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Lesson 27: Integration by Substitution (Section 041 handout)
Lesson 27: Integration by Substitution (Section 041 handout)
Lesson 27: Integration by Substitution (Section 041 handout)
Lesson 27: Integration by Substitution (Section 041 handout)
Lesson 27: Integration by Substitution (Section 041 handout)
Lesson 27: Integration by Substitution (Section 041 handout)
Lesson 27: Integration by Substitution (Section 041 handout)
Lesson 27: Integration by Substitution (Section 041 handout)
Lesson 27: Integration by Substitution (Section 041 handout)
Lesson 27: Integration by Substitution (Section 041 handout)
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Lesson 27: Integration by Substitution (Section 041 handout)

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Integration by substitution is the chain rule in reverse. …

Integration by substitution is the chain rule in reverse.

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  • 1. V63.0121.041, Calculus I Section 5.5 : Integration by Substitution December 13, 2010 Notes Section 5.5 Integration by Substitution V63.0121.041, Calculus I New York University December 13, 2010 Announcements ”Wednesday”, December 15: Review, Movie Monday, December 20, 12:00pm–1:50pm: Final Exam Announcements Notes ”Wednesday”, December 15: Review, Movie Monday, December 20, 12:00pm–1:50pm: Final Exam V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 2 / 37 Objectives Notes Given an integral and a substitution, transform the integral into an equivalent one using a substitution Evaluate indefinite integrals using the method of substitution. Evaluate definite integrals using the method of substitution. V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 4 / 37 1
  • 2. V63.0121.041, Calculus I Section 5.5 : Integration by Substitution December 13, 2010 Outline Notes Last Time: The Fundamental Theorem(s) of Calculus Substitution for Indefinite Integrals Theory Examples Substitution for Definite Integrals Theory Examples V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 5 / 37 Differentiation and Integration as reverse processes Notes Theorem (The Fundamental Theorem of Calculus) 1. Let f be continuous on [a, b]. Then x d f (t) dt = f (x) dx a 2. Let f be continuous on [a, b] and f = F for some other function F . Then b f (x) dx = F (b) − F (a). a V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 6 / 37 Techniques of antidifferentiation? Notes So far we know only a few rules for antidifferentiation. Some are general, like [f (x) + g (x)] dx = f (x) dx + g (x) dx Some are pretty particular, like 1 √ dx = arcsec x + C . x x2 − 1 What are we supposed to do with that? V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 7 / 37 2
  • 3. V63.0121.041, Calculus I Section 5.5 : Integration by Substitution December 13, 2010 No straightforward system of antidifferentiation Notes So far we don’t have any way to find 2x √ dx x2 + 1 or tan x dx. Luckily, we can be smart and use the “anti” version of one of the most important rules of differentiation: the chain rule. V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 8 / 37 Outline Notes Last Time: The Fundamental Theorem(s) of Calculus Substitution for Indefinite Integrals Theory Examples Substitution for Definite Integrals Theory Examples V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 9 / 37 Substitution for Indefinite Integrals Notes Example Find x √ dx. x2 + 1 Solution Stare at this long enough and you notice the the integrand is the derivative of the expression 1 + x 2 . V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 10 / 37 3
  • 4. V63.0121.041, Calculus I Section 5.5 : Integration by Substitution December 13, 2010 Say what? Notes Solution (More slowly, now) Let g (x) = x 2 + 1. Then g (x) = 2x and so d 1 x g (x) = g (x) = √ dx 2 g (x) x2 + 1 Thus x d √ dx = g (x) dx x2 + 1 dx = g (x) + C = 1 + x2 + C . V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 11 / 37 Leibnizian notation FTW Notes Solution (Same technique, new notation) √ Let u = x 2 + 1. Then du = 2x dx and 1 + x2 = u. So the integrand becomes completely transformed into 1 √ x dx 2 du √ 1 √ du = = x2 + 1 u 2 u 1 −1/2 = 2u du √ = u+C = 1 + x2 + C . V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 12 / 37 Useful but unsavory variation Notes Solution (Same technique, new notation, more idiot-proof) √ Let u = x 2 + 1. Then du = 2x dx and 1 + x 2 = u. “Solve for dx:” du dx = 2x So the integrand becomes completely transformed into x x du 1 √ dx = √ · = √ du x2 + 1 u 2x 2 u 1 −1/2 = 2u du √ = u+C = 1 + x2 + C . Mathematicians have serious issues with mixing the x and u like this. However, I can’t deny that it works. V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 13 / 37 4
  • 5. V63.0121.041, Calculus I Section 5.5 : Integration by Substitution December 13, 2010 Theorem of the Day Notes Theorem (The Substitution Rule) If u = g (x) is a differentiable function whose range is an interval I and f is continuous on I , then f (g (x))g (x) dx = f (u) du That is, if F is an antiderivative for f , then f (g (x))g (x) dx = F (g (x)) In Leibniz notation: du f (u) dx = f (u) du dx V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 14 / 37 A polynomial example Notes Example Use the substitution u = x 2 + 3 to find (x 2 + 3)3 4x dx. Solution If u = x 2 + 3, then du = 2x dx, and 4x dx = 2 du. So (x 2 + 3)3 4x dx = u 3 2du = 2 u 3 du 1 1 = u 4 = (x 2 + 3)4 2 2 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 15 / 37 A polynomial example, by brute force Notes Compare this to multiplying it out: (x 2 + 3)3 4x dx = x 6 + 9x 4 + 27x 2 + 27 4x dx = 4x 7 + 36x 5 + 108x 3 + 108x dx 1 = x 8 + 6x 6 + 27x 4 + 54x 2 2 Which would you rather do? It’s a wash for low powers But for higher powers, it’s much easier to do substitution. V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 16 / 37 5
  • 6. V63.0121.041, Calculus I Section 5.5 : Integration by Substitution December 13, 2010 Compare Notes We have the substitution method, which, when multiplied out, gives 1 (x 2 + 3)3 4x dx = (x 2 + 3)4 2 1 8 = x + 12x 6 + 54x 4 + 108x 2 + 81 2 1 81 = x 8 + 6x 6 + 27x 4 + 54x 2 + 2 2 and the brute force method 1 (x 2 + 3)3 4x dx = x 8 + 6x 6 + 27x 4 + 54x 2 2 Is there a difference? Is this a problem? V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 17 / 37 A slick example Notes Example sin x Find tan x dx. (Hint: tan x = ) cos x Solution V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 18 / 37 Outline Notes Last Time: The Fundamental Theorem(s) of Calculus Substitution for Indefinite Integrals Theory Examples Substitution for Definite Integrals Theory Examples V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 21 / 37 6
  • 7. V63.0121.041, Calculus I Section 5.5 : Integration by Substitution December 13, 2010 Substitution for Definite Integrals Notes Theorem (The Substitution Rule for Definite Integrals) If g is continuous and f is continuous on the range of u = g (x), then b g (b) f (g (x))g (x) dx = f (u) du. a g (a) Why the change in the limits? The integral on the left happens in “x-land” The integral on the right happens in “u-land”, so the limits need to be u-values To get from x to u, apply g V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 22 / 37 Example π Notes Compute cos2 x sin x dx. 0 Solution (Slow Way) First compute the indefinite integral cos2 x sin x dx and then evaluate. Let u = cos x. Then du = − sin x dx and cos2 x sin x dx = − u 2 du = − 3 u 3 + C = − 1 cos3 x + C . 1 3 Therefore π π 1 1 2 cos2 x sin x dx = − cos3 x =− (−1)3 − 13 = . 0 3 0 3 3 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 23 / 37 Definite-ly Quicker Notes Solution (Fast Way) Do both the substitution and the evaluation at the same time. Let u = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = − 1. So π −1 1 cos2 x sin x dx = −u 2 du = u 2 du 0 1 −1 1 1 3 1 2 = u = 1 − (−1) = 3 −1 3 3 The advantage to the “fast way” is that you completely transform the integral into something simpler and don’t have to go back to the original variable (x). But the slow way is just as reliable. V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 24 / 37 7
  • 8. V63.0121.041, Calculus I Section 5.5 : Integration by Substitution December 13, 2010 An exponential example Notes Example √ ln 8 Find √ e 2x e 2x + 1 dx ln 3 Solution V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 25 / 37 Another way to skin that cat Notes Example √ ln 8 Find √ e 2x e 2x + 1 dx ln 3 Solution V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 30 / 37 A third skinned cat Notes Example √ ln 8 Find √ e 2x e 2x + 1 dx ln 3 Solution V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 31 / 37 8
  • 9. V63.0121.041, Calculus I Section 5.5 : Integration by Substitution December 13, 2010 A Trigonometric Example Notes Example Find 3π/2 θ θ cot5 sec2 dθ. π 6 6 Before we dive in, think about: What “easy” substitutions might help? Which of the trig functions suggests a substitution? V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 32 / 37 Solution Notes V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 33 / 37 Graphs 3π/2 θ θ π/4 Notes cot5 sec2 dθ 6 cot5 ϕ sec2 ϕ dϕ π 6 6 π/6 y y θ ϕ 3π π ππ 2 64 The areas of these two regions are the same. V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 35 / 37 9
  • 10. V63.0121.041, Calculus I Section 5.5 : Integration by Substitution December 13, 2010 Graphs π/4 1 −5 Notes 6 cot5 ϕ sec2 ϕ dϕ √ 6u du π/6 1/ 3 y y ϕ u ππ 1 1 64 √ 3 The areas of these two regions are the same. V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 36 / 37 Summary Notes If F is an antiderivative for f , then: f (g (x))g (x) dx = F (g (x)) If F is an antiderivative for f , which is continuous on the range of g , then: b g (b) f (g (x))g (x) dx = f (u) du = F (g (b)) − F (g (a)) a g (a) Antidifferentiation in general and substitution in particular is a “nonlinear” problem that needs practice, intuition, and perserverance The whole antidifferentiation story is in Chapter 6 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 37 / 37 Notes 10

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