The Definite Integral

Chapter 5 – The Definite Integral

5.1 Estimating with Finite Sums

ex. A train is moving along a track at a steady rate of 75 miles per hour from 7:00 AM to 9:00 AM. What is the total distance traveled by the train?

v 퐷 = 푅푇 = 75(2) = 150 miles

Suppose we look at the graph of the function representing the trains speed at any given time. 75 . Note the shaded rectangle. What is its area? 75 2  150 . This is not a coincidence. The total distance and the area of the o rectangle are found by multiply the same numerical values. . This same connection can be made no matter how fast the train is going or how long or short the time interval was.

What if the train had a velocity v that changed or varied t as a function of time? 7 8 9 v

The graph to the left is no longer a horizontal line, so

the region under the graph (the area) would no longer be a rectangle.

Would the area of this region give us the total distance traveled over the time interval? . Newton, Leibniz, and others thought that it would, and that is why they were interested in a calculus for finding areas under curves. o They imagined the time interval being partitioned into many tiny subintervals . Each one so small that the velocity over that interval would literally be constant. . This would geometrically be the same as slicing the t a b region into narrow strips, each being a narrow rectangle. Gottfried Wilhelm Leibniz (b. 1646, d. 1716) was a German philosopher, mathematician, and logician who is probably most well known for having invented the differential and integral calculus (independently of Sir Isaac Newton). Ex. A particle starts at x = 0 and moves along the x-axis with velocity v(t) = t2 for time t  0. Where is the particle at t = 3?  The graph to the right shows the graph of the velocity function over the interval [0,3] v  The region is divided into subintervals of length

¼ and have curved tops that slope upward.  To find the area of the strip: o We will use the rectangle whose “top” passes through the midpoint of the interval.

o Note the two arrows . The area of the two “triangular” regions are very close to equal, so the area of the rectangle

would be a good approximation of the area of the strip. o The smaller the interval (base of the rectangle, the more accurate the area.

 So here is a table to help us out: t

1 1 1 1 3 3 Subinterval [0, 4 ] [ 4 , 2 ] [ 2 , 4 ] [ 4 ,1] 1 3 5 7 Midpoint (m) 8 8 8 8 2 1 9 25 49 Height (m ) 64 64 64 64 1 2 1 9 25 49 Area ( 4 m ) 256 256 256 256  Continuing this table, we would derive the area of each of the 12 subintervals to be: 1 9 25 49 81 121 169 225 289 361 441 529 2300 256  256  256  256  256  256  256  256  256  256  256  256  256  8.98  We can now conclude that the particle has moved approximate 9 units in 3 seconds.

Rectangular Approximation Method (RAM in the book):  The above example we used the Midpoint Rectangular Approximation Method (MRAM) o We determine the height using the midpoint of the subinterval.  We could use the left endpoint of the subinterval as the height of the rectangle (LRAM) o Suppose we used the same problem with only 6 intervals (the size 1 of the interval would be 2  x ). o This would result in the following sum: 2 1 1 2 1 2 1 3 2 1 2 1 5 2 1 0 ( 2 )  ( 2 ) ( 2 ) 1 ( 2 )  ( 2 ) ( 2 )  2 ( 2 )  ( 2 ) ( 2 )  6.875 1 2 3  We could use the right endpoint of the subinterval as the height of the rectangle (RRAM) o This would result in the following sum: 1 2 1 2 1 3 2 1 2 1 5 2 1 2 1 ( 2 ) ( 2 ) 1 ( 2 )  ( 2 ) ( 2 )  2 ( 2 )  ( 2 ) ( 2 )  (3) ( 2 ) 11.375  Using the MRAM, we would have the following result: 1 2 1 3 2 1 5 2 1 7 2 1 9 2 1 11 2 1 1 2 3 ( 4 ) ( 2 )  ( 4 ) ( 2 )  ( 4 ) ( 2 )  ( 4 ) ( 2 )  ( 4 ) ( 2 )  ( 4 ) ( 2 )  8.9375

1  Note that the result is not as accurate as when the interval was 4 .

 Note: You can determine the size of the interval using the following formula:

x2  x1 where xn represents the interval [x1, x2] and ni is the # of intervals ni  The following table shows how the number of intervals make a difference:

n LRAMn MRAMn RRAMn 6 6.875 8.9375 11.375 12 7.90625 8.984375 10.15625 24 8.4453125 8.99609375 9.5703125 48 8.720703125 8.999023438 9.283203125 100 8.86545 8.999775 9.13545 1000 8.9865045 8.99999775 9.0135045

 So as the number of intervals increase, LRAM, RRAM, and MRAM get closer and closer to the actual area under the curve.

5.2 Definite Integrals Riemann Sums:  In the previous section, we obtained the area as a sum of areas of strips o The terms in the sums were obtained by multiplying selected function values by the lengths of the intervals. o What happens as we make the intervals smaller and smaller and increase the number of intervals?

Def: Sigma Notation – enables us to express a large sum in a compact way. o We use the Greek symbol sigma 

n a  a  a  a  ...  a  a  k 1 2 3 n1 n k 1

o The Greek Letter  stands for sum in mathematics. The index k tells us where to begin (the number below sigma) and where to end (on top of sigma). . If the symbol ∞ appears above the  , then it indicates that terms go indefinitely. o We will be looking at Riemann Sums (pronounced “ree-mahn”) after George Friedrich Riemann. . LRAM, MRAM, and RRAM are all examples of Riemann sums  They are considered Riemann sums because of the way were constructed.

Homework: Pg. 269-270 QR #1-10, Ex. 1-6, 15, 16 Pg. 282-283 QR #1-10 5.2 Definite Integrals (con’t) Suppose we have a continuous function f(x) defined on a closed interval [a, b]. Look at the diagram below:

a b

o Note the graph can have negative values as well as positive values o Now we partition the interval [a, b] into n subintervals by choosing n–1 points, called x1, x2, x3, …, xn-1, between a and b using only the following condition: a < x1 < x2 < … < xn-1 < b o To make things more consistent, we change a to x0 and b to xn.

The set P is called the partition of [a, b] and is written P = [x0, x1, x2, …, xn]

o This partition P determines n closed subintervals. st nd . The 1 subinterval is [x0, x1], the 2 is [x1, x2], etc… th . The k subinterval would be [xk–1, xk]

 The length of this subinterval would be xk  xk  xk 1

. So the partition P divides [a, b] into n subintervals of length x1,x2 ,...,xn . o In each subinterval, we select some number; call it ck, for 1  k  n. o On each subinterval, we stand a vertical rectangle that reaches from the x-axis to touch the curve (above or below the x-axis) at (ck, f(ck))

Note: The ck value is an arbitrary point in the interval.

c1 c2 It is not a specific a value for the ck cn b interval.

푓(푐) ∙ ∆푥 is the area of the rectangle of the kth subinterval

o On each subinterval, we form the product f (ck )  xk . This product can be negative, positive, or zero (depending on f(ck)). Finally, we find the sum of these products: 푎푟푒푎푠 n Sn   f (ck )  xk k1

o This partition is a Riemann Sum for f on the interval [a, b]. o As the number of partitions increases (or becomes finer and finer), the sum of the areas of the rectangles would approach the area of the region. o So the lengths of the subintervals must approach zero. . We can make sure of this if we have the longest subinterval length (called the norm of the partition and denoted by P ) o This leads us to the definition of an Integral

Def: The Definite Integral as a Limit of Riemann Sums

Let f be a function defined on the closed interval [a, b]. For any partition P of [a, b], let the numbers ck be chosen arbitrarily in the subintervals [xk–1, xk]. If there exists a number I such that As the length of the n norm approaches 0, then all of the lim f (ck ) xk  I P 0  subintervals will k 1 approach 0 no matter how P and the ck’s are chosen, then f is integrable on [a, b] and f is the definite integral of f over [a, b].

o This works only if f is continuous.

Theorem: The Existence of Definite Integrals

All continuous functions are integrable. That is, if a function f is continuous on an interval [a, b], then its definite integral over [a, b] exists.

The Definite Integral of a Continuous Function on [a, b]

Let f be continuous on [a, b] and let [a, b] be partitioned into n subintervals of equal ba length x  n . Then the definite integral of f over [a, b] is given by:

lim f (ck )x n  k1 th where each ck is chosen arbitrarily in the k subinterval.

Terminology and Notation of Integration dy  Leibniz chose the notation for the derivative, dx because it had the advantage of retaining an Leibniz – identity as a “fraction” even though both numerator and denominator had approached zero The “father” of (tend to zero). Calculus o Though not really fractions, derivative can behave like fractions. dy dy du Ex. Chain rule:   dx du dx . Note the du would cancel normally with fractions.  Leibniz introduced the notation for the definite integral. o First in derivative notation: y dy lim  x0 x dx

where “” represents “difference” which is switched to a “d” for differential. o Then in his integral notation (the Greek Letters become Roman Letters)

n The Greek Letter Σ (sigma) b f ( c k )  x  f ( x ) dx corresponds to the Roman lim  a n   k 1 Letter S

. The difference, x goes to zero, becoming a differential dx. . The Greek “” has become an elongated Roman “S”  This is so that the integral can continue to be a “sum”.

Def: “The Integral from a to b of f of x with respect to x”

Upper limit of integration Also called the Upper BOUND b x is the variable of Integral f (x)dx integration a

Lower limit of integration Also called the Lower BOUND  The function is the integrand

o When you find the value of the integral, you have evaluated the integral. o The value of the definite integral of a function over a particular interval depends on the function and not the letter we choose to represent its independent variable. o You can find integrals with respect to other variables as well: b b b f (t)dt or f (u)du instead of f (x)dx a a a

4 Ex. The interval [–1,3] is partitioned into n subintervals of equal lengths x  n . Let mk denote the midpoint of the kth subinterval. Express the limit

2 lim (3(mk )  2mk  5)x n 

as an integral. So, you basically replace the following: 1. lim Σ with an integral → 3 2. All mk become x 2 2 3. ∆x becomes dx lim (3(mk )  2mk  5)x  (3x  2x  5)dx n   1 4. The interval become the upper and lower bounds

Definite Integral and Area  If an integrable function y  f (x) is non-negative throughout an interval [a,b], then each

non-zero term f (ck )xk is the area of the rectangle reaching from the x-axis up to the curve y  f (x) .

o The Riemann sum  f (ck )xk , which is the sum of the areas of these rectangles, gives an estimate for the area of the region between the curve and the x-axis from a to b.

Def: Area Under a Curve (as a Definite Integral)

If y  f (x) is non-negative and integrable over a closed interval [a, b]., then the area under the curve y = f(x) from a to b is the integral of f from a to b.

b A  f (x)dx a a b

2 Ex. Evaluate the integral: 4  x 2 dx What does this function look like?? Top half of a circle 2 f (x)  4  x2 f (x)  4  x 2 is a function whose graph is a semicircle of radius 2. y  4  x2

We can now use ½ the area of a circle to find the integral: y 2  4  x2

2 2 1 2 1 2 x  y  4 Area = 2 r  2 (2)  2 It is the top half of the Therefore circle with C(0,0) and r = 2 2 4  x 2 dx  2 2

o If the curve is below the x-axis, then you can determine the area between the curve and the x-axis. It will be a negative value. To make it positive, simply multiply by –1. . Or another way of looking at it: b A   f (x)dx when f(x)  0 a

o If the curve is both above and below the x-axis, then the value of the integral is the area above the x-axis minus the area below.

b f (x)dx  (area above the x-axis) – (area below the x-axis) a

 Ex. Given sin xdx  2 , find: (use the graph to the left) 0

2  1. sin xdx 5. 2sin xdx  0 Extending the graph, you get Green hump – Since the integral is another hump below x-axis. a limit, the coefficient can move Therefore = 2 out front. Therefore 2x2 = 4 2  2 2. sin xdx 6. sin(x  2)dx 0 2 One hump above (+2) and one Note the bounds. If you plug them into x hump below (2). Put together is  2, you get 0 and π. So this is the hump 0. shifted 2 units right. Answer is 2     3. 2 sin xdx 7. sin udu 0  x= splits the hump in half. Note the original graph has the entire graph for this problem. One hump above (+2) and Answer is 1 one hump below (2). Answer is 0   4. (2  sin x)dx 8. cos xdx 0 0 Note the red curve (this is 2 + sin x). Think about the relationship of the sin and cos The hump is still 2 over a rectangle curves. Sketch it….you will see it is 0. that is 2 x π. So it is 2 + 2휋

Th: The Integral of a Constant

If f(x) = c, where c is a constant, on the interval [a,b], then b b f ( x )dx  cdx  c (b  a ) z a

Ex. A train moves along a track at a constant rate of 75 mph from 7:00 AM to 9:00AM. Express the total distance traveled as an integral. Evaluate the integral using the theorem above. The bounds are the times. Since the train is moving at a constant 75, the speed function would be f(x) = 75. 9 Distance = 75dt  75(9  7)  75(2) 150 7 Integrals on TI 84+ o Using the MATH key you can find a definite integral. o The 9th function down is fnInt which stands for function integral. o The syntax for this:

fnInt(f(x), x, a, b)

where f(x) is the function, x is x, a is the lower bound, and b is upper bound

ex. Find the following integrals using the calculator

5 2 1 4  x 2 1. x sin xdx 2. dx 3. e dx 1 0 1 x 2 0

2.052759…. 3.1415926…. (π) 0.8662269……

Homework: Day 1: Pg 282-283 #1-6, 7-12 (without calculator), 15-17,19,23,25,27 Day 2: Pg 282-283 #29 -46 5.3 Definite Integrals and Antiderivatives

Properties of Definite Integrals: b a 1. Order of Integration:  f (x)dx   f (x)dx Switch the bounds, change the sign of the integral!! a b a 2. Zero:  f (x)dx  0 If the bounds are the same, then the integral = 0 a b b 3. Constant Multiple:  kf (x)dx  k f (x)dx for any value k a a b b Constant multipliers can be moved out front of the   f (x)dx   f (x)dx integral a a b b b 4. Sum and Difference: [ f (x)  g(x)]dx   f (x)dx   g(x)dx a a a Integral of a sum/difference is the sum/difference of the integrals (integrate each term) b c c 5. Additivity: f (x)dx  f (x)dx  f (x)dx    If two integrals have the same integrand (function) are added a b a together with a common bound as the upper on one and lower on the other, the result is an integral with unique upper/lower 6. Max-min Inequality: bounds. (Common bounds cancel out) If max f and min f are the maximum and minimum value of f on [a,b], then b The value of an integral will be between the lowest value the min f  (b  a)  f (x)dx  max f  (b  a) function is on the interval times the length of the interval and  the highest value of the function times the length of the interval. a You can use this to justify where an integral falls between b b 7. Domination: f (x)  g(x) on [a, b]   f (x)dx   g(x)dx a a b Since the integral is the area under the curve. If one curve f (x)  0 on [a, b]  f (x)dx  0 is above the other, then the area under the higher curve is  greater than the area under the lower curve. (DUH! – lol) a

ex. Suppose: 1 4 1  f (x)dx  5  f (x)dx  2  h(x)dx  7 1 1 1

Find each of the following integrals, if possible:: Sum/Difference Property 1 4 1 Constant Multiplier Property (a)  f (x)dx (b)  f (x)dx (c) [2 f (x)  3h(x)]dx 4 1 1 Bounds switched: 2 This is the Additivity Property. Take the first two integrals and put together. 2 푓(푥)푑푥 + 3 ℎ(푥)푑푥 = 2(5) + 3(7) = 31 Therefore, this is 5 + (2) = 3 1 2 4 (d)  f (x)dx (e)  h(x)dx (f) [ f (x)  h(x)]dx 0 2 1 A popular mistake on this one is that Same as (d) in theory. If you Using the sum/difference property to people see half the interval, therefore double/triple/… the interval, you don’t separate the integral. half the value. WRONG!!! You have double/triple/… the value

no knowledge about the function. Problem is we can’t do ℎ(푥)푑푥 Cannot Be Determined (CBD) ∫ Cannot Be Determined (CBD) Cannot Be Determined (CBD) The Average Value of a Function

 The average of n numbers is the sum of the numbers divided by n  To find the average value of a function, we use the following definition:

Def: The Average (Mean) Value

If f is integrable on [a,b], its average (mean) value on [a,b] is The rectangle has the same area as the shaded 1 b region under the curve av( f )  f (x)dx  b  a a

ex. Find the average value of f (x)  4  x 2 on [0,3]. Does f ever equal this value in the interval?

1 b av( f )   f (x)dx b  a a You can use fnInt() on 3 your calculator (for 1 2   (4  x )dx now) 3  0 0 1   3 3 1

Now to see if f ever is equal to the average value: f (x)  4  x 2 1  4  x 2 Since 3 is in the x 2  3 interval [0,3], the x   3 answer is yes

The Mean Value Theorem for Definite Integrals

 The example above shows that the average value is achieved in the interval of a continuous function.  This concept is known as the Mean Value Theorem for Definite Integrals

Theorem: The Mean Value Theorem for Definite Integrals

This is the same idea as the Mean If f is continuous on [a, b], then at some point c on [a, b], Value Theorem from derivatives, except with integrals.

b Proof of this can be done once the next 1 section is done (The Fundamental f (c)  f (x)dx Theorem of Calculus)  b  a a

Connecting Derivatives and Integrals  The derivative with respect to x of the integral of f from a to x is simply f  Or in other words: x In essence, this relates derivatives and integrals as inverse d functions to each other. f (t)dt  f (x)  dx a  This means that the integral is an antiderivative of f  This leads to: o If F is an antiderivative of f, then x  f (t)dt  F(x)  C where C is a constant a

Setting x = a, we get: a  f (t)dt  F(a)  C a 0  F(a)  C C  F(a)

Substituting back into the previous equation: x f (t)dt  F(x)  F(a)  a

Where F(x) is any antiderivative of f.

Some Antiderivatives: Below is a table of Derivatives and corresponding Antiderivative:

Derivative Antiderivative d xn1 x n  nx n1 is antiderivative of xn dx n 1 d sin x  cos x sin x is antiderivative of cos x dx d cos x  sin x –cos x is antiderivative of sin x dx d 1 1 ln x  ln x is the antiderivative of dx x x

and so on…

Homework: Day 1: Pg. 290-291 Q.R. 1-10, Exercises #1-6, 7, 8, 10, 13, 15, 19, 20, 21, 24, 25, 28 Day 2: Pg. 290-291 Exercises #9, 12, 14, 18, 22, 23, 26, 27, 29, 31

5.4 Fundamental Theorem of Calculus

. Newton and Leibniz came up with a connection between differentiation and integration. o The helped start the scientific revolution o This is also considered the most important computational discovery in the history of mathematics.

Th: The Fundamental Theorem of Calculus (Part I)

If f is continuous on [a, b], then the function x F(x)   f (t)dt a Note the bounds of the integral: has a derivative at every point x in [a, b], and The upper bound is just an x dF d x  f (t)dt  f (x) The lower bound is a constant (any dx dx  constant, does not matter what a value)

Proof: Using the formal definition of the derivative: dF F(x  h)  F(x)  lim dx h0 h Using a Property of Integration: xh x f (t)dt  f (t)dt c b c   f (x)dx  f (x)dx  f (x)dx a a     lim a a b h0 h c b c xh f (x)dx  f (x)dx  f (x)dx f (t)dt     a a b  lim x h0 h 1 xh  This is the Average Value Theorem  lim f (t)dt b h0 h  1  x  av( f )   f (x)dx b a a

Therefore: with a= x and b = x+h dF 1 xh  lim f (t)dt h0  dx h x  lim f (c) h0 where c is between x and x + h

As h 0, then f(c)  f(x)

dF So………  f (x) dx

Ex. Using the Fundamental Theorem, find:

d x d x 1 costdt dt 1.  2.  2 dx  dx 0 1t 1 cos x 1 + 푥

Have you noticed that it is simply the function with the t changed to an x???

Yes, for these, it is that simple!

x2 Ex. If y  costdt , then find dy Ok, this one is different. Note the upper bound is 푥 , not x. You have to do it  dx differently when the bound is not just an x 1

You use the Chain Rule: Substitute the 푥in for t, but then multiply by the derivative of 푥.

푑푦 = cos 푥 (2푥) = 2푥 cos 푥 푑푥

Th: The Fundamental Theorem of Calculus (Part 2)

If f is continuous at every point [a,b], and if F is any antiderivative of f on [a, b], then

b  f (x)dx  F(b)  F(a) a

This part of the Fundamental Theorem is also called the Integral Evaluation Theorem.

3 Ex. Evaluate: (x3 1)dx using an antiderivative. 1

The antiderivative of 푓(푥) = 푥 + 1 is 퐹(푥) = + 푥. By the theorem above: 퐹(3) = + 3 = 퐹(−1) = + (−1) = − So ∫ 푥 + 1 푑푥 = − = = 24 Ex. Find the area of the region between the curve y  4  x 2 on the interval [0,3] and the x-axis.

The region is shaded in the diagram to the left. The area under the curve is defined: 푥 3 0 4−푥 푑푥 = 4푥 − = 4(3) − − 4(0) − 3 3 3 4−푥 푑푥 = (12−9) − (0−0) =3

How to find Total Area: To find the area between the graph of y = f(x) and the x-axis over [a, b]: Always graph the function 1. partition [a,b] with the zeros of f (find where f crosses x–axis) on your calculator if you 2. integrate f over each subinterval can. You want to see if the curve crosses the xaxis. 3. add the absolute values of each of the integrals

ex. Find the area of the region between y  x 2 16 and the x–axis on the interval [0,5] Note the graph is below the xaxis from 0 to 4. So you do the integrals in each section: ∫ 푥 − 16 푑푥 = − 16푥 = − 64 − [0 − 0] = − →

∫ 푥 − 16 푑푥 = − 16푥 = −16(5) − −64 = −80− +64 ∫ 푥 − 16 푑푥 = −16= →

Answer = = 47

ex. Find the area of the region bounded by the x–axis and y = xcos 2x on the interval [–3,3].

You can’t do the antiderivative for this problem (well…yet). This is a calculator question. You need to find where the zeroes To find the Total Area are. using a calculator: 푥 cos(2푥) = 0 when x = 0 and cos2푥 = 0 → 2푥 = ⋯, ,− , , ,… |푓(푥)| 푑푥 푥=⋯, ,− , , ,…

푥 cos2푥 푑푥 = −0.999017 푥 cos 2푥 푑푥 = 0.142699

푥cos2푥 푑푥 = 1.57079 푥cos2푥 푑푥 = −1.57079

푥 cos 2푥 푑푥 = − 0.142699 푥 cos 2푥 푑푥 = 0.999017 Adding all the absolute values of the integrals: 5.425

Homework: Pg 302-303 Q.R. #1-9odd, Exercises #1-9odd, 12, 15, 16, 25-27, 29-35 5.5 Trapezoidal Rule

. Some functions do not have antiderivatives that elementary functions (simple to find) . Some examples are: cos x 3 x 1 x x cos x sin x 2 x . You can’t use the Fundamental Theorem to evaluate integrals of these functions. . So you have to approximate them using LRAM, RRAM, or MRAM (MRAM is the most accurate). . There is another way of approximating these integrals. It is called the Trapezoidal Rule o We will divide the interval into n trapezoids of equal height, see picture to the right: o We assume that f is continuous and positive on the interval; [a,b]. This results in the integral: 1 2 3 b f (x)dx  a . This integral represents the area of the region bounded by f and the x-axis from x=a to x=b. ba . First we partition the interval into n equal subintervals each of which is x  n , such that

a  x0  x1  x2  ...  xn  b . We then form a trapezoid for each subinterval. (x1, f(x1)) . The area of this first trapezoid is: Area of Trapezoid:

 f (x0 )  f (x1 ) b  a  b2 b1  b2    A  h  2  n  2 (x0, f(x0))

b . So the sum of the areas of all the trapezoids will be the 1 h approximation of the area under the curve. This will x0 x1

look like this:

 f (x0 )  f (x1 ) b  a   f (x1 )  f (x2 ) b  a   f (xn1 )  f (xn ) b  a  Area          ...      2  n   2  n   2  n 

 b  a  f (x0 )  f (x1 ) f (x1 )  f (x2 ) f (xn1 )  f (xn )      ...    n  2 2 2    b  a    f (x0 )  f (x1 )  f (x1 )  f (x2 )  ...  f (xn1 )  f (xn )   2n   b  a    f (x0 )  2 f (x1 )  2 f (x2 )  ...  2 f (xn1 )  f (xn )   2n 

. This is the Trapezoidal Rule.

Theorem: The Trapezoidal Rule

b Let f be continuous on [a, b]. The Trapezoidal Rule for approximating  f (x)dx is given a by: b  b  a  f (x)dx  f (x )  2 f (x )  2 f (x ) ... 2 f (x )  f (x )   0 1 2 n1 n  a  2n  or b  b  a  f (x)dx  y  2y  2y ... 2y  y   0 1 2 n1 n  a  2n 

And as n → ∞, then the right-side of the equation approaches the left-side.

. Note the coefficients in the Trapezoidal Rule has the following pattern: 1 2 2 2 … 2 2 1

ex. Use the Trapezoidal Rule to approximate  sin xdx Compare the results for n = 4 and n = 8. 0 n = 4: ∆푥 = = = 푏 − 푎 휋 휋 3휋 휋 √2 √2 sin 푥푑푥 ≈ 푓(0) +2푓 +2푓 +2푓 + 푓(휋) = 0+2 +2(1) +2 +0 2푛 4 2 4 8 2 2 휋 = 2√2 +2 8 sin 푥푑푥 ≈ 1.89611

n = 8: ∆푥 = = = 푏 − 푎 휋 휋 3휋 휋 5휋 3휋 7휋 sin 푥푑푥 ≈ 푓(0) +2푓 +2푓 +2푓 +2푓 +2푓 +2푓 +2푓 + 푓(휋) 2푛 8 4 8 2 8 4 8 휋 sin 푥푑푥 ≈ [0 + 2(0.38268) + 2(0.7071) + 2(0.92388) + 2(1) + 2(0.92388) + 2(0.7071) + 2(0.38268) + 0] 16 휋 sin 푥푑푥 ≈ [0 + 2(0.38268) + 2(0.7071) + 2(0.92388) + 2(1) + 2(0.92388) + 2(0.7071) + 2(0.38268) + 0] 16 휋 sin 푥 푑푥 ≈ (10.05464) = 1.97422 16 2  x 2 dx Using n = 4 1 n = 4: ∆ 푥 = = = 푏 − 푎 5 3 7 1 25 9 49 푥2푑푥 ≈ 푓(1) +2푓 +2푓 +2푓 + 푓(2) = 1+2 +2 +2 +4 2푛 4 2 4 8 16 4 16

Error: (See next page) 1 푦 = 푥 푏 − 푎 = (18.75) = 2.34375 |퐸| ≤ ℎ 푀 푦 = 2푥 8 12 푦 =2 1 1 |퐸 | ≤ (2) 12 4 1 1 |퐸 | ≤ (2) 12 16

|퐸| ≤ 0.010416666666 Actual Value = 2.333333 Simpson’s Rule Notice the coefficients for Simpson’s Rule – b To approximate  f (x)dx, use 1 4 2 4 …….. 2 4 1 a h S  y  4y  2y  4y ... 2y  4y  y  3 0 1 2 3 n2 n1 n ba Where [a, b] is partitioned into an even number of n of subintervals of equal length h  n .

2 Ex. Use Simpson’s Rule with n = 4 to approximate 5x4dx 0

n = 4: ℎ = = = ℎ 1 3 1 1 81 5푥 푑푥 ≈ 푓(0) +4푓 +2푓(1) +4푓 + 푓(2) = 5∙1+4 5∙ +2(5∙1) +4 5∙ + 5∙16 3 2 2 6 16 16

1 = (192.5) = 32.08333 Error: 6 푦 = 5푥 푏 − 푎 |퐸| ≤ ℎ 푀() 푦 = 20푥 180 2−0 1 푦 = 60푥 |퐸 | ≤ (120) 푦 = 120푥 180 2 1 1 푦 = 120 |퐸 | ≤ (120) 90 16

|퐸| ≤ 0.08333333 Actual Value = 32

Error Analysis o If you use the Trapezoidal Rule or Simpson’s rule, then there will be a slight error since thee are approximations. . This error can be predicted from the concavity of the curve . Suppose the exact value can not be found but a value within 0.001 is ok.  How do you know if our approximation is with 0.001? o The Trapezoidal Rule approximates curves with straight lines . Then error depends on how curvy the graph is. o The error in trapezoidal approximations (ET) satisfies the following inequality: b  a E  h2M T 12 f " where h is the length of the subinterval and Mf” is the maximum value of f" on [a, b]

o The error in Simpson approximations (ES) depends on h and the fourth derivative. It satisfies the following inequality:

b  a 4 E  h M ( 4) S 180 f (4) where h is the length of the subinterval and M ( 4) is the maximum value of f f on [a, b], provided it is continuous

Homework: Pg. 312-313 #1-7, 10 (Part a and b – use calculator), 13, 16, 18, 19

5.5 Trapezoidal Rule and Simpson’s Rule (alt)

 Earlier we found the area under the curve y = f(x) using Riemann Sums (LRAM, MRAM, and RRAM) o LRAM and RRAM were the least accurate unless there the number of subintervals was very high making Δx very small. o This is an appropriate method, but tends to be a bit inaccurate.  Instead of using rectangles, suppose we connect the points on the curve at each subinterval (see below)

(x4, f(x4))

(x3, f(x3))

(x2, f(x2))

b5 b4

b3 (x1, f(x1))

(x0, f(x0)) b1

x0 h x1 h x2 h x3 h x4

a b  Each region is now a trapezoid. The “gap” is no longer as large was the gaps formed using rectangle. We also do not need to find midpoints is we were using MRAM. 1  The area under the curve would now be the areas of each of the trapezoids ( A  h(B  b) , B and b are 2 the lengths of the bases, the parallel sides, and h is the distance between the bases).  This leads to the following expression:

1 1 1 1 A  2 h(b1  b2 )  2 h(b2  b3 )  2 h(b3  b4 )  2 h(b4  b5 )  1 h[b  b  b  b  b  b  b  b ] 2 1 2 2 3 3 4 4 5 1  2 h[b1  2b2  2b3  2b4  b5 ] 1  2 h[ f (x0 )  2 f (x1 )  2 f (x 2 )  2 f (x3 )  f (x 4 )]

 Would this pattern happen if we had more subintervals?? o Yes -This is called the Trapezoidal Rule

Theorem: The Trapezoidal Rule b Let f be continuous on [a, b]. The Trapezoidal Rule for approximating  f (x)dx is given a by: b  b  a  f (x)dx  f (x )  2 f (x )  2 f (x ) ... 2 f (x )  f (x )   0 1 2 n1 n  a  2n  or b  b  a  f (x)dx  y  2y  2y ... 2y  y   0 1 2 n1 n  a  2n 

And as n → ∞, then the right-side of the equation approaches the left-side.

. Note the coefficients in the Trapezoidal Rule has the following pattern: 1 2 2 2 … 2 2 1

ex. Use the Trapezoidal Rule to approximate  sin xdx Compare the results for n = 4 and n = 8. 0 n = 4: ∆푥 = = = 푏 − 푎 휋 휋 3휋 휋 √2 √2 휋 sin 푥푑푥 ≈ 푓(0) +2푓 +2푓 +2푓 + 푓(휋) = 0+2 +2(1) +2 +0 = 2√2 +2 2푛 4 2 4 8 2 2 8 sin 푥푑푥 ≈ 1.89611

2  x 2 dx Using n = 4 1 n = 4: ∆푥 = = = 푏 − 푎 5 3 7 1 25 9 49 푥2푑푥 ≈ 푓(1) +2푓 +2푓 +2푓 + 푓(2) = 1+2 +2 +2 +4 2푛 4 2 4 8 16 4 16

Error: (See next page) 1 = (18.75) = 2.34375 푏 − 푎 푦 = 푥 8 |퐸| ≤ ℎ 푀 12 푦 = 2푥 1 1 |퐸 | ≤ (2) 푦 =2 12 4 1 1 |퐸 | ≤ (2) 12 16

2 Ex. Use Simpson’s Rule with n = 4 to approximate 5x4dx 0 n = 4: ℎ = = = ℎ 1 3 1 1 81 5푥푑푥 ≈ 푓(0) +4푓 +2푓(1) +4푓 + 푓(2) = 5∙1+4 5∙ +2(5∙1) +4 5∙ + 5∙16 3 2 2 6 16 16

1 = (192.5) = 32.08333 6 Error: 푦 = 5푥 푏 − 푎 |퐸| ≤ ℎ 푀() 푦 = 20푥 180 2−0 1 푦 = 60푥 |퐸 | ≤ (120) 푦 = 120푥 180 2 1 1 푦 = 120 |퐸 | ≤ (120) 90 16

|퐸| ≤ 0.08333333 Actual Value = 32

o The error in Simpson approximations (ES) depends on h and the fourth derivative. It satisfies the following inequality:

b  a 4 E  h M ( 4) S 180 f (4) where h is the length of the subinterval and M ( 4) is the maximum value of f f on [a, b], provided it is continuous o

Homework: Pg. 312-313 #1-7, 10 (Part a and b – use calculator), 13, 16, 18, 19

Alternate Notes for

Section 5.2 5.2 Definite Integrals (con’t) – Short Version Suppose we have a continuous function f(x) defined on a closed interval [a, b]. Look at the diagram below:

a b

The set P is called the partition of [a, b] and is written P = [x0, x1, x2, …, xn]

o This partition P determines n closed subintervals. st nd . The 1 subinterval is [x0, x1], the 2 is [x1, x2], etc… th . The k subinterval would be [xk–1, xk]

 The length of this subinterval would be xk  xk  xk 1

Note: The ck value is an arbitrary point in the interval. c c 1 2 It is not a specific a ck cn b value for the interval. 푓(푐) ∙ ∆푥 is the area of the rectangle of the kth subinterval

o On each subinterval, we form the product f (ck )  xk . This product can be negative, positive, or zero (depending on f(ck)). Finally, we find the sum of these products: n 푎푟푒푎푠 Sn   f (ck )  xk k1

Def: The Definite Integral as a Limit of Riemann Sums

Let f be a function defined on the closed interval [a, b]. For any partition P of [a, b], let the numbers ck be chosen arbitrarily in the subintervals [xk–1, xk]. If there exists a number I such that As the length of the n norm approaches 0, lim f (ck )  xk  I then all of the P 0  subintervals will k 1 approach 0 no matter how P and the ck’s are chosen, then f is integrable on [a, b] and I is the definite integral of f over [a, b].

o This works only if f is continuous.

Theorem: The Existence of Definite Integrals

All continuous functions are integrable. That is, if a function f is continuous on an interval [a, b], then its definite integral over [a, b] exists.

Def: “The Integral from a to b of f of x with respect to x”

Upper limit of integration The Greek Letter Σ (sigma) Also called the Upper BOUND corresponds to the Roman b Letter S x is the variable of Integral f (x)dx integration a The function is the Lower limit of integration integrand Also called the Lower BOUND

Definite Integral and Area  If an integrable function y  f (x) is non-negative throughout an interval [a,b], then each

non-zero term f (ck )xk is the area of the rectangle reaching from the x-axis up to the curve y  f (x) .

o The Riemann sum  f (ck )xk , which is the sum of the areas of these rectangles, gives an estimate for the area of the region between the curve and the x-axis from a to b.

Def: Area Under a Curve (as a Definite Integral)

If y  f (x) is non-negative and integrable over a closed interval [a,b]., then the area under the curve y = f(x) from a to b is the integral of f from a to b.

b A  f (x)dx a a b

What does this function look like?? Top half of a circle 2 Ex. Evaluate the integral: 4  x 2 dx 2 f (x)  4  x 2 f (x)  4  x 2 is a function whose graph is a semicircle of radius 2. y  4  x 2

2 2 We can now use ½ the area of a circle to find the integral: y  4  x 2 2 1 2 1 2 x  y  4 Area = 2 r  2 (2)  2 It is the top half of the Therefore circle with C(0,0) and r = 2 2 4  x 2 dx  2 2

o If the curve is both above and below the x-axis, then the value of the integral is the area above the x-axis minus the area below.

b f (x)dx  (area above the x-axis) – (area below the x-axis) a

 Ex. Given sin xdx  2 , find: (use the graph to the left) 0

Extending the graph, you get 2 another hump below x-axis. Green hump – Since the integral is 1. sin xdx Therefore = 2   a limit, the coefficient can move 5. 2sin xdx out front. Therefore 2x2 = 4 0 2 One hump above (+2) and one 2. sin xdx 0 hump below (2). Put together is  2 0. 6. sin(x  2)dx 2 Note the bounds. If you plug them into x   2, you get 0 and π. So this is the hump 2 3. sin xdx shifted 2 units right. Answer is 2    0 x= splits the hump in half. 7. sin udu Answer is 1  Note the original graph has the entire graph  for this problem. One hump above (+2) and 4. (2  sin x)dx one hump below (2). Answer is 0 0  8. cos xdx Note the red curve (this is 2 + sin x). 0 The hump is still 2 over a rectangle Think about the relationship of the sin and cos that is 2 x π. So it is 2 + 2휋 curves. Sketch it….you will see it is 0.

Th: The Integral of a Constant

If f(x) = c, where c is a constant, on the interval [a,b], then b b f ( x )dx  cdx  c (b  a ) z a

Integrals on TI 84+ o Using the MATH key you can find a definite integral. o The 9th function down is fnInt which stands for function integral. o The syntax for this:

fnInt(f(x), x, a, b)

where f(x) is the function, x is x, a is the lower bound, and b is upper bound

ex. Find the following integrals using the calculator

5 2 1 4  x 2 1. x sin xdx 2. dx 3. e dx 1 0 1 x 2 0

2.052759…. 3.1415926…. (π) 0.8662269……

Homework: Day 1: Pg 282-283 #7-12 (without calculator), 15-17,19,23,25,27 Day 2: Pg 282-283 #29-46

5.2 Definite Integrals (con’t) - condensed version

The Definite Integral of a Continuous Function on [a, b]

Let f be continuous on [a, b] and let [a, b] be partitioned into n subintervals of equal ba length x  n . Then the Definite Integral of f over [a, b] is given by:

lim f (ck )x n  k 1 th where each ck is chosen arbitrarily in the k subinterval.

Note: The ck value is an arbitrary point in the interval. c1 c2

It is not a specific a ck cn b value for the interval. 푓(푐) ∙ ∆푥 is the area of the rectangle of the kth subinterval

 Because the rectangles could be below the x-axis, the area of a rectangle could be negative.  You can estimate the integral by using a Riemann Sum (left, right, or midpoint).

But there is a way to find the exact value:

Def: “The Integral from a to b of f of x with respect to x” The Greek Letter Σ (sigma) corresponds to the Roman Upper limit of integration Letter S Also called the Upper BOUND b x is the variable of Integral f (x)dx integration a

Lower limit of integration Also called the Lower BOUND  The function is the integrand

Definite Integral and Area  If an integrable function y  f (x) is non-negative throughout an interval [a,b], then each

non-zero term f (ck )xk is the area of the rectangle reaching from the x-axis up to the curve y  f (x) .

o The Riemann sum  f (ck )xk , which is the sum of the areas of these rectangles, gives an estimate for the area of the region between the curve and the x-axis from a to b.

Def: Area Under a Curve (as a Definite Integral)

If y  f (x) is non-negative and integrable over a closed interval [a,b]., then the area under the curve y = f(x) from a to b is the integral of f from a to b.

b A  f (x)dx a a b

What does this function look like?? Top half of a circle 2 Ex. Evaluate the integral: 4  x 2 dx 2

f (x)  4  x 2 is a function whose graph is a semicircle of radius 2.

We can now use ½ the area of a circle to find the integral: f (x)  4  x2

1 r 2  1 (2) 2  2 y  4  x2 Area = 2 2 y 2  4  x 2 Therefore 2 2 2 x  y  4 4  x 2 dx  2 2 It is the top half of the circle with C(0,0) and r = 2

o If the curve is both above and below the x-axis, then the value of the integral is the area above the x-axis minus the area below.

b f (x)dx  (area above the x-axis) – (area below the x-axis) a

 Ex. Given sin xdx  2 , find: (use the graph below) 0

 

2  1. sin xdx 5. 2sin xdx Green hump – Since the integral is  Extending the graph, you get 0 a limit, the coefficient can move another hump below x-axis. out front. Therefore 2x2 = 4 Therefore = 2

2  2 2. sin xdx 6. sin(x  2)dx Note the bounds. If you plug them into x 0 One hump above (+2) and one 2 hump below (2). Put together is  2, you get 0 and π. So this is the hump 0. shifted 2 units right. Answer is 2

 2  3. sin xdx x= splits the hump in half. 7. sin udu 0  Note the original graph has the entire graph Answer is 1 for this problem. One hump above (+2) and one hump below (2). Answer is 0   4. (2  sin x)dx 8. cos xdx 0 Note the red curve (this is 2 + sin x). 0 The hump is still 2 over a rectangle Think about the relationship of the sin and cos that is 2 x π. So it is 2 + 2휋 curves. Sketch it….you will see it is 0.

Th: The Integral of a Constant

If f(x) = c, where c is a constant, on the interval [a,b], then b b f ( x )dx  cdx  c (b  a ) a a

Examples:

Ex. A train moves along a track at a constant rate of 75 mph from 7:00 AM to 9:00AM. Express the total distance traveled as an integral. Evaluate the integral using the theorem above.

9 Distance = 75dt  75(9  7)  75(2) 150 7

Integrals on TI 84+ o Using the MATH key you can find a definite integral. o The 9th function down is fnInt which stands for function integral. o The syntax for this: old OS Newer OS

fnInt(f(x), x, a, b)

where f(x) is the function, x is x, a is the lower bound, and b is upper bound

ex. Find the following integrals using the calculator

5 2 1 4  x 2 1. x sin xdx 2. dx 3. e dx 1 0 1 x 2 0

2.052759…. 3.1415926…. (π) 0.8662269……

4. ∫ 푥 ln푥 푑푥

551.41057

5. Find the area between the x-axis and 푦=푥 +푥 −푥+8 on the interval [0, 4]

푥 + 푥 − 푥 + 8 푑푥

푥 푥 푥 + − + 8푥 4 3 2 256 64 16 + − +8(4) − [0+0−0+0] 4 3 2

64 64 264 + 64 328 64 + −8+32=88+ = = 3 3 3 3

Homework: Day 1: Pg 282-283 #1-6, 7-12 (without calculator), 15-17,19,23,25,27 Day 2: Pg 282-283 #29 -46