Quick Review Of Differentiation and Integration From Calculus I and II

In this review we will cover: Differentiation rules and examples Integration of inde nite and de nite integrals Improper integrals Integration by substitution and integration by parts Differentiation

Derivatives provide us with the rate of change of a function.

General differentiation rules you need to know:

- 1 d Power Rule: d n n Derivative of a constant: c = 0 x = nx dx dx d d Derivative of a constant times a function: cf (x) = c f (x) or simply (cf ) = cf dx dx keep the constantʹ as isʹ and multiply by the derivative of a function

Product Rule: = Quotient Rule: f f g − g f ( f g) f g + g f ( ) = 2 g ʹ ʹ g ʹ ⋅ ʹ ʹ⋅ ʹ⋅ ⋅ ⋅ dy dy du d Chain Rule: = or [ f ( g ( x ))] = f ( g( x )) g (x) dx du dx dx ʹ ʹ ⋅ Differentiate the outer function ⋅keeping the inner function the same and multiply by the derivatives of the inner function

Some function-speci c derivatives: 1 1 d x x d x x d d e = e a = a ln a ln x = loga x = dx dx dx x dx x ln a d d d 2 sinx = cosx cos x = −sinx tanx = sec x dx dx dx d 2 d d cotx = −csc x secx = secx tanx cscx = −cscx cotx dx dx dx Exercise 1: Find the derivatives of

4 a) f (x) = x sin 2x cosx b) g(x) = ln x 2 2 c) f (x)= 5 e x + x − + 3 x 3 d) h (x ) = ln (tan(e x)) Integration

a) Inde nite integral is a function (also called antiderivative), written as

f (x )dx = F ( x) + C where F '(x ) = f (x ) ∫ 4 For example: 3 = x + ∫ x dx 4 C

b b) De nite integral ∫ f (x) dx = F (b) − F (a) is the area under y = f (x) from x = a to x = b a where F '( x ) = f ( x ) and f is continous on [ a, b ]. 5 5 2 2 2 2 8 For example: 2 dx = − =− − − = ∫1 x x 1 5 ( 1 ) 5

Some basic integration formulas you may need: + 1 x n kdx = kx +C x ndx= + C, ( n ≠ -1 ) ∫ ∫ n +1 ∫ sin(x)dx = − cos(x)+C ∫ cos(x) dx= sin(x) + C 2 ∫ sec ( x) dx= tan(x)+C ∫ csc2( x)dx =− cot(x) +C ∫ sec(x) tan(x)dx=sec(x) + C ∫ csc(x)cot(x)dx =− csc(x) +C a x ex dx = ex + C a x dx = + C ∫ ∫ ln(a ) 1 1 1 dx = ln x +C = − 1 x + ∫ x 2 2 dx tan C ∫ a +x a ( a ) 1 − 1 dx= sin ( x) + C ∫ 2 = + + 1−x ∫ secx dx ln secx tanx C ∫ cscx dx = ln | cscx − cotx | + C

Exercise 2: Determine the following inde nite integrals.

Two rules you must know: ∫ c f (x) dx = c ∫ f (x) dx ∫ ( f ( x ) + g( x) ) dx = ∫ f (x ) dx + ∫ g ( x ) dx

2 − 3 4 −2 5 a) ∫ ( x 8 + x ) dx 3 2 b) ∫ ( x + 5) dx c) ∫ 3e−x dx 11 − 2 2x + 5x + x3 d) 3 dx ∫ x 7 4 e) 3x x − 3 dx ∫ ( x ) 1 f) dx ∫ 2x+ 3 2 −4 5 g) ∫ (sin( x) cos( x ))dx 3 + 5 6 h) ∫ ( x ) dx 1 i) + csc2 x dx ∫ ( sec x ) 3x − 2 2 x + 3 − x j) e e e ∫ 2 dx e x Exercise 3: Evaluate the following de nite integrals. 10 2 a) ( 50x − 6 x ) dx −∫1 16 − b) x x 3 dx ∫4 3 x 0.25 c) 2e t (e t+ 1 ) dt ∫0 Improper integrals

There are two kinds of improper integrals:

a) Integrals with in nite limits of integration For example: ∞ b −2 −2 1 − 2 b 1 −2 1 − 2 1 1 e t dt = lim e t dt = lim − e t = lim − e b− − e (0) = 0 + = ∫ ∫ 2 2 2 2 2 0 b ∞ 0 b ∞ 0 b ∞ ( )

b) Integrals with discontinuous→ integrands→ → For example: 1 1 1 1 1 1 1 1 dx = lim dx = lim − = lim − − − = ∞ ∫ 2 0 ∫ 2 0 0 1 ( ) 0 x b + b x b + x b b + ⎡ b ⎤ ⎣⎢ ⎦⎥ There are also improper→ integrals that→ require splitting→ and using two independent limits:

∞ 0 ∞ 0 b − 2 − 2 −2 − 2 −2 ∫ e t dt = ∫ e t dt +∫ e t dt = lim ∫ e t dt + lim ∫ e t dt −∞ −∞ 0 a ∞ a b ∞ 0 → → Exercise 4: Evaluate each integral. 1 1 a) dx ∫0 x ∞ 1 b) ∫ dx 2 x ∞ − 3 c) ∫ −3e x dx 0 Not all integrals are as simple as the examples we’ve seen so far. What if we have to evaluate the following integrals?

2 3 x 2 3 x x + 5dx 2 5 dx x sin(2x )dx ∫ ∫ (20− x ) ∫ For such integrals, we need to do the chain rule in reverse, also known as the substitution method. Integration by substitution

This method can be used whenever the integrand function is recognized to be of the form f ( g ( x )) g ( x ) The following guideline summarizes the steps involved in u-substitution method:⋅ ʹ Integration by Substitution Choose a new variable . Usually try choosing to be some inner function of the integrand whose derivative is also in the uintegrand. u Compute du. Replace all terms in the original integrand with expressions involving u and du. Evaluate the resulting integrand. (If you can’t, you may need to try a different or a different method of integration.) u u

Replace u with the corresponding expression in x.

For example, to evaluate x , choose = 2 + 2 since its derivative is 2 and we can write the 2 dx u x x ∫ x +2 entire expression in terms of u and du. 2 u = x + 2 du du= 2xdx = xdx 2 When substituting we should get the entire expression in terms of u and du. du x 2 1 1 1 dx = = du = ln u + C = ln(x2 + 2) +C ∫x2+2 ∫ u 2 ∫ u 2 2 Exercise 5: Evaluate the following integrals:

2 10 a) ∫ 2 x (x +4 ) dx b) 2 3 ∫ x x + 5 dx x c) 2 5 dx ∫ (20−x ) 2 2 3 d) ∫ x sin( x ) dx 3 e) xdx ∫ 2 4x +5 2 f) ∫ xe x dx Less apparent substitution Exercise 6: Evaluate

2 a) ∫ x x − 1 dx b) xdx ∫ 2x+1 x c) dx ∫ x−2 Substitution for definite integrals

Method 1: Use u-substitution to nd an antiderivative, (i.e. solve the inde nite integral) and then use the Fundamental Theorem to evaluate F ( b) − F (a) Method 2: Change the limits of integration as you change variables from x to u. This method tends to be more ef cient because you do not need to change the function of u back to function of x. Exercise 7: Evaluate 4 x a) dx ∫ 2 0 x + 1 1 2 b) ∫ ( x +1) 2x+ x dx 0 π2 4 c) cos x sinxdx ∫0 If the integration by substitution does not work and the integral consists of a product of functions (or contains inverse trig functions or logarithmic functions), try integration by parts. Integration by parts Integration by Parts ∫ udv= uv − ∫ vdu To choose , you may use LIATE which tells you the order of preference for choosing . Each letter represents ua type of function (L-logarithmic, I-inverse trig, A-arithmetic, T-trigonometric,u E-exponential). Find du by differentiating u and nd v by integrating dv. Substitute u, dv, v, du in the above formula. If the resulting integral in the second term cannot be evaluated, see if you need to do integration by parts again or if any other method can be used for that term.

For example, to evaluate xsin3xdx u =x dv=sin3xdx ∫ 1 du= dx v =− cos3x Applying the formula, 3

1 1 1 1 1 1 1 3 =− 3 − − 3 =− 3 + 3 =− 3 + 3 + C ∫xsin xdx 3xcos x ∫ 3cos xdx 3x cos x 3∫cos xdx 3 xcos x 3 ( 3 sin x ) 1 1 =− 3 + 3 +C 3xcos x 9 sin x Exercise 8: Evaluate

2 a) ∫x lnxdx 2 2 b) ∫ (x +1)e xdx 2 c) ∫ xsec xdx − 1 d) ∫ sin xdx Integration by parts for definite integrals For de nite integrals that require integration by parts, use the following formula:

b b b udv =uv − vdu ∫ a ∫ a a 1 = = t For example, to evaluate t u t dv e dt ∫ te dt = = t Applying the formula, 0 du dt v e 1 1 t t 1 t t 1 t 1 1 0 1 0 te dt =te 0 − e dt = te 0 −e 0 =(1e − 0e )−(e −e ) = e − e + 1 = 1 ∫0 ∫0 Exercise 9: Evaluate e/2 a) ln2xdx 1∫/2 1 3 b) xe xdx ∫0

c) π 2x cosxdx ∫0 ANSWERS: ln(17) 3 4 3. a) 473 7. a) 1. a) 4x sin2x +2x cos 2x 1 2 − sinxlnx− cosx b) 11/96 b) b) x 3 2 c) 104.605 (lnx) c) 1/5 2 1 2 4. a) 2 c) 10 x xe + + 2 3 1 2 b) diverges 8. a) x ln x 3 x x − x + C 3 2 3 3 9 3 x x c) - 1 d) e sec (e ) 2 2 2 2 3 x e x xe x 3e x tan(e x) b) − + + C 2 2 4 4 1 2 +4 11 2. a) 9 − 2 + 15 3 +C 5. a) (x ) x x x + C c) + + C 9 11 x tan x ln cos x 1 5 2 b) 7 4 25 3 +5 3/2 +C − 1 2 x + x + x + C b) (x ) d) x sin x + 1− x +C 7 2 9 − 1 c) −3e x c) +C 8 20− 2 4 9. a) 1/2 2 5 ( x ) d) 9 − 4 1 x − x + x + C 2 3 +C 1 2 3 9 4 d) − cos( x ) b) e 6 + 1 9 9 9 − 1 3 e) + 12 + C 2 x x e) 4 + 5 +C c) -4 3 4 x 1 2 f) ln 2x + 3 +C ex 2 f) +C 1 4 2 g) 2 − 5 +C − cos x sin x 2 7/2 4 5/2 2 3/2 2 5 6. a) (x−1) + (x−1) + (x −1) +C 7 7 5 3 (3x+5) h) +C 1 1 b) 2 +1 3/2 − 2 +1 1/2 + C 21 6 ( x ) 2 ( x ) i) sin x − cot x + C c) x + 2 ln x − 2 + C j) x−2 − − 3x +C e x e Call: 905.721.8668 ext. 6578 Email: [email protected] Student Learning Centre Website: ontariotechu.ca/studentlearning This document is licensed under Attribution-NonCommercial 4.0 International (CC BY-NC 4.0). For more information visit creativecommons.org/licenses/by-nc/4.0/