x2−8 16. 5 ln √x + √ + C 8 8 √ √ √ 17. x2 − 11 − 11 sec−1(x/ 11) + C ( ) 1 −1 + x + 18. 2 tan x x2+1 C √ 19. x2 − 3 + C (Trig. Subst. is not needed) √ 1 −1 − 1 − 2( − 2) + 20. 8 sin x 8 x 1 x 1 2x C 21. − √ 1 + C (Trig. Subst. is not needed) x2+9

√ √ x2−10 22. 5 x x2 − 10 + 25 ln √x + √ + C 2 10 10 ( ) 1 x+2 + 1 −1 x+2 + 23. 18 x2+4x+13 54 tan 2 C 24. √ x − sin−1 x + C 1−x2 ( √ ) − 2 √ 1 − 5 x − −1( / ) + 25. 7 x sin x 5 C √ √ 2 26. 1 x x2 + 3 − 3 ln √x +3 + √x + C 2 2 3 3 27. π/2 √ √ 28. 16 3 − 8 ln(2 + 3) √ √ 29. 2 2 + 2 ln(1 + 2) 30. π/4 + 1/2) √ 31. 9 sin−1(1/3) + 8 Note: the new lower bound is θ = sin−1(−1/3) and the new upper bound is θ = sin−1(1/3). The ﬁnal answer comes with recognizing that − − sin (1(−1/3) = )− sin 1(1/3) and that ) √ cos sin−1(1/3) = cos sin−1(−1/3) = 8/3. 32. π/8 99

■ 7.5 Paral Fracon Decomposion ■ In this secon we invesgate the anderivaves of raonal funcons. Recall that Understand this material and raonal funcons are funcons of the form f(x) = p(x) , where p(x) and q(x) are be able to work simple q(x) examples. polynomials and q(x) = . Such funcons arise in many contexts, one of which ̸ It is important to understand is the solving of certain fundamental diﬀerenal equaons. that now, in theory, you can We begin with an example that demonstrates the movaon behind this integrate about rational secon. Consider the integral dx. We do not have a simple formula function. x Big people use a CAS. for this (if the denominator were∫ x +−, we would recognize the anderivave as being the arctangent funcon). It can be solved using Trigonometric Subs- tuon, but note how the integral is easy to evaluate once we realize: / / = . x x − x + − −

Thus

/ / dx = dx dx x x − x + ∫ − ∫ − ∫ = ln x ln x + + C. | − | − | |

This secon teaches how to decompose

/ / into . x x − x + − −

p(x) We start with a raonal funcon f(x) = q(x) , where p and q do not have any common factors and the degree of p is less than the degree of q. It can be shown that any polynomial, and hence q, can be factored into a product of linear and irreducible quadrac terms. The following Key Idea states how to decompose a raonal funcon into a sum of raonal funcons whose denominators are all of lower degree than q.

Note If the degree of the numerator is greater than or equal to that of the denominator, divide.

Example

x4 x2+ 1 x3 - + 1  = 1 x2+ 1 polynomial division So x4 ∫ x2+ 1 dx x3 - + 1  = ∫ 1 x2+ 1 dx x4

= 4 - x + arctanx + C 100

Key Idea 7.. Paral Fracon Decomposion p(x) Let be a raonal funcon, where the degree of p is less than the q(x) degree of q.

. Linear Terms: Let (x a) divide q(x), where (x a)n is the highest − − power of (x a) that divides q(x). Then the decomposion of p(x) − q(x) will contain the sum A A A + + + n . (x a) (x a) ··· (x a)n − − − . Quadrac Terms: Let x + bx + c divide q(x), where (x + bx + c)n is the highest power of x + bx + c that divides q(x). Then the p(x) decomposion of q(x) will contain the sum B x + C B x + C B x + C + + + n n . x + bx + c (x + bx + c) ··· (x + bx + c)n

To find the coefficients Ai, Bi and Ci: . Mulply all fracons by q(x), clearing the denominators. Collect like terms. . Equate the resulng coefficients of the powers of x and solve the resulng system of linear equaons.

x4 Example ∫ x2+ 1 dx x4 x2 - 1 By polynomial division x2+ 1 = 1 + x2+ 1 x2 - + 1  = ∫ 1 x2+ 1 dx x3 = 3 - x + arctan x + C

The following examples will demonstrate how to put this Key Idea into prac-ce. Example 7.. stresses the decomposion aspect of the Key Idea.

Example 7.. Decomposing into paral fracons f(x) = without solving Decompose (x + )(x )(x + x + )(x + x + ) − for the resulng coeﬃcients.

S The denominator is already factored, as both x + x + and x + x + cannot be factored further. We need to decompose f(x) properly. Since (x + ) is a linear term that divides the denominator, there will be a

A x + 101

term in the decomposion. As (x ) divides the denominator, we will have the following terms in the − decomposion: B C D , and . x (x ) (x ) − − − Ex + F The x + x + term in the denominator results in a term. x + x + Finally, the (x + x + ) term results in the terms Gx + H Ix + J and . x + x + (x + x + )

All together, we have A B C D = + + + + (x + )(x )(x + x + )(x + x + ) x + x (x ) (x ) − − − − Ex + F Gx + H Ix + J + + x + x + x + x + (x + x + )

Solving for the coeﬃcients A, B ... J would be a bit tedious but not “hard.”

Example 7.. Decomposing into paral fracons Perform the paral fracon decomposion of . x − S The denominator factors into two linear terms: x = − (x )(x + ). Thus − A B = + . x x x + − − To solve for A and B, ﬁrst mulply through by x = (x )(x + ): − − A(x )(x + ) B(x )(x + ) = − + − x x + − = A(x + ) + B(x ) − = Ax + A + Bx B − Now collect like terms.

= (A + B)x + (A B). − The next step is key. Note the equality we have:

= (A + B)x + (A B). − 102

Techniques of Andiﬀerenaon

For clarity’s sake, rewrite the le hand side as

x + = (A + B)x + (A B). − On the le, the coefficient of the x term is ; on the right, it is (A + B). Since both sides are equal, we must have that = A + B. Likewise, on the le, we have a constant term of ; on the right, the constant term is (A B). Therefore we have = A B. − − We have two linear equaons with two unknowns. This one is easy to solve by hand, leading to A + B = A = / . A B = ⇒ B = / − − Thus / / = . x x − x + − − Example■ ..7. Integrang using paral fracons Use paral fracon decomposion to integrate dx. (x )(x + ) ∫ − S We decompose the integrand as follows, as described by Key Note: Equaon7. offers a direct route to Idea 7..: finding the values of A, B and C. Since the A B C = + + . equaon holds for all values of x, it holds (x )(x + ) x x + (x + ) in parcular when x = . However, when − − x = , the right hand side simplifies to To solve for A, B and C, we mulply both sides by (x )(x + ) and collect like − A( + ) = A. Since the le hand side terms: is sll , we have = A. Hence A = /. Likewise, the equality holds when x = = A(x + ) + B(x )(x + ) + C(x ) (7.) ; this leads to the equaon = C. − − − − Thus C = /. = Ax + Ax + A + Bx + Bx B + Cx C − − − Knowing A and C, we can find the value of = (A + B)x + (A + B + C)x + (A B C) B by choosing yet another value of x, such − − as x = , and solving for B. We have

x + x + = (A + B)x + (A + B + C)x + (A B C) − − leading to the equaons

A + B = , A + B + C = and A B C = . − − These three equaons of three unknowns lead to a unique soluon:

A = /, B = / and C = /. − − 103

Thus / / / dx = dx + − dx + − dx. (x )(x + ) x x + (x + ) ∫ − ∫ − ∫ ∫ Each can be integrated with a simple substuon with u = x or u = x+ − (or by directly applying Key Idea .. as the denominators are linear funcons). The end result is dx = ln x ln x + + + C. (x )(x + ) | − | − | | (x + ) ∫ −

Example 7.. Integrang using paral fracons x Use paral fracon decomposion to integrate dx. (x )(x + ) ∫ − S Key Idea 7.. presumes that the degree of the numerator is less than the degree of the denominator. Since this is not the case here, we begin by using polynomial division to reduce the degree of the numerator. We omit the steps, but encourage the reader to verify that Note: The values of A and B can be quickly x x + found using the technique described in = x + + . (x )(x + ) (x )(x + ) the margin of Example 7... − − Using Key Idea .., we can rewrite the new raonal funcon as: x + A B = + (x )(x + ) x x + − − for appropriate values of A and B. Clearing denominators, we have

x + = A(x + ) + B(x ) − = (A + B)x + (A B). − This implies that:

= A + B = A B. − Solving this system of linear equaons gives

/ = A / = B. 104

We can now integrate.

x / / dx = x + + + dx (x )(x + ) x x + ∫ − ∫ ( − ) x = + x + ln x + ln x + + C. | − | | |

Example 7.. Integrang using paral fracons x + x + Use paral fracon decomposion to evaluate dx. (x + )(x + x + ) ∫ S The degree of the numerator is less than the degree of the denominator so we begin by applying Key Idea ... We have:

x + x + A Bx + C = + . (x + )(x + x + ) x + x + x +

Now clear the denominators.

x + x + = A(x + x + ) + (Bx + C)(x + ) = (A + B)x + (A + B + C)x + (A + C).

This implies that:

= A + B = A + B + C = A + C.

Solving this system of linear equaons gives the nice result of A = , B = and C = . Thus − x + x + x dx = + − dx. (x + )(x + x + ) x + x + x + ∫ ∫ ( ) The ﬁrst term of this new integrand is easy to evaluate; it leads to a ln x+ | | term. The second term is not hard, but takes several steps and uses substuon techniques. x The integrand − has a quadrac in the denominator and a linear x + x + term in the numerator. This leads us to try substuon. Let u = x +x+, so du = (x + ) dx. The numerator is x , not x + , but we can get a x + − 105

term in the numerator by adding in the form of “ .” − x x + − = − − x + x + x + x + x + = . x + x + − x + x + We can now integrate the ﬁrst term with substuon, leading to a ln x+x+ | | term. The ﬁnal term can be integrated using arctangent. First, complete the square in the denominator: = . x + x + (x + ) +

An anderivave of the laer term can be found using Theorem .. and substuon: x + dx = tan− + C. x + x + √ √ ∫ ( ) Let’s start at the beginning and put all of the steps together.

x + x + x dx = + − dx (x + )(x + x + ) x + x + x + ∫ ∫ ( ) x + = dx + dx dx x + x + x + − x + x + ∫ ∫ ∫ x + = ln x + + ln x + x + tan− + C. | | | | − √ √ ( ) As with many other problems in calculus, it is important to remember that one is not expected to “see” the ﬁnal answer immediately aer seeing the problem. Rather, given the inial problem, we break it down into smaller problems that are easier to solve. The ﬁnal answer is a combinaon of the answers of the smaller problems.

Paral Fracon Decomposion is an important tool when dealing with raonal funcons. Note that at its heart, it is a technique of algebra, not calculus, as we are rewring a fracon in a new form. Regardless, it is very useful in the realm of calculus as it lets us evaluate a certain set of “complicated” integrals. The next secon introduces new funcons, called the Hyperbolic Funcons. They will allow us to make substuons similar to those found when studying Trigonometric Substuon, allowing us to approach even more integraon problems. 106

Exercises x x Terms and Concepts . − dx (x + )(x )(x ) ∫ − − . Fill in the blank: Paral Fracon Decomposion is a method of rewring funcons. x + x + . dx x + x ∫ − . T/F: It is somemes necessary to use polynomial division before using Paral Fracon Decomposion. x . dx x x ∫ − − . Decompose without solving for the coefficients, as x x x x + done in Example− ... . − dx x x + ∫ − x . Decompose − without solving for the coefficients, as x . dx − x + x + x done in Example ... ∫ x x + x + . Decompose − without solving for the coefficients, as . dx x x + x + done in Example− ... ∫ x + x + x + . dx . Decompose without solving for the coefficients, as (x + )(x + x ) x + x ∫ − done in Example 7... x + x . − dx (x )(x + x + ) Problems ∫ − x + x + . dx In Exercises – , evaluate the indefinite integral. (x + )(x + ) ∫ x + x x . dx . − − dx x + x (x )(x + x + ) ∫ − ∫ − x . − dx x x + x + x . − dx ∫ (x )(x x + ) ∫ − − . − dx x + x + x . dx ∫ − (x + )(x + x + ) ∫ x + . dx In Exercises – , evaluate the definite integral. x + x + ∫ x + x + . dx . dx (x + ) (x + )(x + ) ∫ ∫ x x + . dx . − − dx (x + ) (x + )(x + ) ∫ ∫ x + x + x + x . − dx . dx x(x + ) (x )(x + x + ) ∫ ∫− − x x + x . − − dx . dx (x )(x + )( x) (x + )(x + x + ) ∫ − − ∫ 107

Solutions 7.5

1. raonal 2. T A + B 3. x x−3 A + B 4. x−3 x+3 5. A√ + B√ x− 7 x+ 7 A + Bx+C 6. x x2+7 7. 3 ln |x − 2| + 4 ln |x + 5| + C 8. 9 ln |x + 1| − 2 ln |x| + C 1 ( | + | − | − |) + 9. 3 ln x 2 ln x 2 C 10. ln |x + 1| + ln |3x + 1| + C | + | − 2 + 11. ln x 5 x+5 C − 4 − | + | + 12. x+8 3 ln x 8 C 5 + | | + | + | + 13. x+1 7 ln x 2 ln x 1 C 14. − ln |2x − 3| + 5 ln |x − 1| + 2 ln |x + 3| + C − 1 | − | + 2 | − | + 3 | + | + 15. 5 ln 5x 1 3 ln 3x 1 7 ln 7 x 3 C 16. x + ln |x − 1| − ln |x + 2| + C 2 x + + 125 | − | + 64 | + | − 35 + 17. 2 x 9 ln x 5 9 ln x 4 2 C 18. 2x + C ( √ ( )) − + 19. 1 − ln x2 + 2x + 3 + 2 ln |x| − 2 tan 1 x√ 1 + C 6 2 ( ) − tan 1 x√+2 − 3 2 + + + + √ 6 + 20. 2 ln x 4x 10 x C 6 21. ln 3x2 + 5x − 1 + 2 ln |x + 1| + C 22. 2 ln |x − 3| + 2 ln |x2 + 6x + 10| − 4 tan−1(x + 3) + C () − 23. 9 ln x2 + 9 + 1 ln |x + 1| − 4 tan 1 x + C 10( 5 15 3 ( )) 1 2 + + − | − | + −1 x+1 + 24. 2 3 ln x 2x 17 4 ln x 7 tan 4 C ( ) √ ( ) − 25. 3 ln x2 − 2x + 11 + ln |x − 9| + 3 2 tan−1 √x 1 5 10 + C 1 √ ( ) − 2 ln x2 + 10x + 27 + 5 ln |x + 2| − 6 2 tan 1 x√+5 + C 26. 2

27. ■26. ln(2000/243) ≐ 2.108 28. ■27. 5 ln(9/4) − ln(17/2) ≐ 3.3413 − 29. ■28. −π/4 + tan 1 3 − ln(11/9) ≐ 0.263 30. ■29. 1/8 108

b In the definition of definite integral , x 7.6 Improper Integrals ∫ a f( )dx, it was assumed that the integrand was bounded (not infinite) on the interval of integration. Also it was assumed that the limits of integration a and b were real numbers, not the extended reals, -∞ or +∞. In applications, these restrictions are unnecessary and undesirable.

y = f(x)

X a b

y = 1 x2 25

20 Caution : If f(x) is infinite at an interior point x of 15 a x b the interval, < < , break up the interval so that 10 the infinity occurs at endpoints. 5

-1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 X 1 Example Evaluate dx . ∫-1 x2 1 x 1 = - -1 1 1 = - 1 - (- -1 ) = -2.

Wrong! The answer should be positive since f(x) > 0 on the interval -1 ≤ x ≤ 1.

Normally do not integrate across a point where the integrand is infinite. Correct: 1 0- 1 ∫ dx ≈ ∫ dx + ∫ dx -1 x 2 -1 x 2 0+ x 2 1 - 1 x 0 x 1 hyperreal arithmetic = - -1 + - 0+ 1 1 1 1 = (- 0- - - -1 ) + (- 1 - - 0+ ) = (+∞ - 1) + (-1 + ∞) = 2(+∞) - 2 = +∞. Note: We allow +∞ as an answer because in all applications this answer is meaningful. If this is a ‘find the area under the curve’ problem, it would cost infinite many $'s to buy the paint to coat it. 109

Type I Improper Integrals: infinite limits of integration Type II Improper Integrals: infinite integrands With hyperreal methods, these categories are not very relevant. Mixed Type I, Type II Integrals }

Possible Outcomes: * An extended real number * Does not exist

Note in your Apex readings that the author uses limit methods. The limit method is more prone toward making errors. Hyperreal methods also seem more natural. e-x

1 Example Type I

X 0 1 2 3 4 +∞ -x e-x + ∞ ∫ e dx = - 0 0 Use the more natural hyperreal e-∞ e0 = - - (- ) notation when doing improper = 0 + 1 integrals: = 1 + 0+ = dx, dx > 0 1 . - x2+ 1 0 = -dx, dx > 0 Example Type I 1

X -3 -2 -1 0 1 2 3 +∞ dx + ∞ ∫ = arctan x -∞ x2+ 1 - ∞ = arctan(+∞) - arctan(-∞) π π = 2 - (- 2 ) = π

1/x ln x 5 3

4 2

3 1 Example Mixed Type 0 X 2 1 2 3 4 5 6 -1 1 -2 X 0 1 2 3 4 5 -3 +∞ dx + ∞ ∫ = ln x + 0 x 0 = ln(+∞) - ln(0+) = +∞ - (-∞) = {∞ + ∞} = +∞

+∞ x ∫ dx Exercise Show that -∞ x2+ 1 does not exists because it is the indeterminate form {∞ - ∞}. Answer: See the front cover or the next page

e i x+ e- i x x = cos 2 ↑ ex ↓ ex+ e-x cosh x = 2 110

Improper Integraon

We begin this secon by considering the following deﬁnite integrals:

• dx ., y + x ≈ ∫ 1 • dx ., + x ≈ ∫ , . 0 5 • dx .. + x ≈ ∫ Noce how the integrand is /( + x) in each integral (which is sketched x in Figure 7..). As the upper bound gets larger, one would expect the “area ...... 5 10 under the curve” would also grow. While the definite integrals do increase in value as the upper bound grows, they are not increasing by much. In fact, 1 Figure 7.6.: Graphing f(x) = consider: 1 + x 2 b b = − = − − = − . dx tan x tan b tan tan b + x − ∫ As b , tan− b π/. Therefore it seems that as the upper bound b grows, → ∞ → b ≐ the value of the definite integral dx approaches π/ .. This + x should strike the reader as being∫ a bit amazing: even though the curve extends “to infinity,” it has a finite amount of area underneath it. b When we defined the definite integral f(x) dx, we made two spulaons: ∫a . The interval over which we integrated, [a, b], was a finite interval, and . The funcon f(x) was connuous on [a, b] (ensuring that the range of f was finite). In this secon we consider integrals where one or both of the above condions do not hold. Such integrals are called improper integrals.

Note: when using hyperreal methods, the only time you should break up the interval of integration is when the the integrand has an infinite value in the interior of the interval of integration.

Exercise above solution +∞ xdx x ∫ x2 -∞ 1 + x2 + 1 1 1 +∞ 2x dx = u = 1 + x2 2 ∫-∞ 1 + x2 1 +∞ = ln 1+ x2 2 |-∞ X 1 1 -10 -5 5 10 = ln (+∞) - ln (+∞) 2 2 = ln (+∞) - ln (+∞) = {∞ - ∞}, indeterminate

The integral does not exist. 111

Improper Integrals with Inﬁnite Bounds

Definion 7.6. Improper Integrals with Infinite Bounds; Converge, Diverge . Let f be a connuous funcon on [a, ). Define ∞ b ∞ f(x) dx to be lim f(x) dx. b a →∞ a Limit Talk. ∫ ∫ . Let f be a connuous funcon on ( , b]. Define Grain of salt material. −∞ b b f(x) dx to be lim f(x) dx. a a ∫−∞ →−∞ ∫ . Let f be a connuous funcon on ( , ). Let c be any real num- −∞ ∞ ber; define

c b ∞ f(x) dx to be lim f(x) dx + lim f(x) dx. a a b c ∫−∞ →−∞ ∫ →∞ ∫ An improper integral is said to converge if its corresponding limit exists; otherwise, it diverges. The improper integral in part converges if and only if both of its limits exist.

We prefer not to use the y Example 7.6. Evaluang improper integrals term 'improper'. Evaluate the following improper integrals.

f(x) = ∞ x x . dx . e dx x .5 ∫ ∫−∞ ∞ ∞ . dx . dx x + x ∫ ∫−∞ x S ..... 5 b b ∞ Figure 7.6.: A graph of f(x) = in Ex- = = x . dx lim dx lim − ample 7.6.. - x b x b x ∫ →∞ ∫ →∞ = lim − + b b →∞ = . A graph of the area deﬁned by this integral is given in Figure 7.6.. 112

b ∞ . dx = lim dx y x b x ∫ →∞ ∫ 1 b = lim ln x 1 b | | f(x) = →∞ x = lim ln(b) b →∞ 0.5 = + . ∞ ∞ The limit does not exist, hence the improper integral dx diverges. x ∫ x Compare the graphs in Figures 7.6. and 7.6.; noce how the graph of ..... 1 5 10 f(x) = /x is noceably larger. This diﬀerence is enough to cause the improper integral to diverge. Figure 7.6.: A graph of f(x) = x in Example 7.6.. . ex dx = lim ex dx a a ∫−∞ →−∞ ∫ = lim ex a a →−∞ y = lim e ea a 1 →−∞ − = . f(x) = ex A graph of the area deﬁned by this integral is given in Figure 7.6..

. We will need to break this into two improper integrals and choose a value of c as in part of Deﬁnion 7.6.. Any value of c is ﬁne; we choose c = .

x b ∞ ..... 10 5 1 dx = lim dx + lim dx − − − a b + x a + x + x x ∫−∞ →−∞ ∫ →∞ ∫ Figure 7.6.: A graph of f(x) = e b in Ex-ample 7.6.. = lim tan− x + lim tan− x a a b →−∞ →∞ = lim tan− tan− a + lim tan− b tan− a − b − →−∞ →∞ ( π π ) ( ) y = − + . − − ( ) ( ) f(x) = Each limit exists, hence the original integral converges and has value: + x

= π.

A graph of the area deﬁned by this integral is given in Figure 7.6.. x 5 5 − − ..... Figure 7.6.: A graph of f(x) = +x in Example 7.6..

arctan x π Hyperreally 2

+∞ dx +∞ π π ∞ ∞  π ∫ x2+1 = arctanx -∞ = arctan(+ ) - arctan(- ) = 2 - (- 2 = -∞ X -2 -1 1 2 0 0 0 0

-π 2 113

The previous secon introduced l’Hôpital’s Rule, a method of evaluang limits that return indeterminate forms. It is not uncommon for the limits resulng from improper integrals to need this rule as demonstrated next.

Example 7.6. Improper i ntegraon and l’Hôpital’s Rule Be aware of this ∞ ln x y Evaluate the improper integral dx. x ∫ .4 ln x S This integral will require the use of Integraon by Parts. Let f(x) = x u = ln x and dv = /x dx. Then . b ∞ ln x ln x dx = lim dx x b x →∞ x ∫ ∫ ln x b b 5 = lim + dx ..... b − x x →∞ ( ∫ ) _ln x Figure 7.6.: A graph of f(x) = in Ex- b x ln x ample 7.6.. = lim b − x − x →∞ ( )

ln b = lim ( ln ) . b − b − b − − − →∞ ( ) ln b The /b and ln terms go to , leaving lim + . We need to evaluate b − b ln b →∞ lim with l’Hôpital’s Rule. We have: b b →∞

ln b by LHR /b lim = lim b b b →∞ →∞ = .

Thus the improper integral evaluates as:

∞ ln x dx = . x ∫

Improper Integrals with Inﬁnite Range

We have just considered definite integrals where the interval of integraon was infinite. We now consider another type of improper integraon, where the range of the integrand is infinite. 114

Definion 7.6. Improper Integraon with Infinite Range Let f(x) be a connuous funcon on [a, b] except at c, a c b, where Note: In Definion 7.6., c can be one of ≤ ≤ the endpoints (a or b). In that case, there x = c is a vercal asymptote of f. Define is only one limit to consider as part of the b t b definion. f(x) dx = lim f(x) dx + lim f(x) dx. t c t c+ ∫a → − ∫a → ∫t

Example 7.6. Improper integraon of funcons with inﬁnite range Evaluate the following improper integrals: y

10 . . . dx dx √x x ∫ ∫− S 1 5 f(x) = √x . A graph of f(x) = /√x is given in Figure 7.6.. Noce that f has a vercal asymptote at x = ; in some sense, we are trying to compute the area of a region that has no “top.” Could this have a finite value? ..... x 0.5 1 dx = lim dx + √x a a √x Figure 7.6.: A graph of f(x) = in Ex- ∫ → ∫ √x ample 7.6.. = lim √x a + a → = lim √ √a a + − → ( ) y c = . It turns out that the region does have a finite area even though it has no upper bound (strange things can occur in mathemacs when considering f(x) = the infinite). x 5 . The funcon f(x) = /x has a vercal asymptote at x = , as shown in Figure 7.6., so this integral is an improper integral. Let’s eschew using limits for a moment and proceed without recognizing the improper nature of the integral. This leads to: ..... x .5 .5 − − dx = Figure 7.6.: A graph of f(x) = _in Ex- x − x x ∫− − ample 7.6.. = () − − = . (!) − Clearly the area in queson is above the x-axis, yet the area is supposedly negave! Why does our answer not match our intuion? To answer this, evaluate the integral using Definion 7.6.. t dx = lim dx + lim dx + x t − x t t x ∫− → ∫− → ∫ t = lim + lim + t − − x t − x t → − → = lim + lim + t − t − t + − t → − → + + . ⇒ ∞ − − ∞ ( ) ( ) Neither limit converges hence the original improper integral diverges. The nonsensical answer we obtained by ignoring the improper nature of the integral is just that: nonsensical. 115

It is good practice not to integrate across an infinite discontinuity. However, this can be done if the antiderivative is continuous at the discontinuity of the integrand.

Example 1 F (x) f(x) = an antiderivative x23 4 5

4 2

3 X -2 -1 1 2 2 -2 1

X -4 -3 -2 -1 0 1 2 3

NOTE The details of finding F(x) are somewhat complicated and are omitted. Try it with a CAS or Wolfram Alpha. 1 dx Clearly, ∫ 2/3 = F(1) - F(-1) = 3 - (-3) = 6 -1 x

Understanding Convergence and Divergence Be aware of this topic

Oenmes, but not often, we are interested in knowing simply whether or not an improper integral converges, and not necessarily the value of a convergent integral. We provide here several tools that help determine the convergence or divergence of improper integrals without integrang. Our ﬁrst tool is to understand the behavior of funcons of the form . x p Example 7.6. Improper integraon of /xp y ∞ Determine the values of p for which p dx converges. 1 x f(x) = ∫ x q S We begin by integrang and then evaluang the limit. b < < ∞ p 1 q dx = lim dx x p b x p ∫ →∞ ∫ b p 1 = lim x− dx (assume p = ) f(x) = b ̸ x p →∞ ∫ ..... x b p+ 1 = lim x− b p + →∞ − Figure 7.6.: Plong funcons of p p p = lim b − − . the form /x in Example 7.6.. b p − →∞ − ( )

When does this limit converge – i.e., when is this limit not ? This limit con- ∞ verges precisely when the power of b is less than : when p < < p. − ⇒ Note: if p = 1, we get the indeterminate form {0/0} and so the integral DNE. 116

∞ Our analysis shows that if p > , then p dx converges. When p < x the improper integral diverges; we showed∫ in Example 7.. that when p = the integral also diverges. Figure 7.6. graphs y = /x with a dashed line, along with graphs of y = /xp, p < , and y = /xq, q > . Somehow the dashed line forms a dividing line between convergence and divergence.

The result of Example 7.6. provides an important tool in determining the convergence of other integrals. A similar result is proved in the exercises about improper integrals of the form p dx. These results are summarized in the x following Key Idea. ∫

∞ Key Idea 7.6. Convergence of Improper Integrals dx and dx. x p x p ∫ ∫ ∞ . The improper integral dx converges when p > and diverges when p . x p ≤ ∫ . The improper integral dx converges when p < and diverges when p . x p ≥ ∫

A basic technique in determining convergence of improper integrals is to compare an integrand whose convergence is unknown to an integrand whose Note: We used the upper and convergence is known. We oen use integrands of the form /x p to compare lower bound of “” in Key Idea 7.6. for to as their convergence on certain intervals is known. This is described in the conve-nience. It can be replaced by any a following theorem. where a > .

Theorem 7.6. Direct Comparison Test for Improper Integrals Let f and g be connuous on [a, ) where f(x) g(x). ∞ ≤ ≤

∞ ∞ . If g(x) dx converges, then f(x) dx converges. ∫a ∫a ∞ ∞ . If f(x) dx diverges, then g(x) dx diverges. ∫a ∫a 117

Example 7.6. Determining convergence of improper integrals Determ ine the convergence ofthe following improper integrals. y ∞ x ∞ . e− dx . dx √x x ∫ ∫ − f(x) = x S

. ( ) = x x . The funcon f x e− does not have an anderivave expressible in f(x) = e− terms of elementary funcons, so we cannot integrate directly. It is com- x 2 parable to g(x) = /x , and as demonstrated in Figure 7.6., e− < 1/x ∞ x on [, ). We know from Key Idea 7.6. dx converges, ..... ∞ x ∫ ∞ x x hence e− dx also converges. Figure 7.6.: Graphs of f(x) = e− and f(x) = /x in Example 7... ∫

. Note that for large values of x, ≐ = . We know from Key √x x √x x − ∞

Idea 7.6. and the subsequent note that to dx diverges, so we seek ∫ x compare the original integrand to 1/x. y x > x = √x > √x x f(x) = It is easy to see that when , we have .5BLJOH √x x − .4 − reciprocals reverses the inequality, giving

< . . f(x) = x √x x x −

Using Theorem 7.., we conclude that since ∞ ∞ x dx diverges, dx 4 6 x √x x ..... ∫ ∫ − diverges as well. Figure 7.. illustrates this. __- Figure 7.6.: Graphs of f(x) = /√x x ___ − and f(x) = /x in Example 7... Being able to compare “unknown” integrals to “known” integrals is very useful in determining convergence. However, some of our examples were a lile “too nice.” For instance, it was convenient that < , but what if the x √x x “ x” were replaced with a “+x + ”? That is, what can we− say about the con- − ∞ vergence of dx? We have > , so we cannot √x + x + x √x + x + use Theorem∫ 7... In cases like this (and many more) it is useful to employ the following theorem. 118

Theorem 7.6. Limit Comparison Test for Improper Integrals Let f and g be connuous funcons on [a, ) where f(x) > and g(x) > ∞ for all x. If f(x) lim = L, < L < , x →∞ g(x) ∞ then ∞ ∞ f(x) dx and g(x) dx ∫a ∫a either both converge or both diverge.

Example 7.6. Determining convergence of improper integrals ∞ Determine the convergence of dx. √x + x + ∫ S As x gets large, the denominator of the integrand will begin to behave much like y = x. So we compare to with the Limit √x + x + x Comparison Test:

/√x + x + x lim = lim . x x →∞ /x →∞ √x + x + The immediate evaluaon of this limit returns / , an indeterminate form. ∞ ∞ Using l’Hôpital’s Rule seems appropriate, but in this situaon, it does not lead to useful results. (We encourage the reader to employ l’Hôpital’s Rule at least once to verify this.) y The trouble is the square root funcon. To get rid of it, we employ the fol- f(x) = lowing fact: If lim f(x) = L, then lim f(x) = L. (This is true when either c or L √x + x + 5 x c x c → → is .) So we consider now the limit . ∞ x lim . f(x) = x x →∞ x + x + x This converges to , meaning the original limit also converged to . As x gets 5 5 very large, the funcon looks very much like . Since we know that ..... √x + x + x ∞ ∞ Figure 7.6.: Graphing f(x) = dx diverges, by the Limit Comparison Test we know that dx √x+x+ x √x + x + ∫ ∫ and f(x) = in Example 7.6.. also diverges. Figure 7.6. graphs f(x) = /√x + x + and f(x) = /x, x il-lustrang that as x gets large, the f uncons become indisnguishable. 119

Both the Direct and Limit Comparison Tests were given in terms of integrals over an infinite interval. There are versions that apply to improper integrals with an infinite range, but as they are a bit wordy and a lile more difficult to employ, they are omied from this text.

This chapter has explored many integraon techniques. We learned Subs- tuon, which “undoes” the Chain Rule of differenaon, as well as Integraon by Parts, which “undoes” the Product Rule. We learned specialized techniques for handling trigonometric funcons and introduced the hyperbolic funcons, which are closely related to the trigonometric funcons. All techniques effecvely have this goal in common: rewrite the integrand in a new way so that the integraon step is easier to see and implement. As stated before, integraon is, in general, hard. It is easy to write a funcon whose anderivave is impossible to write in terms of elementary funcons, and even when a funcon does have an anderivave expressible by elementary funcons, it may be really hard to discover what it is. The powerful computer algebra system Mathemaca® has approximately , pages of code dedicated to integraon. Do not let this difficulty discourage you. There is great value in learning integraon techniques, as they allow one to manipulate an integral in ways that can illuminate a concept for greater understanding. There is also great value in understanding the need for good numerical techniques: the Trapezoidal and Simpson’s Rules are just the beginning of powerful techniques for approximating the value of integraon.

The next chapter stresses the uses of integraon. We generally do not ﬁnd anderivaves for anderivave’s sake, but rather because they provide the soluon to some type of problem. The following chapter introduces us to a number of diﬀerent problems whose soluon is provided by integraon.

Note: The Apex author and most mathematicians use the terminology converge or diverge. We prefer the terms exists as an extended real number or does not exist because of concreteness and applications. In applications an infinite answer is always meaningful.

Note: Work a few assigned problem both by the hyperreal method and the limit method. See if there are any for which the hyperreal method does not apply, but for which the limit method works. 120

Exercises 8.6 ∞ Terms and Concepts . dx x ∫ − . The deﬁnite integral was deﬁned with what two spulaons? . dx x b ∫ ∞ − . If lim f(x) dx exists, then the integral f(x) dx is b →∞ said to ∫ . ∫ . dx x ∫− ∞ . If f(x) dx = , and g(x) f(x) for all x, then we ≤ ≤ ∫ . dx ∞ x know that g(x) dx . ∫ − ∫ π ∞ . sec x dx . For what values of p will dx converge? xp ∫ ∫ ∞ . For what values of p will dx converge? . dx xp x ∫ ∫− | | √ ∞ x . For what values of p will dx converge? . xe− dx xp ∫ ∫

∞ x . xe− dx Problems ∫

In Exercises – , evaluate the given improper integral. ∞ x . xe− dx ∫−∞ ∞ x . e − dx ∫ ∞ . x x dx e + e− ∞ ∫−∞ . dx x ∫ . x ln x dx ∞ . x− dx ∫ ∫ ∞ . x ln x dx . dx x + ∫ ∫−∞ ∞ ln x . dx x x . dx ∫ ∫−∞ x . ln x dx . dx ∫ ∫−∞ ( ) ∞ ln x ∞ x . dx . dx x x + ∫ ∫−∞ ∞ ln x ∞ . dx . dx x √x ∫ − ∫

∞ ∞ x . dx . e− sin x dx (x ) ∫ − ∫

∞ x . dx . e− cos x dx (x ) ∫ − ∫ 121

∞ x+x+ In Exercises – , use the Direct Comparison Test or the . e− dx Limit Comparison Test to determine whether the given def- ∫ inite integral converges or diverges. Clearly state what test ∞ √x is being used and what funcon the integrand is being com- . dx pared to. ex ∫

∞ ∞ . dx . dx √x + x x + sin x ∫ − ∫

∞ ∞ x . dx . dx √x x x + cos x ∫ − ∫

∞ √x + ∞ . dx . dx x √x x + x + x + e ∫ − ∫

∞ x ∞ . e− ln x dx . dx ex x ∫ ∫ −

Solutions 7.7

1. 0/0, ∞/∞, 0 · ∞, ∞ − ∞, 00, 1∞, ∞0 2. F 3. F 4. The base of an expression is approaching 1 while its power is growing without bound. 5. derivaves; limits 6. Answers will vary. 7. Answers will vary. 31. 0 8. Answers will vary. 32. 0 9. 3 33. ∞ 10. −5/3 34. ∞ 11. −1 35. ∞ √ 12. − 2/2 36. 0 13. 5 37. 0 14. 0 38. e 15. 2/3 39. 1 16. a/b 40. 1 17. ∞ 41. 1 18. 1/2 42. 1 19. 0 43. 1 20. 0 44. 0 21. 0 45. 1 22. ∞ 46. 1 23. ∞ 47. 1 24. ∞ 48. 1 25. 0 49. 2 50. 1/2 26. 2 51. −∞ 27. −2 52. 1 28. 0 53. 0 29. 0 54. 3 30. 0 122

7.7 Numerical Integraon

The Fundamental Theorem of Calculus gives a concrete technique for finding y the exact value of a definite integral. That technique is based on compung anderivaves. Despite the power of this theorem, there are sll situaons where x y = e− we must approximate the value of the definite integral instead of finding its exact value. The first situaon we explore is where we cannot compute the anderivave of the integrand. The second case is when we actually do not know .5 the funcon in the integrand, but only its value when evaluated at certain points.

An elementary funcon is any funcon that is a combinaon of polynomial, th x n root, raonal, exponenal, logarithmic and trigonometric funcons. We can ...... 5 compute the derivave of any elementary funcon, but there are many elementary funcons of which we cannot compute an anderivave. For example, the (a) following funcons do not have anderivaves that we can express with elementary funcons: y

x sin x 1 e− , sin(x ) and . y = sin(x3) x x . The simplest way to refer to the anderivaves of e− is to simply write 0 5 x e− dx. This secon outlines three common methods of approximang the value of ∫ x deﬁnite integrals. We describe each as a systemac method of approximang 1 1 − area under a curve. By approximang this area accurately, we ﬁnd an accurate

0.5 approximaon of the corresponding deﬁnite integral. .... − We will apply the methods we learn in this secon to the following deﬁnite integrals: (b) π π y x sin(x) e− dx, sin(x ) dx, and dx, π . x 1 ∫ ∫− ∫

sin x as pictured in Figure 7.7.. y = x 0.5 The Le and Right Hand Rule Methods Earlier we addressed the problem of evaluang definite integrals by approximang the area under the curve using rectangles. We revisit those ideas here before introducing other x 5 10 15 methods of approximang definite integrals...... We start with a review of notaon. Let f be a connuous funcon on the b (c) interval [a, b]. We wish to approximate f(x) dx. We paron [a, b] into n a ∫ b a Figure 7.7.: Graphically represenng equally spaced subintervals, each of length ∆x = − . The endpoints of these three definite integrals that cannot be n evaluated using anderivaves. 123

subintervals are labeled as

x = a, x = a + ∆x, x = a + ∆x,..., x = a + (i )∆x,..., x = b. i − n+ Key Idea .. states that to use the Le Hand Rule we use the summaon n n

f(xi)∆x and to use the Right Hand Rule we use f(xi+)∆x. We review i= i= ∑the use of these rules in the context of examples. ∑

Example 7.7. Approximang deﬁnite integrals with rectangles x Approximate e− dx using the Le and Right Hand Rules with equally spaced subintervals.∫ y S We begin by paroning the interval [, ] into equally x y = e− spaced intervals. We have ∆x = − = / = ., so

x = , x = ., x = ., x = ., x = ., and x = . . Using the Le Hand Rule, we have:

n x . . . .8 f(xi)∆x = f(x) + f(x) + f(x) + f(x) + f(x) ∆x ..... i= ∑ ( ) = f() + f(.) + f(.) + f(.) + f(.) ∆x (a)

≐( + . + . + . + .)(.)) y ≐ (.. x y = e− Using the Right Hand Rule, we have:

. n

f(xi+)∆x = f(x) + f(x) + f(x) + f(x) + f(x) ∆x = ∑i ( ) x = f(.) + f(.) + f(.) + f(.) + f() ∆x ...... 8 ≐(. + . + . + . + .))(.) ≐ (.. (b)

x Figure 7.7. shows the rectangles used in each method to approximate the − Figure 7.7.: Approximang e dx in deﬁnite integral. These graphs show that in this parcular case, the Le Hand Example 7.7.. Rule is an over approximaon and the Right Hand Rule is an ∫ under approximaon. To get a beer approximaon, we could use more rectangles, as we did 124

Chapter 7 Integraon

xi Exact Approx. sin(xi ) x π/ . . earlier. We could also average the Le and Right Hand Rule results together, − − − x π/ . . giving − − − x π/ . . . + . − − − = .. x π/ . − − x π/ . . x π/ . . x π/ . . x π/ . . x π/ . . x π/ . . x π/ . . − S We begin by finding ∆x: Figure 7.7.: Table of values used to π b a π/ ( π/) π approximate π sin(x ) dx. − = − − = ≐ .. 40 − n ∫ It is useful to write out the endpoints of the subintervals in a table; in Figure 7.7., we give the exact values of the endpoints, their decimal approximaons, and decimal approximaons of sin(x) evaluated at these points. Once this table is created, it is straighorward to approximate the definite y integral using the Le and Right Hand Rules. (Note: the table itself is easy to 1 create, especially with a standard spreadsheet program on a computer. The y = sin(x3) last two columns are all that are needed.) The Le Hand Rule sums the first . 0 5 values ofi sin3 x ) and mulplies the sum by ∆x; the Right Hand Rule sums the last .

1 1 π − 0.5 Le Hand Rule: sin(x ) dx ≐ (.)(.) = .. − π ..... − π Right Hand Rule: sin(x) dx ≐ (.)(.) = .. (a) π ∫− Average of the Le and Right Hand Rules: .. y The actual answer, accurate to places aer the decimal, is .. Our ap- y = sin(x3) 1 proximaons were once again fairly good. The rectangles used in each approximaon are shown in Figure 7.7.. It is clear from the graphs that using more

0.5 rectangles (and hence, narrower rectangles) should result in a more accurate approximaon.

x 1 1 The Trapezoidal Rule −

0.5 − x ..... In Example 7.7. we approximated the value of e− dx with rectangles of equal width. Figure 7.7. shows the rectangles∫ used in the Le and Right (b) Figure 7.7. Approximang : π π sin(x ) dx in Example 7.7.. − ∫ 125

Hand Rules. These graphs clearly show that rectangles do not match the shape of the graph all that well, and that accurate approximaons will only come by using lots of rectangles. Instead of using rectangles to approximate the area, we can instead use x y trapezoids. In Figure .., we show the region under f(x) = e− on [, ] approximated with trapezoids of equal width; the top “corners” of each trape- x zoid lies on the graph of f(x). It is clear from this ﬁgure that these trapezoids y = e− more accurately approximate the area under f and hence should give a beer x approximaon of e− dx. (In fact, these trapezoids seem to give a great approximaon of the area!) . ∫ The formula for the area of a trapezoid is given in Figure ... We approxi- x mate e− dx with these trapezoids in the following example. x ∫ . . . .8 Example 7.7. Approximang deﬁnite integrals using trapezoids ..... x x e− dx − Use trapezoids of equal width to approximate . Figure 7.7.: Approximang e dx using trapezoids of equal widths. ∫ ∫ S To compute the areas of the trapezoids in Figure .., it will again be useful to create a table of values as shown in Figure ... The lemost trapezoid has legs of length and . and a height of .. Thus, by our formula, the area of the lemost trapezoid is:

+ . b Area = a+b h (.) = .. a 2 . Moving right, the next trapezoid has legs of length . and . and a height h of .. Thus its area is: Figure 7.7.: The area of a trapezoid. . + . (.) = ..

The sum of the areas of all trapezoids is: x xi e− i + . . + . . + . (.) + (.) + (.)+ . . . + . . + . . . (.) + (.) = .. . . . . . x We approximate e− dx ≐ .. x ∫ Figure 7.7: A table of values of e− . There are many things to observe in this example. Note how each term in the ﬁnal summaon was mulplied by both / and by ∆x = .. We can factor these coeﬃcients out, leaving a more concise summaon as: (.) (+.)+(.+.)+(.+.)+(.+.)+(.+.) . [ ] 126

Now noce that all numbers except for the ﬁrst and the last are added twice. Therefore we can write the summaon even more concisely as . + (. + . + . + .) + . . [ ] b This is the heart of the Trapezoidal Rule, wherein a deﬁnite integral f(x) dx a is approximated by using trapezoids of equal widths to approximate the∫ corresponding area under f. Using n equally spaced subintervals with endpoints x, b a x , ..., x , we again have ∆x = − . Thus: n+ n

n b f(x ) + f(x ) f(x) dx i i+ ∆x ≈ ∫a i= ∑ n ∆x = f(xi) + f(xi+ ) i= ∑ ( n ) ∆x = f(x) + f(xi) + f(xn+) . = [ ∑i ]

Example 7.7. Using the Trapezoidal Rule π Revisit Example 7.7. and approximate sin(x) dx using the Trapezoidal Rule π ∫− and equally spaced subintervals.

S We refer back to Figure 7.7. for the table of values of sin(x). Recall that ∆x = π/ ≐ .. Thus we have:

π . . + . + ( .) + ... + . + ( .) π sin(x ) dx ≐ − − − − ∫− [ ( ) ] = ..

Noce how “quickly” the Trapezoidal Rule can be implemented once the table of values is created. This is true for all the methods explored in this secon; the real work is creang a table of xi and f(xi) values. Once this is completed, approximang the definite integral is not difficult. Again, using technology is wise. Spreadsheets can make quick work of these computaons and make using lots of subintervals easy. Also noce the approximaons the Trapezoidal Rule gives. It is the average of the approximaons given by the Le and Right Hand Rules! This effecvely 127

renders the Le and Right Hand Rules obsolete. They are useful when first learning about definite integrals, but if a real approximaon is needed, one is generally beer off using the Trapezoidal Rule instead of either the Le or Right Hand Rule.

How can we improve on the Trapezoidal Rule, apart from using more and more trapezoids? The answer is clear once we look back and consider what we have really done so far. The Le Hand Rule is not really about using rectangles to approximate area. Instead, it approximates a funcon f with constant funcons on small subintervals and then computes the definite integral of these constant funcons. The Trapezoidal Rule is really approximang a funcon f with a linear funcon on a small subinterval, then computes the definite integral of this linear funcon. In both of these cases the definite integrals are easy to compute in geometric terms. So we have a progression: we start by approximang f with a constant funcon and then with a linear funcon. What is next? A quadrac funcon. By approximang the curve of a funcon with lots of parabolas, we generally get an even beer approximaon of the definite integral. We call this process Simp- son’s Rule, named aer Thomas Simpson (-), even though others had used this rule as much as years prior. y Simpson’s Rule Given one point, we can create a constant funcon that goes through that

point. Given two points, we can create a linear funcon that goes through those points. Given three points, we can create a quadrac funcon that goes through those three points (given that no two have the same x–value). Consider three points (x, y), (x, y) and (x, y) whose x–values are equally spaced and x < x < x. Let f be the quadrac funcon that goes through these ..... x three points. It is not hard to show that x x x f(x) dx = − y + y + y . (7.) Figure 7.7.: A graph of a funcon f and ∫x a parabola that approximates it well on ( ) Consider Figure ... A funcon f goes through the points shown and the [, ]. parabola g that also goes through those points is graphed with a dashed line. Using our equaon from above, we know exactly that g(x) dx = − + () + = . ∫ ( ) Since g is a good approximaon for f on [, ], we can state that f(x) dx ≐ . ∫ 128

Noce how the interval [, ] was split into two subintervals as we needed points. Because of this, whenever we use Simpson’s Rule, we need to break the x xi e− i interval into an even number of subintervals. b . . In general, to approximate f(x) dx using Simpson’s Rule, subdivide [a, b] a . . into n subintervals, where n is even∫ and each subinterval has width ∆x = (b . . − a)/n. We approximate f with n/ parabolic curves, using Equaon (.) to com- . pute the area under these parabolas. Adding up these areas gives the formula: (a) b

y ∆ x f(x)+f(x)+f(x)+f(x)+. ..+f(xn )+f(xn)+f(xn+) . a f(x) dx ≐ − ∫ _ [ ] x Note how the coeﬃcients of the terms in the summaon have the paern , , y = e− , , , , ..., , , . Let’s demonstrate Simpson’s Rule with a concrete example. .5 Example 7.7. Using Simpson’s Rule x − x Approximate e dx using Simpson’s Rule and equally spaced subintervals. . . . ..... 5 5 75 ∫ (b) Figure 7.7.: A table of values S We begin by making a table of values as we have in the past, as shown in Figure ..(a). Simpson’s Rule states that

x . e− dx ≐ + (.) + (.) + (.) + . = .. ∫ [ ] Recall in Example 7.7. we stated that the correct answer, accurate to places aer the decimal, was .. Our approximaon with Simpson’s Rule, with subintervals, is beer than our approximaon with the Trapezoidal Rule using ! xi sin(xi ) x . . Figure 7.7.(b) shows f(x) = e− along with its approximang parabolas, − − . . demonstrang how good our approximaon is. The approximang curves are − − . . nearly indisnguishable from the actual funcon. − − . − . . Example 7.7. Using Simpson’s Rule π . . . . Approximate sin(x) dx using Simpson’s Rule and equally spaced inter- π . . ∫− . . vals. . . . . S Figure .. shows the table of values that we used in the − past for this problem, shown here again for convenience. Again, ∆x = (π/ + Figure ..: Table of values used to π π/)/ ≐ .. approximate π sin(x ) dx in Example − ... ∫ 129

y Simpson’s Rule states that 1 π y = sin(x3) . (x) dx ≐ ( .) + ( .) + ( .) + ... π sin − − − ∫− [ 0.5 ... + (.) + (.) + ( .) − = . ] x 1 1 − Recall that the actual value, accurate to decimal places, is .. Our ap- th 0.5 proximaon is within one / of the correct value. The graph in Figure 7.7. − ..... shows how closely the parabolas match the shape of the graph. Figure ..: Approximang π Summary π sin(x ) dx in Example .. with − ∫Simpson’s Rule and equally spaced We summarize the key concepts of this secon thus far in the following Key intervals. Idea.

Key Idea 7.7. Numerical Integraon b a Let f be a connuous funcon on [a, b], let n be a posive integer, and let ∆x = − . n Set x = a , x = a + ∆x, . . ., xi = a + (i )∆x, xn+ = b. b − Consider f(x) dx. ∫ a b Le Hand Rule: f(x) dx ∆x f(x) + f(x) + ... + f(xn) . ≈ ∫a b [ ] Right Hand Rule: f(x) dx ∆x f(x) + f(x) + ... + f(xn+) . a ≈ ∫ [ b ∆x ] Trapezoidal Rule: f(x) dx f(x) + f(x) + f(x) + ... + f(xn) + f(xn+) . a ≈ ∫ [ b ∆x ] Simpson’s Rule: f(x) dx f(x ) + f(x ) + f(x ) + ... + f(xn) + f(xn+ ) (n even). ≈ ∫a [ ]

You probably *)*/#1 /*" /1 -4/)0( -$'$)/ "-/$*).ǹ*(+0/ -.) *)0( -$' $)/ "-/$*)/*.()4 $('+' ..4*02$.#ǹ# - $.!- *)Ȏ'$) .$/ , Wolfram Alpha, for exampleǹ

For programmable calculators or computers - simply sum each term below from 1 to n for every rectangle. The formulas above are for hand calculations. Hopefully you do this more than once.

Δ Δ i LH Rule A i = x y -1 Y

Δ i y = f(x) Trapezoidal Rule ΔA i = x ( y - 1 + yi ) 2

Δ ΔA = Δx yi Ai RH Rule i

Δx Δ ( y -1 + 4 y i + y Simpson's Rule A i = 3 i i+1 ) Δx Δx Δx x x xi xi x xn X o 1 - 1 N-1 = a = b 130

Exercises 7.7 Terms and Concepts . √x + dx ∫ . T/F: Simpson’s Rule is a method of approximang an- π derivaves. . x sin x dx ∫ . What are the two basic situaons where approximang the value of a definite integral is necessary? π/ . √cos x dx ∫ . Why are the Le and Right Hand Rules rarely used? . Simpson’s Rule is based on approximang porons of a . ln x dx funcon with what type of funcon? ∫ . dx sin x + Problems ∫− In Exercises – , a definite integral is given. . dx sin x + (a) Approximate the definite integral with the Trapezoidal ∫ Rule and n = . (b) Approximate the definite integral with Simpson’s Rule and n = . (c) Find the exact value of the integral.

. x dx ∫−

. x dx ∫ π . sin x dx ∫

. √x dx ∫

. (x + x x + ) dx − ∫ In Exercises – , a region is given. Find the area of the

region using both the Trapezoid's and Simpson’s Rule: . x dx (a) where the measurements are in cenmeters, taken in ∫ cm increments, and π . cos x dx (b) where the measurements are in hundreds of yards, ∫ taken in yd increments.

. √ x dx . − ∫− In Exercises – , approximate the deﬁnite integral with the Trapezoidal Rule and Simpson’s Rule, with n = . . . .9

. . . cos x dx . .1 ∫ ( ) . ex dx

. − . ∫ . . . NOTE #25 and #26 are especially good questions if you are going to build a free form swimming pool. 131 Solutions 7.7

1. F 2. When the anderivave cannot be computed and when the integrand is unknown. 3. They are superseded by the Trapezoidal Rule; it takes an equal amount...... of work and is generally more accurate. 4. A quadrac funcon (i.e., parabola) 5. (a) 3/4 (b) 2/3 (c) 2/3 6. (a) 250 (b) 250 (c) 250 √ 7. (a) 1 (1 + 2)π ≐ 1.896 4 √ 1 ( + )π≐ . (b) 6 1 2 2 2 005 (c) 2 √ √ 8. (a) 2 + 2 + 3 5.15 √ ≐ √ (b) 2/3(3 + 2 + 2 3) ≐ 5 .25 (c) 16/3 ≐ 5.33 9. (a) 38.5781 (b) 147/4 ≐ 36 .75 (c) 147/4 ≐ 36 .75 10. (a) 0.2207 (b) 0.2005 (c) 1/5 11. (a) 0 (b) 0 (c) 0 √ 12. (a) 9/2(1 + 3) 12.294 √ ≐ (b) 3 + 6 3 ≐ 13 .392 (c) 9π/2 ≐ 14 .137 13. Trapezoidal Rule: 0.9006 Simpson’s Rule: 0.90452 14. Trapezoidal Rule: 3.0241 Simpson’s Rule: 2.9315 15. Trapezoidal Rule: 13.9604 Simpson’s Rule: 13.9066 16. Trapezoidal Rule: 3.0695 Simpson’s Rule: 3.14295 17. Trapezoidal Rule: 1.1703 Simpson’s Rule: 1.1873 18. Trapezoidal Rule: 2.52971 Simpson’s Rule: 2.5447 19. Trapezoidal Rule: 1.0803 Simpson’s Rule: 1.077 20. Trapezoidal Rule: 3.5472 Simpson’s Rule: 3.6133 132 Chapter 8 Applications of Integration

Preview Differentials are important in discovering the fundamental law governing a quantity Q; over a short interval its behavior may be quite simple. If you want to know its growth rate, you divide by dt or perhaps dx depending on whether the quantity changes in time or space. In the applications of this chapter, the total amount or the change in the amount Q is desired; it is the integral of dQ t 2 ∫ x 2 obtained by summing the differentials of Q obtaining ∫ t 1 dQ or x 1 dQ. In this chapter we examine the process of first determining the differential law and then obtaining its integral, the total amount. We will call this the Five Step Procedure. The steps are: I. Draw a picture illustrating the quantity. Label all quantities used in the problem. II. Show a typical differential region. Label. III. Find a simple formula for the differential element which is asymptotically equal its exact amount. IV. Integrate. V. Evaluate.

This procedure should be used on all relevant exercises in this chapter. Using end integrals formulas often lead to mistakes. Also you want to be fluent with this procedure so that you can readily use it in your area of application. In the the sections from Apex Calculus, five steps are also required. In exams each step is worth 20% of the total points awarded the problem. Note that we will assume that the bounding functions are continuous so that dx an infinitesimal implies dy is an infinitesimal. We will begin with an example from Chapter 5, finding the area under a curve. ≤ ≤ π Example Find the area under the curve y = sin x, 0 x 2 .

Y Y

1 y = sin x 1 y = sin x

(x, y) (x, y) ● I, II dA dA

dx dx X X x π ⋯ x ⋯ π 2 2

Method 1 Method 2 Sin x is continuous. So dx an infinitesimal Properly magnified, sin x implies dy an infinitesimal. III appears constant, so dA ≈ sin(x)dx dA ≈ sin(x)dx Here, because the curve is continuous, Here, because the element looks exact, dx an infinitesimal implies dy is an implies an ≈ is justified. In applications infinitesimal, implies an ≈ is justified. such intuitive reasoning is often justified.

π/ A = 2 sin x dx IV ∫ 0

π / 2 V = -cos x 0 = 1. 133

8. Area Between Curves

y We are oen interested in knowing the area of a region. Forget momentarily that we addressed this already in Secon . and approach it instead using the technique described in Key Idea 8... f(x) Let Q be the area of a region bounded by connuous funcons f and g. If we break the region into many subregions, we have an obvious equaon:

g(x) Total Area = sum of the areas of the subregions. The issue to address next is how to systemacally break a region into subregions. A graph will help. Consider Figure 8.. (a) where a region between two curves is x shaded. While there are many ways to break this into subregions, one parcularly a b eﬃcient way is to “slice” it vercally, as shown in Figure 8.. (b), into n equally ..... (a) spaced slices. y We now approximate the area of a slice. Again, we have many options, but using a rectangle seems simplest. Picking any x-value ci in the i-th slice, we set

f(x) the height of the rectangle to be f(ci ) − g(c i ), the difference of the corresponding y-values. The width of the rectangle is a small difference in x-values, which we represent with ∆x. Figure 8.1.1(c) shows sample points c i chosen in each g(x) subinterval and appropriate rectangles drawn. (Each of these rectangles represents a differental element.) Each slice has an area approximately equal to

[f(ci) g(ci)] ∆x; hence, the total area is approximately the Riemann Sum x − a b n ..... Q = f(ci) g(ci) ∆x. (b) − i= y ∑ ( ) b Taking the limit as n gives the exact area as f(x) g(x) dx. → ∞ a − f(x) ∫ ( ) Theorem 8.. Area Between Curves g(x) (restatement of Theorem ..) Let f(x) and g(x) be connuous funcons deﬁned on [a, b] where f(x) ≥ g(x) for all x in [a, b]. The area of the region bounded by the curves x

...... b (c) f(x) g(x) dx. − ∫a Figure 8..: Subdividing a region into ( ) vercal slices and approximang the areas with rectangles. Example .. Finding area enclosed by curves Find the area of the region bounded by f(x) = sin x + , g(x) = cos(x) − x = and x = π, as shown in Figure 8...

Note again from the above graphs that clearly I c NOTE Often only one typical differential c dA ≈ [f(x) - g(x)]dx. II element is shown with x c chosen to be at the left Integrating from a to b gives the area. hand side of the The text argument is complicated, hard dx element. The differential to remember and no more rigorous. element dx should also (But still, sometimes a detailed be shown. discussion is useful.) x

III dA ≈ [f(x) - g(x)]dx b [f(x) - g(x)] IV A = ∫a dx V = . . . 134

S The graph veriﬁes that the upper boundary of the region is y given by f and the lower bound is given by g. Therefore the area of the region is the value of the integral π π f(x) f(x) g(x) dx = sin x + cos(x) dx − − − ∫ ∫ ( ) ( ( π )) = cos x sin(x) + x x − − 5 4π = π ≐ . units.

g(x) ..... Example 8.. Finding total area enclosed by curves − Find the total area of the region enclosed by the funcons f(x) = x + and − Figure 8..: Graphing an enclosed region g(x) = x x + x as shown in Figure 8... in Example 8... − − S A quick calculaon shows that f = g at x = , and . One can proceed thoughtlessly by compung f(x) g(x) dx, but this ignores − ∫ the fact that on [, ], g(x) > f(x). (In fact, the( thoughtless) integraon returns y /, hardly the expected value of an area.) Thus we compute the total area by − breaking the interval [, ] into two subintervals, [, ] and [, ] and using the proper integrand in each. x Total Area = g(x) f(x) dx + f(x) g(x) dx − − ∫ ∫ ( ) ( ) − = x x + x dx + x + x x + dx − − − − ∫ ∫ − = /( + / ) ( ) .... . = / = . units. Figure 8..: Graphing a region enclosed by two funcons in Example ...

The previous example makes note that we are expecng area to be posive.

When ﬁrst learning about the deﬁnite integral, we interpreted it as “signed area y under the curve,” allowing for “negavet area .” Tha doesn’t apply here; area is √ y = (x ) + to be posive. y = x + − − The previous example also demonstrates that we oen have to break a given region into subregions before applying Theorem 8... The following example shows another situaon where this is applicable, along with an alternate view of applying the Theorem. Example 8.. Finding area: integrang with respect to_y _ Find the area of the region enclosed by the funcons y = √x + , y = (x ) ..... x − − + and y = , as shown in Figure 8... Figure 8..: Graphing a region for Exam-ple ... 135

y S We give two approaches to this problem. In the ﬁrst ap-

√ proach, we noce that the region’s “top” is defined by two different curves. x = (y ) x = y + − − On [, ], the top funcon is y = √x + ; on [, ], the top funcon is y = (x ) + . Thus we compute the area as the sum of two integrals: − − Total Area = √x + dx + (x ) + dx − − − − ∫ ∫ (( ) ) (( ) ) = / + / = /...... x The second approach is clever and very useful in certain situaons. We are Figure 8..: The region used in Example used to viewing curves as funcons of x; we input an x-value and a y-value is re- 8.. with boundaries relabeled as turned. Some curves can also be described as funcons of y: input a y-value and func-ons of y. an x-value is returned. We can rewrite the equaons describing the boundary by solving for x: _ y = √-x + x = (y ) y = (x ) + x = y + 1 . ⇒ − , − − _⇒ − Figure 8.. shows the ■region with the boundaries relabeled. A differenal element, a horizontal rectangle,√ is also pictured. The width of the rectangle is a , small change in y: ∆y. The height of the rectangle is a difference in x-values. The “top” x-value is the largest value, i.e., the rightmost. The “boom” x-value is the smaller, i.e., the lemost. Therefore the height of the rectangle is

y + (y ). − − − The area is found by integrang(√ the) above funcon with respect to y with the appropriate bounds. We determine these by considering the y-values the region occupies. It is bounded below by y = , and bounded above by y = . That is, both the “top” and “boom” funcons exist on the y interval [, ]. Thus

Total Area = y + (y ) dy − − − ∫ (√ / ) = ( y) + y (y ) − − − − ( ) = /.

This calculus–based technique of ﬁnding area can be useful even with shapes that we normally think of as “easy.” Example 8.. computes the area of a triangle. While the formula “ _ base height” is well known, in arbitrary triangles it × × can be nontrivial to compute the height. Calculus makes the problem simple. 136

Example 8.. Finding the area of a triangle Compute the area of the regions bounded by the lines y y = x + , y = x + and y = x + , as shown in Figure 8... − − y = x + y = x + 7 S Recognize that there are two “top” funcons to this region, − causing us to use two deﬁnite integrals. 5 Total Area = (x + ) ( x + ) dx + ( x + ) ( x + ) dx y = x + − − − − − − ∫ ∫ = /(+ / ) ( ) ..... x = /. We can also approach this by converng each funcon into a funcon of y. This Figure 8..: Graphing a triangular region also requires integrals, so there isn’t really any advantage to doing so. We do in Example 8... it here for demonstraon purposes. y The “top” funcon is always x = − while there are two “boom” funcons. Being mindful of the proper integraon bounds, we have y y Total Area = − ( y) dy + − (y ) dy − − − − ∫ ∫ = /(+ / ) ( ) = /. Of course, the ﬁnal answer is the same. (It is interesng to note that the area of all subregions used is /. This is coincidental.)

While we have focused on producing exact answers, we are also able to . make approximaons using the principle of Theorem 8... The integrand in (a) the theo-rem is a distance (“top minus boom”); integrang this distance y funcon gives an area. By taking discrete measurements of distance, we can approximate an area using numerical integraon techniques developed in Secon .. The fol-lowing example demonstrates this. Example 8.. Numerically approximang area . To approximate the area of a lake, shown in Figure 8.. (a), the “length” of the . .

lake is measured at -foot increments as shown in Figure 8.. (b), where the . lengths are given in hundreds of feet. Approximate the area of the lake. ..... x S The measurements of length can be viewed as measuring (b) “top minus boom” of two funcons. The exact answer is found by integrang f(x) g(x) dx, but of course we don’t know the funcons f and g. Our Figure 8..: (a) A sketch of a lake, and (b) − the lake with length measurements. ∫ ( ) discrete measurements instead allow us to approximate. 137

We have the following data points:

(, ), (, .), (, .), (, .), (, .), (, .), (, ).

b a We also have that ∆x = −n = , so Simpson’s Rule gives

Area ≐ + . + . + . + . + . + · · · · · · · = (. units. )

Since the measurements are in hundreds of feet, units = ( ) = , , giving a total area of , . (Since we are approximang, we’d likely say the area was about , , which is a lile more than acres.)

In the next secon we apply our applicaons–of–integraon techniques to ﬁnding the volumes of certain solids. 138 Use the Five Step Procedure in each problem Exercises 8. Terms and Concepts y 2 y = sin x + 1

. T/F: The area between curves is always posive.

. 1 . T/F: Calculus can be used to ﬁnd the area of basic geometric shapes. y = sin x

x . In your own words, describe how to ﬁnd the total area en- π/2 π closed by y = f(x) and y = g(x)...... . Describe a situaon where it is advantageous to ﬁnd an area enclosed by curves through integraon with respect to y instead of x. y Problems 2 2 y = sec x

In Exercises – , ﬁnd the area of the shaded region in the given graph. . 1

y y = sin(4x) y = x + 6 x .... π/8 π/4 . .

y = cos x +

y x ..... π π y = sin x . y . x π y = x + x + π/ π/ π/ π/ − . − y = cos x . − ..... x − y = x + x − ..... −

y y y = 2 x 2 y =

1 . y = 1 . y = x x / π π 2 ..... x . .... . 139

y y

y = √x y = √x + 1 y = √2 x + 1 2 − .5 y = x + − . . x 1 y = 1 .5 − y = x − x ..... − 1 2

In Exercises – , ﬁnd the total area enclosed by the func- 4 ons f and g. y = x + 2

y = x2 . f(x) = x + x , g(x) = x + x . − − 2

. f(x) = x x + , g(x) = x + − −

. f(x) = sin x, g(x) = x/π ..... x 1 1 2 − . f(x) = x x + x , g(x) = x + x − − − − y

. f(x) = x, g(x) = √x 1 x = 1 y + 1 − 2 . f(x) = x + x + x + , g(x) = x + x + x − . 1 2

1 . The funcons f(x) = cos(x) and g(x) = sin x intersect − inﬁnitely many mes, forming an inﬁnite number of re- 1 2 x = 2 y peated, enclosed regions. Find the areas of these regions. 2 −

. The funcons f(x) = cos(x) and g(x) = sin x intersect inﬁnitely many mes, forming an inﬁnite number of re- y peated, enclosed regions. Find the areas of these regions. y = x/

In Exercises – , ﬁnd the area of the enclosed region in two ways: . .5

. by treang the boundaries as funcons of x, and y = √x / − y x . by treang the boundaries as funcons of ...... .5

y y

y = (x ) + y = √x + 1 y = √2 x + 1 − 2 − y = x +

. . y = 1 y = 1

..... x x 1 2 140

In Exercises – , ﬁnd the area triangle formed by the given . Use Simpson’s Rule to approximate the area of the pictured three points. lake whose lengths, in hundreds of feet, are measured in -foot increments. . (, ), (, ), and (, )

. ( , ), (, ), and (, ) − − . (, ), (, ), and (, ) .2

. (, ), (, ), and (, ) . . . . Use the Trapezoidal Rule to approximate the area of the . pictured lake whose lengths, in hundreds of feet, are mea- .9 sured in -foot increments. .9 7.

. . .

Solutions 8.1

1. T 19. All enclosed regions have the same area, with regions being the reﬂecon of adjacent regions.√ One region is formed on T 2. [π/4, 5π/4], with area 2 2. √ 3. Answers will vary. 20. On regions such as [π/6, 5π/6], the√ area is 3 3/2. On regions 4. Answers may vary; one common answer is when the region has such as [−π/2, π/6], the area is 3 3/4. two or more “top” or “boom” funcons when viewing the 21. 1 region with respect to x, but has only 1 “top” funcon and 1 “boom” funcon when viewed with respect to y. The former 22. 5/3 area requires mulple integrals to compute, whereas the laer 23. 9/2 area requires one. 24. 9/4 √ 5. 4π + π2 ≐ 22.436 25. 1/12(9 − 2 2) ≐ 0.514 6. 16/3 26. 4/3 7. π 27. 1 8. π 28. 5 9. 1/2 29. 4 √ 10. 2 2 30. 133/20 2 11. 1/ ln 4 31. 219,000 32. 623,333 2 12. 4/3 13. 4.5 14. 4/3 15. 2 − π/2 16. 8 17. 1/6 18. 37/12 141

8.2 Volumes by Slicing. The Disk Method: 5 Steps Example Find the volume when the region bounded by y = 1 , y = 0, x = 1 and x = 2 is revolved ●x about the X-axis.

(x,y) I, II 1 1/2 y = x — y — X

dx X ■ x 2

When the differential element is rotated about the x-axis, the result is asymptotically a disk.

y2 _1 III dV ≐ π dx = π x 2 dx --

2 dx IV π 2 V = ∫1 x

π - 1  2 V = x 1

π = --2 142

8. Volume by Cross-Seconal Area; Disk and Washer Methods

The volume of a general right cylinder, as shown in Figure 8.., is Area of the base height. We can use this fact as the building block in× ﬁnding volumes of a variety of shapes. Given an arbitrary solid, we can approximate its volume by cung it into n thin slices. When the slices are thin, each slice can be approximated well by a general right cylinder. Thus the volume of each slice is approximately its cross- seconal area thickness. (These slices are the diﬀeren al elements.) × ( ) By orienng a solid along the x-axis, we can let A xi represent the cross- th seconal area of the i slice, and let ∆xi represent the thickness of this slice (the Figure 8..: The volume of a general thickness is a small change in x). The total volume of the solid is approximately: right cylinder n Volume ≐ Area thickness × i= [ ] ∑n

= A(xi)∆xi. = ∑i Recognize that this is a Riemann Sum. By taking a limit (as the thickness of the slices goes to ) we can ﬁnd the volume exactly.

Theorem 8.. Volume By Cross-Seconal Area The volume V of a solid, oriented along the x-axis with cross-seconal area A(x) from x = a to x = b, is

b V = A(x) dx. ∫a

Example 8.. Finding the volume of a solid Find the volume of a pyramid with a square base of side length in and a height of in.

S There are many ways to “orient” the pyramid along the x-

dx axis; Figure 8.. gives one such way, with the pointed top of the pyramid at the origin and the x-axis going through the center of the base. Each cross secon of the pyramid is a square; this is a sample diﬀerenal element. To determine its area A(x), we need to determine the side lengths of Figure 8..: Orienng a pyramid along the x-axis in Example 8.. cc dV ≈ A(x) dx 2 2 c = (2 x) dx = 4 x dx 5 V = 4 ∫ x2dx 0 x 3 = 4 _ 5 3 |0 500_ = 3 143

the square. When x = , the square has side length ; when x = , the square has side length . Since the edges of the pyramid are lines, it is easy to figure that each cross-seconal square has side length x, giving A(x) = (x) = x. If one were to cut a slice out of the pyramid at x = , as shown in Figure 8.., one would have a shape with square boom and top with sloped sides. If the slice were thin, both the boom and top squares would have sides lengths of about , and thus the cross–seconal area of the boom and top would be about in . Leng ∆xi represent the thickness of the slice, the volume of this slice would then be about ∆xiin . Cung the pyramid into n slices divides the total volume into n equally– spaced smaller pieces, each with volume (xi) ∆x, where xi is the approximate locaon of the slice along the x-axis and ∆x represents the thickness of each slice. One can approximate total volume of the pyramid by summing up the volumes of these slices: n Figure 8..: Cung a slice in the pyramid Approximate volume = (xi) ∆x. in Example 8.. at x = . = ∑i Taking the limit as n gives the actual volume of the pyramid; recoginizing this sum as a Riemann→ Sum∞ allows us to find the exact answer using a definite integral, matching the definite integral given by Theorem 8... We have n V = lim (xi) ∆x n →∞ i= ∑ = x dx ∫ = x = in ≐ . in. We can check our work by consulng the general equaon for the volume of a pyramid (see the back cover under “Volume of A General Cone”): area of base height. × × Certainly, using this formula from geometry is faster than our new method, but the calculus–based method can be applied to much more than just cones.

An important special case of Theorem 8.. is when the solid is a solid of revoluon, that is, when the solid is formed by rotang a shape around an axis. Start with a funcon y = f(x) from x = a to x = b. Revolving this curve about a horizontal axis creates a three-dimensional solid whose cross secons 144

are disks (thin circles). Let R(x) represent the radius of the cross-seconal disk at x; the area of this disk is πR(x). Applying Theorem 8.. gives the Disk Method.

Key Idea .8. The Disk Method Let a solid be formed by revolving the curve y = f(x) from x = a to x = b around a horizontal axis, and let R(x) be the radius of the cross-seconal disk at x. The volume of the solid is

b V = π R(x) dx. ∫a

Example 8.. Finding volume using the Disk Method Find the volume of the solid formed by revolving the curve y = /x, from x = to x = , around the x-axis.

S A sketch can help us understand this problem. In Figure 8..(a) the curve y = /x is sketched along with the differenal element – a disk – at x with radius R(x) = /x. In Figure 8.. (b) the whole solid is pictured, along with the differenal element. The volume of the differenal element shown in part (a) of the figure is approximately πR(xi) ∆x, where R(xi) is the radius of the disk shown and ∆x is the thickness of that slice. The radius R(xi) is the distance from the x-axis to the (a) curve, hence R(xi) = /xi. Slicing the solid into n equally–spaced slices, we can approximate the total volume by adding up the approximate volume of each slice: n Approximate volume = π ∆x. x = i ∑i ( ) Taking the limit of the above sum as n gives the actual volume; recognizing this sum as a Riemann sum allows→ us∞ to evaluate the limit with a definite integral, which matches the formula given in Key Idea 8..:

n (b) V = lim π ∆x n x →∞ = i Figure 8..: Sketching a solid in Example ∑i ( ) 8... = π dx x ∫ ( ) = π dx x ∫ 145

= π − x [ ] = π ( ) − − − π[ ] = units.

While Key Idea 8.. is given in terms of funcons of x, the principle involved can be applied to funcons of y when the axis of rotaon is vercal, not horizontal. We demonstrate this in the next example.

Example .. Finding volume using the Disk Method (a) Find the volume of the solid formed by revolving the curve y = /x, from x = to x = , about the y-axis.

S Since the axis of rotaon is vercal, we need to convert the funcon into a funcon of y and convert the x-bounds to y-bounds. Since y = /x defines the curve, we rewrite it as x = /y. The bound x = corresponds to the y-bound y = , and the bound x = corresponds to the y-bound y = /. Thus we are rotang the curve x = /y, from y = / to y = about the y-axis to form a solid. The curve and sample differenal element are sketched in Figure 8.. (a), with a full sketch of the solid in Figure 8.. (b). We integrate to find the volume: (b) V = π dy Figure 8..: Sketching a solid in Example / y 8... ∫ π = − y /

= π units .

We can also compute the volume of solids of revoluon that have a hole in the center. The general principle is simple: compute the volume of the solid irrespecve of the hole, then subtract the volume of the hole. If the outside radius of the solid is R(x) and the inside radius (deﬁning the hole) is r(x), then the volume is

b b b V = π R(x) dx π r(x) dx = π R(x) r(x) dx. − − (a) ∫a ∫a ∫a ( ) One can generate a solid of revoluon with a hole in the middle by revolving a region about an axis. Consider Figure 8..(a), where a region is sketched along

(b)

Figure 8..: Establishing the Washer Method; see also Figure 8... 146

with a dashed, horizontal axis of rotaon. By rotang the region about the axis, a solid is formed as sketched in Figure 8..(b). The outside of the solid has radius R(x), whereas the inside has radius r(x). Each cross secon of this solid will be a washer (a disk with a hole in the center) as sketched in Figure 8... This leads us to the Washer Method.

Key Idea .. The Washer Method Let a region bounded by y = f(x), y = g(x), x = a and x = b be rotated about a horizontal axis that does not intersect the region, forming a solid. Each cross secon at x will be a washer with outside radius R(x) and inside radius r(x). The volume of the solid is Figure 8..: Establishing the Washer Method; see also Figure 8... b V = π R(x) r(x) dx. a − ∫ ( )

Even though we introduced it ﬁrst, the Disk Method is just a special case of the Washer Method with an inside radius of r(x) = .

Example .. Finding volume with the Washer Method Find the volume of the solid formed by rotang the region bounded by y = x x + and y = x about the x-axis. − − S A sketch of the region will help, as given in Figure 8..(a). Rotang about the x-axis will produce cross secons in the shape of washers, as shown in Figure 8..(b); the complete solid is shown in part (c). The outside (a) radius of this washer is R(x) = x + ; the inside radius is r(x) = x x + . As the region is bounded from x = to x = , we integrate as follows− to compute the volume.

V = π (x ) (x x + ) dx − − − ∫ ( ) = π x + x x + x dx − − − ∫ ( ) = π x + x x + x x − − − [ ] = (b) 15 When rotang about a vercal axis, the outside and inside radius funcons must be funcons of y.

(c)

Figure 8..: Sketching the diﬀerenal el-ement and solid in Example 8... 147

Example 8.. Finding volume with the Washer Method Find the volume of the solid formed by rotang the triangular region with verces at (, ), (, ) and (, ) about the y-axis.

S The triangular region is sketched in Figure 8..(a); the differenal element is sketched in (b) and the full solid is drawn in (c). They help us establish the outside and inside radii. Since the axis of rotaon is vercal, each radius is a funcon of y. The outside radius R(y) is formed by the line connecng (, ) and (, ); it is a constant funcon, as regardless of the y-value the distance from the line to the axis of rotaon is . Thus R(y) = . (a) The inside radius is formed by the line connecng (, ) and (, ). The equaon of this line is y = x , but we need to refer to it as a funcon of y. Solving − for x gives r(y) = (y + ). We integrate over the y-bounds of y = to y = . Thus the volume is V = π (y + ) dy − ∫ ( ) ( ) = π y y + dy − − ∫ ( ) = π y y + y − − [ ]

(b) = π ≐ . units. This secon introduced a new applicaon of the definite integral. Our de- fault view of the definite integral is that it gives “the area under the curve.” How- ever, we can establish definite integrals that represent other quanes; in this secon, we computed volume. The ulmate goal of this secon is not to compute volumes of solids. That can be useful, but what is more useful is the understanding of this basic principle of integral calculus, outlined in Key Idea 8..: to find the exact value of some quanty, • we start with an approximaon (in this secon, slice the solid and approximate the volume of each slice), (c) • then make the approximaon beer by refining our original approximaon (i.e., use more slices), Figure 8..: Sketching the solid in Example 8... • then use limits to establish a definite integral which gives the exact value. We pracce this principle in the next secon where we find volumes by slicing solids in a different way. 148 Use the Five Step Procedure in each problem Exercises 8. Terms and Concepts It is an excellent exercise to translate each of the examples of Apex above into the 5 step Method. . T/F: A solid of revoluon is formed by revolving a shape around an axis.

. In your own words, explain how the Disk and Washer Meth- y ods are related. 1 y = √x . Explain the how the units of volume are found in the inte- ..... gral of Theorem ..: if A(x) has units of in, how does y = x . 0.5 A(x) dx have units of in? ∫ . A fundamental principle of this secon is “ can be found by integrang an area funcon.” x 0.5 1

In Exercises – , a region of the Cartesian plane is shaded. Problems Use the Disk/Washer Method to ﬁnd the volume of the solid of revoluon formed by revolving the region about the y- In Exercises – , a region of the Cartesian plane is shaded. axis. Use the Disk/Washer Method to ﬁnd the volume of the solid of revoluon formed by revolving the region about the x- axis. y

y y = x − y = x − . ..... .

x − − ..... x − − y y y = 5x y = 5x

. 5 . 5 .....

x ..... x .5 .5 .5 .5

y y 1 1

y = cos x y = cos x . 0.5 . 0.5 .....

x x 0.5 1 1.5 ..... 0.5 1 1.5 (Hint: Integraon By Parts will be necessary, twice. First let u = arccos x, then let u = arccos x.) 149 y

1 y = √x In Exercises – , a solid is described. Orient the solid along the x-axis such that a cross-seconal area funcon A(x) can y = x . 0.5 be obtained, then apply Theorem .. to ﬁnd the volume of the solid.

. A right circular cone with height of and base radius of . x 0.5 1

In Exercises ..... – , a region of the Cartesian plane is de-

scribed. Use the Disk/Washer Method to ﬁnd the volume of the solid of revoluon formed by rotang the region about each of the given axes.

. Region bounded by: y = √x, y = and x = . Rotate about: . A skew right circular cone with height of and base radius (a) the x-axis (c) the y-axis of . (Hint: all cross-secons are circles.) (b) y = (d) x =

. Region bounded by: y = x and y = . − Rotate about: (a) the x-axis (c) y = − (b) y = (d) x =

. The triangle with verces (, ), (, ) and (, ). . A right triangular cone with height of and whose base is Rotate about: a right, isosceles triangle with side length .

(a) the x-axis (c) the y-axis (b) y = (d) x =

. Region bounded by y = x x + and y = x . − − Rotate about:

(a) the x-axis (c) y = . A solid with length with a rectangular base and triangu- (b) y = lar top, wherein one end is a square with side length and . Region bounded by y = /√x + , x = , x = and the other end is a triangle with base and height of . − the x-axis. Rotate about: (a) the x-axis (c) y = − (b) y = . Region bounded by y = x, y = x and x = . Rotate about: (a) the x-axis (c) the y-axis (b) y = (d) x =

23. Find the volume of water in the tank if it is filled to height h.

—

⟶ — h

5 ———

15 ⟶ 150

Secon 8.2 1. T 2. Answers will vary. 3. Recall that “dx” does not just “sit there;” it is mulplied by A(x) and represents the thickness of a small slice of the solid. Therefore dx has units of in, giving A(x) dx the units of in3. 4. volume √ 5. 48π 3/5 units3 6. 175π/3 units3 7. π2/4 units3 8. π/6 units3 9. 9π/2 units3 10. 35π/3 units3 11. π2 − 2π units3 12. 2π/15 units3 13. (a) π/2 (b) 5π/6 (c) 4π/5 (d) 8π/15 14. (a) 512π/15 (b) 256π/5 (c) 832π/15 (d) 128π/3 15. (a) 4π/3 (b) 2π/3 (c) 4π/3 (d) π/3 16. (a) 104π/15 (b) 64π/15 (c) 32π/5 17. (a) π2/2 (b) π2/2 − 4π sinh−1(1) (c) π2/2 + 4π sinh−1(1) 18. (a) 8π (b) 8π (c) 16π/3 (d) 8π/3 19. Placing the p of the cone at the origin such that the x-axis runs through the center of the circular base, we have A(x) = πx2/4. Thus the volume is 250π/3 units3. 20. The cross–secons of this cone are the same as the cone in Exercise 19. Thus they have the same volume of 250π/3 units3. 21. Orient the cone such that the p is at the origin and the x-axis is perpendicular to the base. The cross–secons of this cone are right, isosceles triangles with side length 2x/5; thus the cross–seconal areas are A(x) = 2x2/25, giving a volume of 80/3 units3. 22. Orient the solid so that the x-axis is parallel to long side of the base. All cross–secons are trapezoids (at the far le, the trapezoid is a square; at the far right, the trapezoid has a top length of 0, making it a triangle). The area of the trapezoid at x is A(x) = 1/2(−1/2x + 5 + 5)(5) = −5/4x + 25. The volume is 187.5 units3. h - 5 375 π 23. V = 15 (h - 5) h(10 - h) + 375 sin-1 + 5 2 151

8.3 Volumes by Cylindrical Shell Method: 5 Steps 1 Find the volume when the region bounded by y = x , y = 0, x = 1 and x = 2 is revolved about the Y-axis.

Y DistanceDDistanc around: 1 c = 2πx

I, II. 1 1/2 y = x y

dx X r 1 x 2

.1 III. dV ≈ 2π x y dx = 2π x· dx = 2π dx _ -x Note that dV is independent of dx. 2 V = 2π dx IV. ∫1 Can you explain this intuitively? V. = 2π x] 2 = 2 π 2 1 I

8. The Shell Method

Oen a given problem can be solved in more than one way. A parcular method may be chosen out of convenience, personal preference, or perhaps necessity. Ulmately, it is good to have opons. The previous secon introduced the Disk and Washer Methods, which computed the volume of solids of revoluon by integrang the cross–seconal area of the solid. This secon develops another method of compung volume, the (a) Shell Method. Instead of slicing the solid perpendicular to the axis of rotaon creang cross-secons, we now slice it parallel to the axis of rotaon, creang “shells.” Consider Figure 8.., where the region shown in (a) is rotated around the y-axis forming the solid shown in (b). A small slice of the region is drawn in (a), parallel to the axis of rotaon. When the region is rotated, this thin slice forms a cylindrical shell, as pictured in part (c) of the figure. The previous secon approximated a solid with lots of thin disks (or washers); we now approximate a solid with many thin cylindrical shells. To compute the volume of one shell, first consider the paper label on a soup can with radius r and height h. What is the area of this label? A simple way of determining this is to cut the label and lay it out flat, forming a rectangle with (b) height h and length πr . Thus the area is A = πr h; see Figure 8..(a). Do a similar process with a cylindrical shell, with height h, thickness ∆x, and approximate radius r. Cung the shell and laying it flat forms a rectangular solid with length πr, height h and depth ∆x. Thus the volume is V πrh∆x; see ≈ Figure 8..(b). (We say “approximately” since our radius was an approximaon.) By breaking the solid into n cylindrical shells, we can approximate the volume of the solid as N .n V ≐ πrihi∆xi, = ∑i th (c) where ri, hi and ∆xi are the radius, height and thickness of the i shell, respec- vely. Figure 8..: Introducing This is a Riemann Sum. Rounding off yields the definite integral. the Shell Method. 152

∆x r r ∆x . . 2πr 2πr .

h h .

. h A = πrh ≈ π ∆ . 2 h . V 2 rh x cut here cut here

(a) (b)

Figure 8..: Determining the volume of a thin cylindrical shell.

Key Idea 8.. The Shell Method Let a solid be formed by revolving a region R, bounded by x = a and x = b, around a vercal axis. Let r(x) represent the distance from the axis of rotaon to x (i.e., the radius of a sample shell) and let h(x) represent the height of the solid at x (i.e., the height of the shell). The volume of the solid is b V = π r(x)h(x) dx. ∫a

Special Cases:

. When the region R is bounded above by y = f(x) and below by y = g(x), then h(x) = f(x) g(x). − y . When the axis of rotaon is the y-axis (i.e., x = ) then r(x) = x. 1 y = 1 2 Let’s pracce using the Shell Method. 1 + x    Example 8.. Finding volume using the Shell Method  h(x)  Find the volume of the solid formed by rotang the region bounded by y = ,   y = /( + x ), x = and x = about the y-axis.     x x 1 S This is the region used to introduce the Shell Method in Fig- ..... | r{z(x) } ure 8.., but is sketched again in Figure 8.. for closer reference. A line is drawn in the region parallel to the axis of rotaon represenng a shell that will Figure 8..: Graphing a region in Example 8... 153

be carved out as the region is rotated about the y-axis. (This is the diﬀerenal element.) The distance this line is from the axis of rotaon determines r(x); as the distance from x to the y-axis is x, we have r(x) = x. The height of this line determines h(x); the top of the line is at y = /( + x), whereas the boom of the line is at y = . Thus h(x) = /( + x) = /( + x). The region is − bounded from x = to x = , so the volume is

x V = π dx. + x ∫

This requires substuon. Let u = + x, so du = x dx. We also change the y bounds: u() = and u() = . Thus we have:

+ x = π du u = ∫ y  .  h(x) = π ln u  = π ln ≐ . units. r(x) | {z } x ..... x Note: in order to ﬁnd this volume using the Disk Method, two integrals would (a) be needed to account for the regions above and below y = /.

With the Shell Method, nothing special needs to be accounted for to compute the volume of a solid that has a hole in the middle, as demonstrated next.

Example .. Finding volume using the Shell Method Find the volume of the solid formed by rotang the triangular region determined by the points (, ), (, ) and (, ) about the line x = .

S The region is sketched in Figure 8..(a) along with the differenal element, a line within the region parallel to the axis of rotaon. In part (b) of the figure, we see the shell traced out by the differenal element, (b) and in part (c) the whole solid is shown. The height of the differenal element is the distance from y = to y = x + , the line that connects the points (, ) and (, ). Thus h(x) = x+ = x. − The radius of the shell formed by the differenal element is the distance from x to x = ; that is, it is r(x) = x. The x-bounds of the region are x = to −

(c)

Figure 8..: Graphing a region in Example 8... 154

x = , giving

V = π ( x)(x) dx − ∫ = π x x) dx − ∫ ( = π x x − ( ) = π ≐ . units .

When revolving a region around a horizontal axis, we must consider the ra- y dius and height funcons in terms of y, not x.

Example 8.. Finding volume using the Shell Method y − = Find the volume of the solid formed by rotang the region given in Example 8.. y x about the x-axis.  h(y)    r(y) S The region is sketched in Figure 8..(a) with a sample dif- | {z }   ferenal element. In part (b) of the figure the shell formed by the differenal   element is drawn, and the solid is sketched in (c). (Note that the triangular re-  x ..... gion looks “short and wide” here, whereas in the previous example the same region looked “tall and narrow.” This is because the bounds on the graphs are (a) different.) The height of the differenal element is an x-distance, between x = y − and x = . Thus h(y) = ( y ) = y+ . The radius is the distance from − − − y to the x-axis, so r(y) = y. The y bounds of the region are y = and y = , leading to the integral

V = π y y + dy − ∫ [ ( )] = π y + y dy − ∫ [ ] (b) = π y + y − [ ] = π − [ ] = π ≐ . units.

(c)

Figure 8..: Graphing a region in Example 8... 155

At the beginning of this secon it was stated that “it is good to have opons.” The next example ﬁnds the volume of a solid rather easily with the Shell Method, but using the Washer Method would be quite a chore.

y Example 8.. Finding volume using the Shell Method Find the volume of the solid formed by revolving the region bounded by y = sin x 1 and the x-axis from x = to x = π about the y-axis.

   S The region and a diﬀerenal element, the shell formed by this    h(x) diﬀerenal element, and the resulng solid are given in Figure 8... The ( ) = ( ) = r(x)  radius of a sample shell is r x x; the height of a sample shell is h x sin x,   = = π  each from x to x . Thus the volume of the solid is  x x π π ..... 2 π z }| { = π . (a) V x sin x dx ∫

This requires Integraon By Parts. Set u = x and dv = sin x dx; we leave it to the reader to ﬁll in the rest. We have:

π π = π x cos x + cos x dx − [ π ∫ ]

= π π + sin x [ ] = π π + [ ] = π ≐ . units . (b)

Note that in order to use the Washer Method, we would need to solve y = sin x for x, requiring the use of the arcsine funcon. We leave it to the reader to verify that the outside radius funcon is R(y) = π arcsin y and the inside − radius funcon is r(y) = arcsin y. Thus the volume can be computed as

π (π arcsin y) (arcsin y) dy. − − ∫ [ ] This integral isn’t terrible given that the arcsin y terms cancel, but it is more onerous than the integral created by the Shell Method. (c) We end this secon with a table summarizing the usage of the Washer and Figure 8..: Graphing a region in Shell Methods. Example 8... 156

Key Idea 8.. Summary of the Washer and Shell Methods Let a region R be given with x-bounds x = a and x = b and y-bounds y = c and y = d. Washer Method Shell Method b d Horizontal π R(x) r(x) dx π r(y)h(y) dy Axis − ∫a ∫c ( ) d b Vercal π R(y) r(y) dy π r(x)h(x) dx Axis − ∫c ∫a ( )

As in the previous secon, the real goal of this secon is not to be able to compute volumes of certain solids. Rather, it is to be able to solve a problem by first approximang, then using limits to refine the approximaon to give the exact value. In this secon, we approximate the volume of a solid by cung it into thin cylindrical shells. By summing up the volumes of each shell, we get an approximaon of the volume. By taking a limit as the number of equally spaced shells goes to infinity, our summaon can be evaluated as a definite integral, giving the exact value. We use this same principle again in the next secon, where we find the length of curves in the plane. 157 Use the Five Step Procedure in each problem

Again, it is an excellent exercise to translate each of the Exercises 8. examples of Apex above into the 5 step Method. Terms and Concepts y 1

. T/F: A solid of revoluon is formed by revolving a shape y = cos x . around an axis. 0.5

. T/F: The Shell Method can only be used when the Washer x Method fails...... 0.5 1 1.5

. T/F: The Shell Method works by integrang cross–seconal y areas of a solid. 1

y = √x . T/F: When ﬁnding the volume of a solid of revoluon that y = x was revolved around a vercal axis, the Shell Method inte- . 0.5 grates with respect to x.

x ..... 0.5 1 Problems

In Exercises – , a region of the Cartesian plane is shaded. In Exercises – , a region of the Cartesian plane is shaded. Use the Shell Method to ﬁnd the volume of the solid of revo- Use the Shell Method to ﬁnd the volume of the solid of revoluon formed by revolving the region about the x-axis. luon formed by revolving the region about the y-axis.

y y

y = x y = x − −

. .

..... x ..... x − − − −

y y

y = 5x

. 5 . 5

..... x ..... x .5 .5 .5 .5 158

y . Region bounded by: y = x and y = . − 1 Rotate about:

(a) x = (c) the x-axis y = cos x (b) x = (d) y = . 0.5 −

. The triangle with verces (, ), (, ) and (, ). Rotate about: x ...... 0 5 1 1 5 (a) the y-axis (c) the x-axis

y (b) x = (d) y =

1 . Region bounded by y = x x + and y = x . y = √x − − Rotate about: y = x . 0.5 (a) the y-axis (c) x = − (b) x =

x ..... 0.5 1 . Region bounded by y = /√x + , x = and the x and y-axes. In Exercises – , a region of the Cartesian plane is de- Rotate about: scribed. Use the Shell Method to ﬁnd the volume of the solid of revoluon formed by rotang the region about each of the (a) the y-axis (b) x = given axes.

. Region bounded by: y = √x, y = and x = . . Region bounded by y = x, y = x and x = . Rotate about: Rotate about:

(a) the y-axis (c) the x-axis (a) the y-axis (c) the x-axis (b) x = (d) y = (b) x = (d) y =

Solutions 8.3 17. (a) 2π ( 2 − 1) 1. T 3 10. 350π/3 units √ − −1 2. F 11. π2/4 units3 (b) 2π(1 2 + sinh (1)) 3. F 12. π/6 units3 18. (a) 16π/3 4. T 13. (a) 4π/5 (b) 8π/3 5. 9π/2 units3 (b) 8π/15 (c) 8π (d) 8π 6. 70π/3 units3 (c) π/2 7. π2 − 2π units3 (d) 5π/6 8. 2π/15 units3 14. (a) 128π/3 √ 9. 48π 3/5 units3 (b) 128π/3 (c) 512π/15 (d) 256π/5 15. (a) 4π/3 (b) π/3 (c) 4π/3 (d) 2π/3 16. (a) 16π/3 (b) 8π/3 (c) 8π 159 8.4 Arc Length

What is the arc length, the length of the curve, y = f(x), a ≤ x ≤ b? Y

y = f(x) ds dy dx

X a x b Theorem Let y = f(x) be differentiable for a ≤ x ≤ b. Then its arc length on that interval is b dy 2 s = 1 +   dx. ∫a dx Derivation f is differentiable on the interval. So f is differentiable a x and therefore locally linear or equivalently 'asymptotically straight' there. Therefore

ds2 ≈ dx2 + dy2 Theorem of Pythagoras Note diagram below  +  dy 2 2 = 1 dx dx b 2 dy 2 ds = 1 +  dy  dx ⟹ s = 1 +   dx. dx ∫a dx

dy 2 +  dy 1 dx ) m = Example Find the length of y = 2x3/2 for 0 ≤ x ≤ 1. dx

3/2 2 y = 2x 1 meaning of slope I., II. 1

ds dy dx

X x 1

y = x3/2 ⇒ dy x1/2 III. 2 dx = 3 2 1 +  dy  + x dx = 1 9

1 IV. s = 1 + 9 x dx ∫0 V. u = 1 + 9x, du = 9dx x = 0 → u = 1 x = 1 → u = 10 10 10 s = 1 u du = 1 2 u3/2 9 ∫1 9 3 1 2 ( = 27 10 10 - 1) 160

8.4 Arc Length and Surface Area Readings y

In previous secons we have used integraon to answer the following quesons: . Given a region, what is its area?

. . Given a solid, what is its volume?

In this secon, we address a related queson: Given a curve, what is its x π π π π length? This is oen referred to as arc length...... Consider the graph of y = sin x on [, π] given in Figure 8..(a). How long (a) is this curve? That is, if we were to use a piece of string to exactly match the y shape of this curve, how long would the string be? As we have done in the past, we start by approximang; later, we will reﬁne our answer using limits to get an exact soluon.

√ The length of straight–line segments is easy to compute using the Distance Formula. We can approximate the length of the given curve by approximang . the curve with straight lines and measuring their lengths. In Figure 8..(b), the curve y = sin x has been approximated with line segments (the interval [, π] has been divided into equally–lengthed subinter- x π π π π vals). It is clear that these four line segments approximate y = sin x very well ..... on the ﬁrst and last subinterval, though not so well in the middle. Regardless, (b) the sum of the lengths of the line segments is ., so we approximate the arc length of y = sin x on [, π] to be .. Figure 8..: Graphing y = sin x on [, π] and approximang the curve with line In general, we can approximate the arc length of y = f(x) on [a, b] in the segments. following manner. Let a = x < x < . . . < xn < xn+ = b be a paron th of [a, b] into n subintervals. Let ∆xi represent the length of the i subinterval [xi, xi+]. Figure 8.. zooms in on the i th subinterval where y = f(x) is approximated by a straight line segment. The dashed lines show that we can view this line seg- y ment as the hypotenuse of a right triangle whose sides have length ∆xi and ∆yi. Using the Pythagorean Theorem, the length of this line segment is ∆xi + ∆yi . yi+1 Summing over all subintervals gives an arc length approximaon √

∆yi n yi L ≐ ∆xi + ∆yi . = ∆xi ∑i √

x As shown here, this is not a Riemann Sum. While we could conclude that x x ..... i i+1 taking a limit as the subinterval length goes to zero gives the exact arc length, we would not be able to compute the answer with a deﬁnite integral. We need th Figure 8..: Zooming in on the i ﬁrst to do a lile algebra. subin-terval [xi, xi+] of a paron of [a, b]. 161

In the above expression factor out a ∆xi term:

n n ∆y ∆x + ∆y = ∆x + i . i i i ∆x = = √ i ∑i √ ∑i ( ) Now pull the ∆xi term out of the square root:

n ∆y = + i ∆x . ∆x i = √ i ∑i This is nearly a Riemann Sum. Consider the ∆yi /∆xi term. The expression ∆yi/∆xi measures the “change in y/change in x,” that is, the “rise over run” of f on the i th subinterval. The Mean Value Theorem of Diﬀerenaon (Theorem th ..) states that there is a ci in the i subinterval where f ′(ci) = ∆yi/∆xi. Thus we can rewrite our above expression as:

n = + f ′(ci) ∆xi. i= ∑ √

This is a Riemann Sum. As long as f ′ is connuous, we can invoke Theorem .. and conclude

b = + f ′(x) dx. a ∫ √

Theorem 8.. Arc Length

Let f be differenable on [a, b], where f ′ is also connuous on [a, b]. Then the arc length of f from x = a to x = b is Note: This is our first use of differena- b bility on a closed interval since Secon L = + f ′(x) dx. .. ∫a √ The theorem also requires that f ′ be connuous on [a, b]; while examples are ar- As the integrand contains a square root, it is oen difficult to use the cane, it is possible for f to be differen- formula in Theorem 8.. to find the length exactly. When exact answers are able yet f ′ is not connuous. difficult to come by, we resort to using numerical methods of approximang definite integrals. The following examples will demonstrate this. 162

y Example 8.. Finding arc length / 8 Find the arc length of f(x) = x from x = to x = .

6 / S We find f ′(x) = x ; note that on [, ], f is differenable and f ′ is also connuous. Using the formula, we find the arc length L as 4

2 L = + x/ dx √ x ∫ ( ) ..... 2 4 = + x dx Figure 8..: A graph of f(x) = x/ ∫ √ / from Example 8... = + x dx ∫ ( ) / = + x · · ( ) / = .units . − ≈ ( ) A graph of f is given in Figure 8...

Example 8.. Finding arc length Find the arc length of f(x) = x ln x from x = to x = . −

S This funcon was chosen speciﬁcally because the resulng integral can be evaluated exactly. We begin by ﬁnding f ′(x) = x/ /x. The − arc length is

x L = + dx √ − x ∫ ( ) x = + + dx − x ∫ √ x = + + dx x ∫ √ x = + dx √ x ∫ ( ) 163

x = + dx y x ∫ ( ) x = + ln x ( ) .5 = + ln ≐ . units. A graph of f is given in Figure 8..; the poron of the curve measured in this x problem is in bold. ..... The previous examples found the arc length exactly through careful choice Figure 8..: A graph of f(x) = x ln x − of the funcons. In general, exact answers are much more diﬃcult to come by from Example 8... and numerical approximaons are necessary.

Example 8.. Approximang arc length numerically Find the length of the sine curve from x = to x = π.

S This is somewhat of a mathemacal curiosity; in Example .. we found the area under one “hump” of the sine curve is square units; now we are measuring its arc length. The setup is straighorward: f(x) = sin x and f ′(x) = cos x. Thus x √ + cos x

π √ L = + cos x dx. π/ / π/ ∫ √ √ π/ / This integral cannot be evaluated in terms of elementary funcons so we will ap- π √ proximate it with Simpson’s Method with n = . Figure .. gives √ + cos x √ evaluated at evenly spaced points in [, π]. Simpson’s Rule then states that Figure 8..: A table of values of π π + cos x dx ≐ − √ + / + () + / + √ y = √ + cos x ∫ · ( ) to evaluate a deﬁnite integral in √ = .. √ √ √ ■Example 8... Using a computer with n = the approximaon is L .; our approxi- ≈ maon with n = is quite good. 164

Surface Area of Solids of Revoluon

We have already seen how a curve y = f(x) on [a, b] can be revolved around an axis to form a solid. Instead of compung its volume, we now consider its surface area. We begin as we have in the previous secons: we paron the interval [a, b] th with n subintervals, where the i subinterval is [xi, xi+]. On each subinterval, we can approximate the curve y = f(x) with a straight line that connects f(xi) y and f(xi+) as shown in Figure 8..(a). Revolving this line segment about the x- axis creates part of a cone (called a frustum of a cone) as shown in Figure 8..(b). The surface area of a frustum of a cone is

. π length average of the two radii R and r. · · x a x i xi+1 b The length is given by L; we use the material just covered by arc length to .... state that (a) L ≐ + f ′(ci) ∆xi

th √ for some ci in the i subinterval. The radii are just the funcon evaluated at the endpoints of the interval. That is,

R = f(xi+) and r = f(xi).

Thus the surface area of this sample frustum of the cone is approximately

( ) + ( ) f xi f xi+ π + f (ci) ∆xi. ′ √ (b) Since f is a connuous funcon, the Intermediate Value Theorem states there f(xi) + f(xi+) is some di in [xi, xi+] such that f(di) = ; we can use this to rewrite Figure 8..: Establishing the formula for surface area. the above equaon as πf(di) + f ′(ci) ∆xi.

Summing over all the subintervals√ we get the total surface area to be approximately n Surface Area ≐ πf(di) + f ′(ci) ∆xi, i= ∑ √ which is a Riemann Sum. Taking the limit as the subinterval lengths go to zero gives us the exact surface area, given in the following theorem. 165

Theorem 8.. Surface Area of a Solid of Revoluon

Let f be diﬀerenable on [a, b], where f ′ is also connuous on [a, b]. . The surface area of the solid formed by revolving the graph of y = f(x), where f(x) , about the x-axis is ≥ b Surface Area = π f(x) + f ′(x) dx. a ∫ √ . The surface area of the solid formed by revolving the graph of y = f(x) about the y-axis, where a, b , is ≥ b Surface Area = π x + f ′(x) dx. a ∫ √

(When revolving y = f(x) about the y-axis, the radii of the resulng frustum are xi and xi+; their average value is simply the midpoint of the interval. In the limit, this midpoint is just x. This gives the second part of Theorem 8...)

Example 8.. Finding surface area of a solid of revoluon Find the surface area of the solid formed by revolving y = sin x on [, π] around the x-axis, as shown in Figure 8...

S The setup is relavely straighorward. Using Theorem .., we have the surface area SA is: π SA = π sin x + cos x dx ∫ √ π = π sinh− (cos x) + cos x + cos x − ( √ ) = π √ + − (( sinh ≐ . units . .( ) The integraon step above is nontrivial, ulizing an integraon method called Trigonometric Substuon. It is interesng to see that the surface area of a solid, whose shape is deﬁned by a trigonometric funcon, involves both a square root and an inverse hyper- Figure 8..: Revolving y = sin x on [, π] bolic trigonometric funcon. about the x-axis.

Example 8.. Finding surface area of a solid of revoluon Find the surface area of the solid formed by revolving the curve y = x on [, ] about the x-axis and the y-axis. 166

S About the x-axis: the integral is straighorward to setup:

SA = π x + (x) dx. ∫ √ Like the integral in Example 8.., this requires Trigonometric Substuon.

π = (x + x) + x sinh− (x) − π ( √ ) = √ sinh− − (a) ( ) ≐ . units. The solid formed by revolving y = x around the x-axis is graphed in Figure 8.. (a). About the y-axis: since we are revolving around the y-axis, the “radius” of the solid is not f(x) but rather x. Thus the integral to compute the surface area is:

SA = π x + (x) dx. ∫ √ This integral can be solved using substuon. Set u = + x; the new bounds are u = to u = . We then have

(b) π = √u du ∫ Figure 8..: The solids used in Example π / 8... = u π = √ − ( ) ≐ . units. The solid formed by revolving y = x about the y-axis is graphed in Figure 8.. (b).

Our ﬁnal example is a famous mathemacal “paradox.”

Example 8.. The surface area and volume of Gabriel’s Horn Consider the solid formed by revolving y = /x about the x-axis on [, ). Find the volume and surface area of this solid. (This shape, as graphed in Figure 8∞.., is known as “Gabriel’s Horn” since it looks like a very long horn that only a supernatural person, such as an angel, could play.)

Figure 8..: A graph of Gabriel’s Horn. 167

S To compute the volume it is natural to use the Disk Method. We have:

∞ V = π dx x ∫ b = lim π dx b x →∞ ∫ b = lim π − b x →∞ ( )

= lim π b − b →∞ ( ) = π units.

Gabriel’s Horn has a finite volume of π cubic units. Since we have already seen that regions with infinite length can have a finite area, this is not too difficult to accept. We now consider its surface area. The integral is straighorward to setup:

∞ SA = π + /x dx. x ∫ √ Integrang this expression is not trivial. We can, however, compare it to other improper integrals. Since < + /x on [, ), we can state that ∞ √ ∞ ∞ π dx < π + /x dx. x x ∫ ∫ √ By Key Idea .., the improper integral on the le diverges. Since the integral on the right is larger, we conclude it also diverges, meaning Gabriel’s Horn has infinite surface area. Hence the “paradox”: we can fill Gabriel’s Horn with a finite amount of paint, but since it has infinite surface area, we can never paint it. Somehow this paradox is striking when we think about it in terms of volume and area. However, we have seen a similar paradox before, as referenced above. We know that the area under the curve y = /x on [, ) is finite, yet ∞ the shape has an infinite perimeter. Strange things can occur when we deal with the infinite.

A standard equaon from physics is “Work = force distance”, when the × force applied is constant. In the next secon we learn how to compute work when the force applied is variable.

NOTE No real paradox here at all. When you paint the area under the curve, the paint thickness is the same for all x. Infinite area ⇒ infinite volume. When you fill the horn with paint, its thickness decreases in two dimensions:

"small x small = very small" 168 Use the Five Step Procedure in each problem

Again, it is an excellent exercise to translate each of Exercises the examples of Apex above into the 5 step Method. Terms and Concepts . f(x) = √ x on [ , ]. (Note: this describes the top − − half of a circle with radius .) . T/F: The integral formula for compung Arc Length was . f(x) = x/ on [ , ]. (Note: this describes the top found by ﬁrst approximang arc length with straight line − − half of an ellipse with a major axis of length and a minor segments. √ axis of length .) . T/F: The integral formula for compung Arc Length includes a square–root, meaning the integraon is probably easy. . f(x) = on [, ]. x

. f(x) = sec x on [ π/, π/]. Problems − In Exercises – , use Simpson’s Rule, with n = , to ap- In Exercises – , ﬁnd the arc length of the funcon on the proximate the arc length of the funcon on the given interval. given interval. Note: these are the same problems as in Exercises –.

. f(x) = x on [, ]. . f(x) = x on [, ].

. f(x) = √x on [ , ]. . f(x) = x on [, ]. −

/ / . f(x) = √x on [, ]. (Note: f ′(x) is not defined at x = .) . f(x) = x x on [, ]. − . f(x) = ln x on [, e]. . f(x) = x + on [, ]. x . f(x) = √ x on [ , ]. (Note: f ′(x) is not defined at − − the endpoints.) . f(x) = x/ √x on [, ]. − . f(x) = x/ on [ , ]. (Note: f ′(x) is not defined − − at the endpoints.) . f(x) = cosh x on [ ln , ln ]. √ − x x . f(x) = on [, ]. . f(x) = e + e− on [, ln ]. x ( ) . f(x) = sec x on [ π/, π/]. − . f(x) = x + on [., ]. x In Exercises – , find the surface area of the described . f(x) = ln sin x on [π/, π/]. solid of revoluon.

( ) . The solid formed by revolving y = x on [, ] about the . f(x) = ln cos x on [, π/]. x-axis. ( ) In Exercises – , set up the integral to compute the arc . The solid formed by revolving y = x on [, ] about the length of the funcon on the given interval. Do not evaluate y-axis. the integral. = [ , ] . The solid formed by revolving y x on about the . f(x) = x on [, ]. x-axis.

. f(x) = x on [, ]. . The solid formed by revolving y = √x on [, ] about the x-axis. . f(x) = √x on [, ]. . The sphere formed by revolving y = √ x on [ , ] − − . f(x) = ln x on [, e]. about the x-axis. 169

Solutions 8.4

1. T 2. F √ 3. 2 4. 6 5. 4/3 6. 6 7. 109/2 8. 3/2 9. 12/5 10. 79953333/400000 ≐ 199.883 √ 11. − ln(2 − 3) _≐ 1.31696 − 12. sinh 1 1 ∫ √ 1 2 13. 0 1 + 4x dx ∫ √ 1 18 14. 0 1 + 100x dx ∫ √ 1 + 1 15. 0 1 4x dx ∫ √ e + 1 16. 1 1 x2 dx ∫ √ 1 + x2 17. −1 1 1−x2 dx ∫ √ 3 + x2 18. −3 1 81−9x2 dx ∫ √ 2 + 1 19. 1 1 x4 dx ∫ √ π/4 + 2 2 20. −π/4 1 sec x tan x dx 21. 1.4790 22. 1.8377 23. Simpson’s Rule fails, as it requires one to divide by 0. However, recognize the answer should be the same as for y = x2; why? 24. 2.1300 25. Simpson’s Rule fails. 26. Simpson’s Rule fails. 27. 1.4058 28. 1.7625 ∫ √ √ 1 29. 2π 0 2x 5 dx = 2π 5 ∫ √ √ 1 2 − 30. 2π 0 x 1 + 4x dx = π/6(5 5 1) ∫ √ √ 1 3 4 − 31. 2π 0 x 1 + 9x dx = π/27(10 10 1) ∫ √ √ √ 1 − 32. 2π 0 x 1 + 1/(4x) dx = π/6(5 5 1) ∫ √ √ 1 − 2 − 2 33. 2π 0 1 x 1 + x/(1 x ) dx = 4π 170

8.5 Water Tank Problems There are many work problem in physics and engineering, often involving advanced scientific concepts.

One easy to understand application is calculating the work required to fill a water tank. It is more complicated in some ways than the previous applications of integrations in that two asymptotic equality approximations are often required. Recall:

Theorem a ≈ A, b ≈ B ⟺ a·A ≈ b·B

The most basic work problem in grade 10 physics is moving a through a straight line distance d by a constant force F is W = F·d

d F

A water tank problem is more complicated. An infinitesimal element of volume dV meter^3 is lifted by a force dF = dm·g where g = 9.08 meter/sec2 is the acceleration of gravity acting through a vertical distance d meters. The infinitesimal work done in lifting the element is then dW = df·h = ( dm·g) h = (δ dV) g h = δ g h dV.

Kg For water, δ = 1000 m3 .

dW = dF · h = (dm g) h = (δ dV) g h = δ g h dV

h 171

Example A water tank is made by rotating the curve y = x2 meters, 0 ≤ x ≤ 2 meters about the Y-axis. How much work is required to fill the tank from a source 3 meters below the bottom of the tank?

dy ( ) x x, y

◼ h◼

X 2

Note that in these problems there are two approximations:

-3 radius of element ≈ x h ≈ y + 3 and we must use Theorem A ≈ B, C ≈ D ⟹ 2 dV ≈ π x dy = πydy AC ≈ BD h ≈ y - (-3) = y + 3 dW ≈ δ g h dV = δg(y + 3)(πydy) = δ g πy2 + 3 y dy

y3 4 4 2 3 y2 ⇒ W = δ g π y + 2 y dy = δ g π + Joules ∫ 0 3 2 0 172

Use the Five Step Procedure in each problem

Exercises Use the grade 10 formula dW = δ g h dV in each problem.

#1. A water tank is 5 meters high and has a square cross section 2 meters on a side. a. How much work is required to fill the tank from a well 10 meters below the bottom of the tank? b. How much work is required to empty the full tank to a height 10 meters above the top of the tank?

#2. A water tank is made by rotating the curve y = x meters, 0 ≤ x ≤ 2 meters about the Y-axis. a. How much work is required to fill the tank from the bottom of the tank? b. How much work is required to empty the full tank over the top of the tank?

#3. A water tank is 5 meters high and has a circular cross section of radius 1 meter. a. How much work is required to fill a half full tank from a well 10 meters below the bottom of the tank? b. How much work is required to half empty a full tank to a height 10 meters above the top of the tank?

#4. A water tank is 10 meters long and has an equilateral triangle cross section of side 2 meters, point down. a. How much work is required to fill the tank from the bottom of the tank? b. How much work is required to empty the full tank out over the top of the tank?

#5. A water tank lying on its side is 5 meters long and has a circular cross section of radius 1 meter. a. How much work is required to fill the tank from a well 4 meters below the bottom of the tank? b. How much work is required to empty a full tank to a height 4 meters above the top of the tank?

#6. A water tank is a sphere of radius 2 meters. a. How much work is required to fill the tank from a well 5 meters below the bottom of the tank? b. How much work is required to empty a full tank to a height 5 meters above the top of the tank?

#7. How much work is required to fill the tank of Exercise 23 of section 2 in this chapter with water pumped in at the bottom of the tank? 173 0 Solutions

#1a.

dy y

X -1 1 2 3 4

5 h = y + 10 W = 4δ g (y + 10) dy ∫ 0 dV = 22 dy = 4 dy = 250 δ g dW = δ g h dV = 4δ g (y +10)dy

#4a.

3 1

y x

X -1 1

3 h = y W = 20 δ g ∫ y2 dy 3 0 By similar triangles = 20 δ g x y = 1 3 y = 3 x dV = 2x · 10 dy = 20x dy = 20 y dy 3 20 2 dW = δ g h dV = δ g y dy 3 174

8.6 Application to Economics. Present and Future Value A dollar at a future time is not worth as much as a dollar now, because a dollar now can be invested at the going interest rate and so will be worth more than one dollar at that future time.

We will assume in the calculations of this section that whether borrowing or investing, money is worth a constant going interest rate r over the period of time considered. The interest rate can include a component that compensates for inflation.

Discrete Investment or Income, Lump Sum Recall the exponential lump sum growth formula r t F = P e .

rt F = P e future value of a present sum Solving for P : P F e-rt = present value of a future sum

Continuous Investments or Income Suppose an investment is made or an income is received continu- $ investment /income stream ously according or an I = I(t) year . In the following derivation, we assume that I(t) is a continuous function and therefore approximately constant on the interval of length dt; then the amount of money received/invested during that interval is approximately 'rate × time' ≈ I(t) dt. . _ .. T- - t Time 0 t t+dt T

Total Money The amount invested/received on the interval dt : dM ≈ I(t) dt. So the total money invested/received is TI t M = ∫ ( ) dt. 0 Present Value The present value of the money on the interval dt at time 0, by the discrete formula dP ≈ dM e-r t = I(t) e-rt dt So the total present value is T I(t) e- r t P = ∫ 0 dt

Future Value The future value of the money on the interval dt at time T, by the discrete formula (1): er(T-t) e r(T-t) dF ≈ dM = I(t) dt So the total future value is T I(t) er(T-t) F = ∫ 0 dt 175

Use the Five Step Procedure in each problem

Exercises

For each of #1 to 7 find: a. The total money (received or invested) b. The present value of this money. c. The future value of this money. #0. You invest $10000 lump sum now at an interest rate of 10% for 40 years. #1. You invest $1000 per year at an interest rate of 10% for 40 years. #2. You invest $100 t per year at an interest rate of 10% for 40 years. #3. You invest $10 t2 per year at an interest rate of 10% for 40 years

For each of $4 to 7 also determine how much should you pay for the annuity. Money is worth 10% interest. #4. You purchase an annuity which pays 10000 $/year for 20 years. #5. You purchase an annuity which pays 10000 $/year forever. #6. You purchase an annuity which pays 1000 t $/year for 20 years. #7. You purchase an annuity which pays 1000 t $/year forever.

Answer each of the following two questions. You wish to give you child $100,000 in 20 years for its education. You invest at 10%/year return. #8. What lump sum should you invest now? #9. At what constant yearly rate should you invest over the next 20 years?

10. Explain why interest implies inflation and why some of the world's major religions have at times forbidden interest. 176

Answers

#1 present value = $10000 total money invested = $10000 ( ) future value = 10 000 e.1 40 = $545982

total money

40  1000 ⅆt 0

present value

40 -0.1 t  1000 ⅇ ⅆt 0

future value

40 0.1 (40-t)  1000 ⅇ ⅆt 0

#2. I[t] = 100 t #4. total money 20 -0.1 t  10 000 ⅇ ⅆt total money 0 40  100 t ⅆt 0

present value

40 -0.1 t  100 t ⅇ ⅆt 0 #5. present value, what you should pay:

+∞ future value -  10 000 0.1 t ⅆt 0 e 40 0.1 (40-t)  100 t ⅇ ⅆt = $100000 0 177

Chapter 9 Generalized Functions

Generalized functions and their special generalized calculus give correct answers in some important areas of application where ordinary calculus fails. One area is writing piecewise defined functions in a form where the Fundamental Theorem of Calculus applies. Another is a way of representing impulse spikes or the density of a point particle in function form.

9.1 Piecewise Defined Functions In applications, functions are often defined by piecing together simpler continuous functions.

Example The Unit Step Function, U(t). This piecewise defined function is very important in applications. For example, it can represent ‘turning on’ an electric potential of 1 Volt exactly at time t = 0.

U (t)

T -3 -2 -1 1 2 3 4

Sectionally Continuous Functions For both reasons of application and mathematics we modify acceptable types of piecewise defined functions.

Definition A sectionally continuous function is a function which is

1. continuous on the real line except at finitely many points in each subinterval. - + 2. at each point of discontinuity xi, f(xi ) and f(xi ) are finite real numbers. + f(xi ) is the limit from the right as x approaches xi - f(xi ) is the limit from the left as x approaches xi 3. f(x) is undefined at each point of discontinuity. 4. f(-∞) = 0 (this is not unduly restrictive because in many application, a quantity is 0 initially or ‘turned on’ at some finite time after the creation). 178

The Unit Step Function is not a sectionally continuous function according to our definition because it is defined at the jump point t = 0. A modification of that function, the Heaviside Generalized Function, is a sectionally continuous function. In applications, it is physically meaningful not to define the function at t = 0 because one does not know and it does not matter whether a quantity is turned on exactly at or just before or just after t = 0!

H (t)

T -3 -2 -1 1 2 3 4

Example The Heaviside function multiplied by 3 and translated 2 to the right is 3 H(t - 2). It can represent ‘turning on’ an electric potential of 3 Volts at t = 2.

3 H (t - 2)

T 1 2 3 4 5 6

Example A sawtooth function. The open circle convention at discontinuities is normally assumed when working with generalized functions and will not usually be shown explicitly.

Sawtooth (x)

X -1 1 2 3 4

Example A more complicated sectionally continuous function is 0, x < 0 c(x) = sin x , 0 ⩽ x < 2π. x - 8, x > 2π

c (x)

X -2 2 4 6 8 179

Example This jump function j(x) is not a sectionally continuous function because it has infinitely many discontinuities on the finite interval 0 < x < 1. (At endpoints of continuous subintervals, j(x) is not defined.) Nevertheless, because the discontinuities occur on an appropriate sequence of points, the theory we will develop in the next sections will also apply to this example.

y = j(x) 1

X 1

Exercises

1. Write the equation for each sectionally continuous function using piecewise notation. a.

Y 1.0 0.5 X -2 2 4 6 8 10 -0.5 -1.0 b.

Y 2.5 2.0 1.5 1.0 0.5 X 1 2 3 4 5 6 c.

X -2 -1 1 2 3 4 5 d.

Y 2 1 X -4 -2 2 4 6 8 10 180

2. Graph each. Which are sectionally continuous functions? Graph each. Modify the ones which are not sectionally continuous, when possible, so they are sectionally continuous.

0, x < -1 a. f(x) =  2, x ⩾ -1

0, x ≤ 0 b. g(x) =  x, x > 0

0, x < 0 c. h(x) = sin (π x), 0 ⩽ x ⩽ 2 0, x > 2

0, x < 0 1, 0 < x < 2 d. k(x) = -1, 2 < x < 4 0, x > 4

x e. l(x) = e

x e. m(x) = e- .

3. Invent four sectionally continuous functions. Graph each.

4. The Square Tooth Function (repeats indefinitely to the right).

ST(x) 1 ⋯ X -2 2 4 6 8

Graph each of the following. State if not a sectional continuous function. a. y = 3ST(x) b. y = -2ST(x) c. y = ST(-x) d. y = ST(2x) x e. y = sin x ST (π ) f. y = 2 ST(x-2) 181

Solutions

0 x < -1 1 - x - 1 < x < 0 0 0 < x < 1 1d. y = 4 -(x - 3)2 1 < x < 5 0 5 > x < 6 1 6 < x < 8 0 x > 8

2a. not sectionally continuous.

f(x) 4

-3 -2 -1 1 2 3 X

-1

2c. sectionally continuous

h(x) 1.0

0.5

X -2 -1 1 2 3

-0.5

-1.0

2e. sectionally continuous ex

-4 -3 -2 -1 1 2 X 182

9.2 Generalized Function Calculus Graphically Generalized functions and their generalized integrals and derivatives give correct answers in many areas of applications where ordinary derivatives or integrals fail.

bf x Generalized Integrals First we note that, by the definition of definite integral, ∫a ( ) dx, does not exist if f(x) is undefined somewhere on the interval a ≤ x ≤ b.

y = c

a b X

b N * * This is because in the definition of integral, ∑i f (xi ) dx ≈> f(x) dx, if for some choice of dx, f(xi ) is =1 ∫a undefined, then the entire sum is undefined (even an infinitesimal defect poisons the whole sum). However, since the area under a point, no matter what's its value, is 0, we generalize the definition of definite integral to ignore isolated points where f(x) is undefined. Such an integral is called a generalized integral of f. bf x So in this graphical example we agree that ∫a ( ) dx = (b-a)·c. Before we continue, recall the following for reference.

Fact:

x f t If f is continuous (except at isolated points) ⇒ ∫a ( ) dt is smooth (differentiable). Proof: DIY

Integrating Derivatives We would like, because of the Fundamental Theorem, the following to be true:

x f t ∫ a ' ( ) dt = f(x) - f(a),

"the integral of the derivative of a function is the function minus its initial value". For sectionally functions, f(-∞) = 0. Then

x f t f x ∫-∞ ' ( ) dt = ( ) An even better Fundamental Theorem!

This holds if f(x) is continuous using generalized integration. However, we will see that if f is not continuous, x f t we will also have to generalize the idea of derivative for ∫ - ∞ ' ( ) dt = f(x) to hold. 183

Example f(x) is continuous. Everything is fine. Verify mentally, using the area interpretation.

x = ( ) f '(t) y f x  f '(t) dt = f(x) -∞ 3 3 3 2 2 2 1 1 1 X -2 -1 1 2 -2 -1 1 2 T -2 -1 1 2 X

The Heaviside Function Let us recall the basic sectionally continuous function, H(x). 0 t < 0 H(t) =  . 1 t > 0 It is undefined at t = 0. It is useful as a multiplier in applications for ‘turning on’ a quantity at time t = 0.

H(t) 3

T -2 -1 1 2 3 -1

Example Let us try differentiating and then integrating H(x). Everything is not fine.

x H(t) H' (t)  H ' (t) dt ≠ H(x) -∞ 3 3 3 2 2 2 1 1 1 T T - - X 2 1 1 2 3 -2 -1 1 2 3 -2 -1 1 2 3 -1 -1 -1

The problem in this example is taking the ordinary derivative of H(t) at a discontinuity. Somehow we lose information at x = 0. We will have to search for a correct 'generalized derivative'. (The generalized integral was used in integrating H '(t)). Let us search for a more useful derivative by considering a modified Heaviside function and its derivative.

x H (t) H H H ϵ '  ϵ' (t) dt = ϵ(x) ϵ(t) -∞

1/ϵ

1 1 ⋮ ⋮

T X ϵ ϵ · · · T ϵ

In more detail: H 184 ϵ'(t)

1/ϵ

0 t < O 1 Hϵ(t) =  O < t < ϵ .◼ ' ϵ ⋮ ⋮ 0 t > ϵ ◼

● ● ● T ϵ

H H H The above graph shows the approximation ϵ(t) to (t) with ϵ an infinitesimal. The area under ϵ '(t) is 1, just what we need. We call the resulting idea, the (Dirac) delta function, named after its inventor. It is written δ(x). The delta function is not a function in the usual sense because at the 'interesting' place x = 0, it is undefined. It is called a distribution and is a generalized function .

Definition The Dirac Delta Function δ(x) is defined analytically by: 1. δ(x) = 0, x ≠ 0 x  t generalized derivative 2. -∞ δ( ) dt = H(x). Conversely the of H(x) is δ(x). Equivalently d H x x dx ( ) = δ ( ) .

Graphically δ(x) is shown by an upward unit arrow with its tail at (0, 0). H The above ϵ '(t) is a perfectly good definition for δ(x) because for every infinitesimal ϵ > 0 the ‘spike’ fits between 0 and ϵ. Also the area of the spike, 1, is independent of ϵ. However, most mathematicians prefer a definition free of infinitesimals. We will use the following definition although some proofs are easier if we H use ϵ ' (t),

The Dirac Delta Function δ(x), a traditional definition 1. δ(x) = 0, x ≠ 0 x d 2. δ(t) dt = H(x) or conversely H(x) = δ(x) ∫-∞ dx

Graphically δ(x) is shown as as a unit arrow with its tail at (0, 0).

δ(x) 3

-2 -1 1 2 3 X

-1

We now with generalized functions, generalized derivatives and generalized integrals have the desired x H (t) derivative of H(x) which satisfies ∫-∞ ' dt = H(x)! 185

x H(t) H ' (t)  H ' (t) dt = H(x) -∞

1 1 1

- - T T X 2 1 1 2 3 -2 -1 1 2 3 -2 -1 1 2 3

sThe method of finding the 'correct' generalized derivative of any generalized function is now clear. Its generalized derivative is just its ordinary derivative plus appropriated shifted delta functions multiplied by the magnitude of the function's jump at each discontinuity. We say that b δ(x-a) is a ‘delta function of strength b at x = a'.

Example Let f(t) be the function graphed below.

f(x)

X -2 2 4 6 8

-2

Its generalized derivative is then:

f ' (x) 4

X -2 2 4 6 8

-2

-4

You can verify, by visual generalized integration of f '(t) from -∞ to x, that the result is f(x).

The Rule for Graphical Generalized Differentiation: 1. Draw the ordinary derivative of the function. 2. Add an arrow with length equal to the jump at each discontinuity. 186

Summary Functions which are 0 for x large and negative and which are continuous and ordinary differentiable except at isolated points and are bounded are called generalized functions. The generalized function calculus consists of the generalized functions together with their generalized derivatives and integrals.

A generalized function with generalized differentiation and generalized integration satisfies the Fundamental Theorem of Calculus! x  f ' (t) dt = f(x) -∞

Exercises Do all quickly. 1. Graph the generalized derivative f '(x). Verify its generalized integral is f(x).

f(x)

X -2 2 4 6

f ' (x) 3 3

2 2

1 1

X X -2 2 4 6 -2 2 4 6 -1 -1

-2 -2

2. Graph the generalized derivative g '(x). Verify its generalized integral is g(x).

g(x)

X -2 2 4 6 8 -1

-2

-3

g ' (x) 3

X -2 2 4 6 8

-1

-2 a1 187

3. Invent your own h(x). Graph the generalized derivative h '(x). Verify its generalized integral is h(x).

h(x) 3 2 1 X -2 2 4 6 8 -1 -2

h ' (x) 3 3 2 2 1 1 X X -2 2 4 6 8 -2 2 4 6 8 -1 -1 -2 -2

d x f(t) 4. Generally 'integrating first and then differentiating' causes no problem: dx ∫a dt = f(x). bf t a. Prove this. Hint: use the Fundamental Theorem of Calculus, ∫a ( ) dt = F(b) - F(a). b. Verify this graphically for the Heaviside function H(x).

H ( x) 4 3 2 1 X -3 -2 -1 1 2 3 4 -1 -2

x  H(t) dt -∞ 4 3 2 1 X -3 -2 -1 1 2 3 4 -1 -2

d x  H(t) dt dx -∞ 4 3 2 1 X -3 -2 -1 1 2 3 4 -1 -2 188

9.3 Generalized Functions Analytically

Review The Heaviside Function H(t) 0 t < 0 H(t) =  1 t > 0 H(t) 3

T -2 -1 1 2 3

-1

Review The Dirac Delta Function δ(x): 1. δ(x) = 0, x ≠ 0 x 2. δ(t) dt = H(x) or conversely d H(x) = δ(x) ∫-∞ dx

Graphically δ(x) is shown by an upward unit arrow with its tail at x = 0 on the X-axis. δ(x) 3

X -2 -1 1 2 3

-1

Using the Heaviside Step function to write equations of generalized functions First we do this for one function segment. Y

y = f(x)

a b X

y = 'turn on f(x) at x = a' and then 'turn off f(x) at x = b' = f(x)H(x-a) - f(x)H(x-b).

For general generalized functions apply the above technique to each function segment. 189

Example

f(x) 4 x - 1 3

X -1 1 2 3 4 5

Its equation is f(x) = (x - 1)H(x-1) - (x - 1)H(x-3).

Generalized Calculus Properties of H and δ 1. δ(x) = 0, x ≠ 0 1. δ(x-a) = 0, x ≠ a x x 2. ∫-∞δ(t) dt = H(x) 2. ∫-∞δ(t - a) dt = H(x-a) 3. H '(x) = δ(x) 3. H '(x-a) = δ(x-a) 4. f(x) δ(x) = f(0) δ(x) 4. f(x) δ(x-a) = f(a) δ(x-a) 'Sifting Property'

The Sifting Property is useful in simplifying expressions involving the delta function.

Proof The only mystery is #4, the Sifting Property. f(x) δ(x) = f(0) δ(x) follows from #1, since the only value of f that 'counts' is at x = 0.

Y y = f(x)

Example Let us look at the previous example. f(x) = (x - 1)H(x-1) - (x - 1)H(x-3). By the Product Rule

f '(x) = 1 H(x-1) + (x - 1)δ(x-1) - 1 H(x-3) - (x - 1)δ(x-3) = H(x-1) - H(x-3) - 2δ(x-3) since by the Sifting Property: (x - 1)δ(x-1) = (1 - 1)δ(x-1) = 0 (x - 1)δ(x-3) = (3 - 1)δ(x-3) = 2δ(x-3) Note that this derivative does not involve a delta function at x = 1. This is because f is continuous there. The graph of the derivative is

f '(x )I

2 1 which looks just right. X -1 1 2 3 4 5 -1

-2 190

Interpretation of δ '(x) The delta function and its derivative are important in applications. Their geometric approximations are useful in understanding their properties.

Here are the approximations of H(x) and δ(x) again and δ '(x). You can verify them by starting at the top with differentiation or starting at the bottom with integration.

Hϵ(x)

X ϵ

δϵ(x)

1 ϵ ⋮

X ϵ

To find the derivative of δϵ(x) we will use another version of it in triangular form which also has area 1.

δϵ(x)

1 ϵ ⋮

X ϵ

δϵ'(x)

1 ϵ2

⋮

X ϵ

⋮ 191

Graph of δ ' (x) A double headed arrow of length 1.

δ '(x)

-2 -1 1 2 3

Physical Interpretations δ(t) is a strong kick forward, 'unit impulse' at t = 0. The area under it is 1. You can calculate that the effect of a force δ(t) when applied to a particle is to produce an instantaneous change in its velocity.

δ '(t) is a very strong kick forwards followed immediately by an equal strong kick backward at t = 0. The area under each spike is +∞ for every non-zero ϵ. We will see that the effect of the force δ ''( t) when applied to a particle is to produce an instantaneous change in its position with no net change of velocity.

Note The Dirac Delta function is a genuine hyperreal based function, not the hyperreal extension of a real function. Clearly we need a hyper-hyperreal calculus based on hyperinfinitesimals x smaller in size than any positive infinitesimal, in particular the ϵ we used in the pre-delta function. However, we can still get by thinking of the derivative as a slope and the integral as an area: there is no need here to develop a full hyper-hyperreal calculus.

Exercises Work some from 1 to 9. Marvel the advanced application appendix.

1. A. Write a formula for f(x) shown below in terms of H analytically. Find f(x) in terms of H and δ and simplify using the Sifting Property. Finally verify analytically that the Fundamental Theorem of Calculus holds. _ B. Repeat part A but this time do graphically. f(x)

X -2 2 4 6

C. Why doesn't f '(x) involve a delta function? 192

2. Repeat parts 1 A and 1 B for the function below.

y = g(x)

X π

3. Repeat parts 1 A and 1 B for the function below.

g(x)

X -2 2 4 6 8 -1

-2

-3

4. Verify each by graphing and/or analytically. x H(t) a. ∫-∞ dt = xH(x) x t H(s) 1 x2 b. ∫-∞∫-∞ ds dt = 2 H(x) 193

5. Verify each Sifting Property of δ '(x). a. f(x) δ '(x) = -f '(0)δ(x) b. f(x) δ '(x-a)= -f '(a)δ(x-a)

6. Graph each. a. δ(x) x δ(t) b. ∫-∞ dt Write a formula for this function. x t δ(s) c. ∫-∞∫-∞ ds dt Write a formula for this function.

7. Criticise the graphical representation of δ ' (x).

8. a. Draw an approximation δϵ'(x) for δ ''(x). b. How would you show the graph of δ ''(x)? c. What is the Sifting Property of δ ''(x)?

+∞δ(x) 9. Show that ∫-∞ dx = 1.

10. A particle with mass 1 initially at rest at x = 0. At t = 0, the particle is subject to the following forces: a. F = H(t) b. F = δ(t) c. F = δ '(t). Use Newton's Law, F = m a, to find the velocity and position as a function of time in each case. Graph 11. a. Find the antiderivatives of the functions f(x) and g(x) in exercises 1 and 2. x Notation: F(x) = ∫ - ∞ f(t) dt. Verify the Fundamental Theorem of Calculus for these functions. 194

Chapter 10 Differential Equations

10.1 First Order Separable Differential Equations

The laws of growth of a quantity - particularly in finance, the natural sciences and engineering - are often expressed by equations where derivatives of the quantity appear. In this chapter we will begin the study of simple first order and second order differential equations.

First order differential equations have the forms

dy dy dx = f(x, y) or F(x, y, dx ) = 0.

Their solutions have one arbitrary constant because in one way or another an indefinite integration is involved. The simple differential equation dy general solution x2 dx = x has the y = 2 + C. The constant C is found by requiring the solution to pass through a point. If the solution goes through x2 (2, 2), then C = 0 and y = 2 .

How does a differential equation determine a solution? Let us look at an example.

dy Example Consider the differential equation dx = x + y. Let us find the solution passing through the point (0, 2). Write the differential equation in its approximate differential form: Δy ≐ (x + y)Δx.

Start at (0, 1). Chose Δx = 1 (very large!). Find Δy. Get the point (0+Δx, 1+Δy). Repeat. (0, 1) ⟹ Δx = 1, Δy = (0 + 1)·1 ⟹ (1, 2) (1, 2) ⟹ Δx = 1, Δy = (1 + 2)·1 ⟹ (2, 5) (2, 5) and so on. See the blue data points. Not bad? It goes roughly in the right direction when com- pared to the exact solution, the black curve.

For a better solution take Δx = 0.1, say. Then you would get the red data. Much better. You could do this by hand in a few minutes with good mental arithmetic and concentration For a very good solution take Δx = 0.01. Then you would get the purple data. The data points barely peek out from the under exact solution on the right. You could do this with some simple programming. For a solution asymptotically equal to the exact solution, take Δx = dx, an infinitesimal, of course! This is how a first order differential determines a solution. 195

12 Exact Solution Δx = 0.01

10 Δx = 0.1

→↑Δy 6 Δx Δx = 1 4

0.5 1.0 1.5 2.0 2.5 X

dy = x + y ex The exact general solution of dx is y = C - x - 1. It is graphed for several values of C below. It is a homework problem to find this solution. The solution passing through (0, 1) is

y = 2ex - x - 1.

dy Solutions of = x + y dx

= C = C C = C = C = Y 12 — —

6 C =

X 0.0 0.5 1.0 1.5 2.0 2.5 196

First order differential equations, Variables Separable

Both mathematicians and students like exact analytic methods which work for a large number of differential equations. One type is the variables separable equations:

dy dx = f(x) g(y) dy g(y) = f(x) dx separating variables dy f x ∫ g(y) = ∫ ( )dx integrating The solution.

dy y Example Find the general solution of dx = x .

dy dx y = x separating variables dy dx ∫ y = ∫ x integrating ln y = ln x + ln C equivalent to C as the constant of integration. Why? ln y - ln x = ln C Property 1 of logs ln(y /x) = ln C y = Cx

x dy = Example Find the solution of the initial value problem  dx y . y(0) = 1 dy x dx = y

y dy = x dx separating variables

∫ y dy = ∫ x dx integrating y2 x2 C2 2 = 2 + 2 looking ahead y2 - x2 = C2 a hyperbola 12 - 02 = C2 initial condition C2 = 1 Y 3 2 1 y2 - x2 = 1 Solution: 0 X -1 -2 -3 -3 -2 -1 0 1 2 3

Since the bottom curve does not pass through the point (0, 1), the correct solution is:

Y 3.0 2.5 2.0

x2 Y 1.5 Solution: y = 1 + 1.0 0.5 0.0 -3 -2 -1 0 1 2 3 X 197

Sometimes separation of variables is not entirely obvious.

dy ex-y Example Find the general solution of dx = .

dy ex-y dx = x y = e e- property of exponents y x e dy = e dx separating variables y x ∫ e dy = ∫e dx integrating y x e = e + C x Note: y = ln e + ln C ex y = ln( + C) = x + D is wrong. Why?

Exercise 10.1.0 Find the exact solution of --dx = x + y. Hint: make the change of variable u = x + y. Exercise 10.1.1 Which of the following equations are separable?

(a) y′ = sin(ty) t y (b) y′ = e e

Exercise 10.1.3 Solve the initial value problem y = tn with y(0) = 1 and n 0. ′ ≥

Exercise 10.1.4 Solve y′ = lnt.

Exercise 10.1.5 Identify the constant solutions (if any) of y′ = t siny.

y Exercise 10.1.6 Identify the constant solutions (if any) of y′ = te .

Exercise 10.1.7 Solve y′ = t/y.

2 Exercise 10.1.8 Solve y′ = y 1. − 3 Exercise 10.1.9 Solve y′ = t/(y 5). You may leave your solution in implicit form: that is, you may stop once you have done the integration,− without solving for y.

1/3 Exercise 10.1.10 Find a non-constant solution of the initial value problem y′ = y , y(0) = 0, using separation of variables. Note that the constant function y(t) = 0 also solves the initial value problem. This shows that an initial value problem can have more than one solution. 198

10.1.2 y = arctant +C Solutions 10.1

tn+1 10.1.2 y = arctant +C 10.1.3 y = + 1 n + 1 tn+1 10.1.4 y = t lnt t +C 10.1.3 y = + 1 − n + 1 10.1.5 y = nπ, for any integer n. 10.1.4 y = t lnt t +C − 10.1.6 none 10.1.5 y = nπ, for any integer n.

10.1.7 y = t2 +C 10.1.6 none ± p 10.1.8 y = 1, y = (1 + Ae2t )/(1 Ae2t) 10.1.7 y = t2 +C ± − ± p 10.1.9 y4/4 5y = t2/2 +C 10.1.8 y = 1, y = (1 + Ae2t )/(1 Ae2t) − ± − 10.1.10 y = (2t/3)3/2 10.1.9 y4/4 5y = t2/2 +C − 10.1.10 y = (2t/3)3/2

10.2 First Order Homogeneous Linear Equations

A simple, but important and useful, type of separable equation is the ﬁrst order homogeneous linear equation:

Deﬁnition 10.1.2.1: First Order Homogeneous Linear Equation

A ﬁrst order homogeneous linear differential equation is one of the form y′ + p(t)y = 0 or equivalently y′ = p(t)y. −

“Homogeneous” refers to the zero on the right side of the equation, provided that y′ and y are on the left. “Linear” in this deﬁnition indicates that both y′ and y appear independently and explicitly; we don’t see y′ or y to any power greater than 1, or multiplied by each other (i.e. y′y). 199

Example 10.2.1: Linear Examples

The equation y′ = 2t(25 y) can be written y′ +2ty = 50t. This is linear, but not homogeneous. The − equation y′ = ky, or y′ ky = 0 is linear and homogeneous, with a particularly simple p(t) = k. 2 − − The equation y′ + y = 0 is homogeneous, but not linear.

Since ﬁrst order homogeneous linear equations are separable, we can solve them in the usual way:

y′ = p(t)y 1 − dy = p(t)dt y − Z Z ln y = P(t) +C | | y = eP(t) ± P t y = Ae ( ), where P(t) is an anti-derivative of p(t). As in previous examples, if we allow A = 0 we get the constant − solution y = 0.

Example 10.2.2 Solving an IVP Solve the initial value problem y′ + ycost = 0, subject to y(0) = 1/2 and y(2) = 1/2.

Solution. We start with P(t) = cost dt = sint, Z − − so the general solution to the differential equation is

sint y = Ae− .

To compute A we substitute: 1 = Ae sin0 = A, 2 − so the solutions is 1 y = e sint. 2 − For the second problem, 1 = Ae sin2 2 − 1 A = esin2 2 so the solution is 1 ♣ y = esin2e sint. 2 − 200

Differential Equations

Example 10.2.2

Solve the initial value problem ty′ + 3y = 0, y(1) = 2, assuming t > 0.

Solution. We write the equation in standard form: y′ + 3y/t = 0. Then 3 P(t) = dt = 3lnt − t − Z and 3lnt 3 y = Ae− = At− . 3 Substituting to ﬁnd A: 2 = A(1)− = A, so the solution is 3 y = 2t− . ♣

Exercises for 10.2

Find the general solution of each equation in the following exercises. y Exercise 10.2.1 y′ + 5y = 0 Exercise 10.2.3 y′ + = 0 1 +t2

2 Exercise 10.2.2 y′ 2y = 0 Exercise 10.2.4 y′ +t y = 0 − In the following exercises, solve the initial value problem.

t Exercise 10.2.5 y′ + y = 0, y(0) = 4 Exercise 10.2.10 y′ + ycos(e ) = 0, y(0) = 0

Exercise 10.2.6 y′ 3y = 0, y(1) = 2 Exercise 10.2.11 ty′ 2y = 0, y(1) = 4 − − − 2 Exercise 10.2.7 y′ + ysint = 0, y(π) = 1 Exercise 10.2.12 t y′ + y = 0, y(1) = 2, t > 0 − t 3 Exercise 10.2.8 y′ + ye = 0, y(0) = e Exercise 10.2.13 t y′ = 2y, y(1) = 1, t > 0

4 3 Exercise 10.2.9 y′ + y 1 +t = 0, y(0) = 0 Exercise 10.2.14 t y′ = 2y, y(1) = 0, t > 0 p Exercise 10.2.15 A function y(t) is a solution of y′ +ky = 0. Suppose that y(0) = 100 and y(2) = 4. Find k and ﬁnd y(t).

k 13 Exercise 10.2.16 A function y(t) is a solution of y′ + t y = 0. Suppose that y(0) = 1 and y(1) = e− . Find k and ﬁnd y(t).

Exercise 10.2.17 A bacterial culture grows at a rate proportional to its population. If the population is one million at t = 0 and 1.5 million at t = 1 hour, ﬁnd the population as a function of time.

Exercise 10.2.18 A radioactive element decays with a half-life of 6 years. If a mass of the element weighs ten pounds at t = 0, ﬁnd the amount of the element at time t. 201

Solutions 10.2

5t 10.2.1 y = Ae−

10.2.2 y = Ae2t

arctant 10.2.3 y = Ae−

t3/3 10.2.4 y = Ae−

t 10.2.5 y = 4e−

3t 3 10.2.6 y = 2e − − 10.2.7 y = e1+cost

2 et 10.2.8 y = e e−

10.2.9 y = 0

10.2.10 y = 0

10.2.11 y = 4t2 (1/t) 1 10.2.12 y = 2e − − 1 t 2 10.2.13 y = e − −

10.2.14 y = 0

t ln5 10.2.15 k = ln5, y = 100e−

10.2.16 k = 12/13, y = exp( 13t1/13) − −

10.2.17 y = 106et ln(3/2)

t ln(2)/6 10.2.18 y = 10e− 202

10.3 First Order Linear Equations

y' + p(t)y = f(t)

A common method for solving such a differential equation is by means of the same integrating factor you used in the previous section.*********************************************************************

Then

P(t) P(t) P(t) e− y′ + e− p(t)y = e− f (t) d (e P(t)y) = e P(t) f (t). dt − −

Integrating both sides gives

P(t) P(t) e− y = e− f (t)dt Z

P(t) P(t)

e f (t) dt y =e Z − t, the solution

Solve dy - y = Example dt 2 6

t t Integrating Factor = e∫-2 dt = e-2

dy e-2 t - y e-2 t = e-2 t dt 2 6 d [y e-2 t] e-2 t dt = 6 t t y e-2 = -3 e-2 + C y integrating both sides e-2 t y = -3 + C

Note: in the exercises, use the procedure of the example. Don't use the red formula. 203

NOTE Big people, when no one is watching, use Wolfram Exercises for 10.3 Alpha or other resources to evaluate difficult integrals.

In the following exercises, ﬁnd the general solution of the equation.

Exercise 10.3.1 y′ + 4y = 8

Exercise 10.3.2 y′ 2y = 6 10.3.11 a. Derive again the general − solution of

Exercise 10.3.3 y′ + ty = 5t y′ + p(t)y = 0 { y(0) = y t t o. Exercise 10.3.4 y′ + e y = 2e − b. Show that the above system can 2 be written in the one-line form preferred in Exercise 10.3.5 y′ y = t − some applications by

y' + p(t)y = y Exercise 10.3.6 2y′ + y = t o δ(t).

Exercise 10.3.7 ty′ 2y = 1/t, t > 0 Important Note − Show it is correct by solving it. When you

Exercise 10.3.8 ty′ + y = √t, t > 0 integrate, use generalized integration and the Sifting Property.

Exercise 10.3.9 y′ cost + ysint = 1, π /2 < t < π /2 The idea here is that all solutions are 0 at minus − infinity and get a displacement at t = 0 because of an impulse force. Exercise 10.3.10 y′ + ysect = tant, π /2 < t < π /2 −

Solutions 10.3

4t 10.3.1 y = Ae− + 2

10.3.2 y = Ae2t 3 − (1/2)t2 10.3.3 y = Ae− + 5

et 10.3.4 y = Ae− 2 − 10.3.5 y = Aet t2 2t 2 − − − t/2 10.3.6 y = Ae− + t 2 − 1 10.3.7 y = At2 − 3t c 2 10.3.8 y = + √t t 3 10.3.9 y = Acost + sint A t 10.3.10 y = + 1 sect + tant − sect + tant 204 Memory work in 1958: all the formulas Derivative Formulas d n n-1 (x ) = n x d x x dx d (e ) = e (ax) = axln a dx dx d 1 d 1 (ln x) = (logax) = dx x dx x ln a d d (sin x) = cos x (cos x) = -sin x dx dx d d (tan x) = sec2x (cot x) = - csc2x dx dx d d (sec x) = sec x tan x (csc x) = - csc x cot x dx dx d d (sin-1x) = 1 (cos-1x) = -1 dx dx 1 - x2 1- x2 d d (tan-1x) = 1 (cot-1x) = -1 dx 1 + x2 dx 1 + x2 d d (sec-1x) = 1 (csc-1x) = -1 dx dx x x2- 1 x x2- 1 d d (sinh x) = cosh x (cosh x) = sinh x dx dx d d (tanh x) = sech2x (coth x) = - csch2x dx dx d d (sech x) = -sech x tanh x (csch x) = - csch x coth x dx dx d d (sinh-1x) = 1 (cosh-1x) = 1 dx dx 1 + x2 x2- 1 d 1 d -1 (tanh-1x) = , |x| < 1 (coth-1x) = , |x| > 1 dx 1 - x2 dx 1- x2 d d (sech-1x) = -1 (csch-1x) = -1 dx dx x2 x 1 - x2 |x| - 1

The above list includes derivatives of all the basic elementary functions. Integral Formulas

un un+1 du ∫ du = + C ∫ u = ln |u| + C au ∫ eudu = eu+ C audu = + C ∫ ln a ∫ cos u du = sin u + C ∫ sin u du = -cos u + C ∫ sec2u du = tan u + C ∫ csc2u du = -cot u + C ∫ sec u tan u du = sec u + C ∫ csc u cot u du = -csc u + C ∫ tan u du = ln |sec u| + C ∫ cot u du = ln |sin u| + C ∫ sec u du = ln |sec u + tan u| + C ∫ csc u du = ln |csc u - cot u| + C ∫ cosh u du = sinh u + C ∫ sinh u du = cosh u + C ∫ sech2u du = tanh u + C ∫ csch2u du = -coth u + C ∫ sech u tanh u du = -sech u + C ∫ csch u coth u du = -csch u + C ∫ du = sin-1u + C ∫ du = cosh-1u + C 1- u2 u2- 1 ∫ du = sinh-1u + C ∫ du = tan-1u + C 1 + u2 1 + u2 ∫ du = tanh-1u + C, |u| < 1 ∫ du = coth-1u + C, |u| > 1 1 - u2 1 - u2 ∫ du = sec-1u + C ∫ du = sech-1u + C u u2- 1 u 1 - u2 ∫ du = csch-1u + C |u| u2+ 1 Integrals of basic inverse functions can always be done by integration by parts. Memory work in 2015: the boldface formulas