Upcoming SlideShare
×

# Lesson 26: Integration by Substitution (handout)

2,162 views

Published on

• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

• Be the first to like this

### Lesson 26: Integration by Substitution (handout)

1. 1. Section 5.5 Integration by Substitution V63.0121.006/016, Calculus I New York University April 27, 2010 Announcements April 29: Movie Day April 30: Quiz 5 on §§5.1–5.4 Monday, May 10, 12:00noon Final Exam: SILV 703 (Section 16) MEYR 121/122 (Section 6) Announcements April 29: Movie Day April 30: Quiz 5 on §§5.1–5.4 Monday, May 10, 12:00noon Final Exam: SILV 703 (Section 16) MEYR 121/122 (Section 6) V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 2 / 38 Resurrection Policy If your ﬁnal score beats your midterm score, we will add 10% to its weight, and subtract 10% from the midterm weight. Image credit: Scott Beale / Laughing Squid V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 3 / 38 Notes Notes Notes 1 Section 5.5 : Integration by SubstitutionV63.0121.006/016, Calculus I April 27, 2010
2. 2. Objectives 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. V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 4 / 38 Outline Last Time: The Fundamental Theorem(s) of Calculus Substitution for Indeﬁnite Integrals Theory Examples Substitution for Deﬁnite Integrals Theory Examples V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 5 / 38 Diﬀerentiation and Integration as reverse processes Theorem (The Fundamental Theorem of Calculus) 1. Let f be continuous on [a, b]. Then d dx x a f (t) dt = f (x) 2. Let f be continuous on [a, b] and f = F for some other function F. Then b a f (x) dx = F(b) − F(a). V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 6 / 38 Notes Notes Notes 2 Section 5.5 : Integration by SubstitutionV63.0121.006/016, Calculus I April 27, 2010
3. 3. Techniques of antidiﬀerentiation? 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 x √ x2 − 1 dx = arcsec x + C. What are we supposed to do with that? V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 7 / 38 No straightforward system of antidiﬀerentiation So far we don’t have any way to ﬁnd 2x √ x2 + 1 dx 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. V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 8 / 38 Outline Last Time: The Fundamental Theorem(s) of Calculus Substitution for Indeﬁnite Integrals Theory Examples Substitution for Deﬁnite Integrals Theory Examples V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 9 / 38 Notes Notes Notes 3 Section 5.5 : Integration by SubstitutionV63.0121.006/016, Calculus I April 27, 2010
4. 4. Substitution for Indeﬁnite Integrals Example Find x √ x2 + 1 dx. Solution Stare at this long enough and you notice the the integrand is the derivative of the expression 1 + x2. V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 10 / 38 Say what? Solution (More slowly, now) Let g(x) = x2 + 1. Then g (x) = 2x and so d dx g(x) = 1 2 g(x) g (x) = x √ x2 + 1 Thus x √ x2 + 1 dx = d dx g(x) dx = g(x) + C = 1 + x2 + C. V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 11 / 38 Leibnizian notation FTW Solution (Same technique, new notation) Let u = x2 + 1. Then du = 2x dx and 1 + x2 = √ u. So the integrand becomes completely transformed into x dx √ x2 + 1 = 1 2du √ u = 1 2 √ u du = 1 2u−1/2 du = √ u + C = 1 + x2 + C. V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 12 / 38 Notes Notes Notes 4 Section 5.5 : Integration by SubstitutionV63.0121.006/016, Calculus I April 27, 2010
5. 5. Useful but unsavory variation Solution (Same technique, new notation, more idiot-proof) Let u = x2 + 1. Then du = 2x dx and 1 + x2 = √ u. “Solve for dx:” dx = du 2x So the integrand becomes completely transformed into x √ x2 + 1 dx = x √ u · du 2x = 1 2 √ u du = 1 2u−1/2 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. V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 13 / 38 Theorem of the Day 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: f (u) du dx dx = f (u) du V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 14 / 38 A polynomial example Example Use the substitution u = x2 + 3 to ﬁnd (x2 + 3)3 4x dx. Solution If u = x2 + 3, then du = 2x dx, and 4x dx = 2 du. So (x2 + 3)3 4x dx = u3 2du = 2 u3 du = 1 2 u4 = 1 2 (x2 + 3)4 V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 15 / 38 Notes Notes Notes 5 Section 5.5 : Integration by SubstitutionV63.0121.006/016, Calculus I April 27, 2010
6. 6. A polynomial example, by brute force Compare this to multiplying it out: (x2 + 3)3 4x dx = x6 + 9x4 + 27x2 + 27 4x dx = 4x7 + 36x5 + 108x3 + 108x dx = 1 2 x8 + 6x6 + 27x4 + 54x2 Which would you rather do? It’s a wash for low powers But for higher powers, it’s much easier to do substitution. V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 16 / 38 Compare We have the substitution method, which, when multiplied out, gives (x2 + 3)3 4x dx = 1 2 (x2 + 3)4 + C = 1 2 x8 + 12x6 + 54x4 + 108x2 + 81 + C = 1 2 x8 + 6x6 + 27x4 + 54x2 + 81 2 + C and the brute force method (x2 + 3)3 4x dx = 1 2 x8 + 6x6 + 27x4 + 54x2 + C Is this a problem? No, that’s what +C means! V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 17 / 38 A slick example Example Find tan x dx. (Hint: tan x = sin x cos x ) Solution Let u = cos x. Then du = − sin x dx. So tan x dx = sin x cos x dx = − 1 u du = − ln |u| + C = − ln | cos x| + C = ln | sec x| + C V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 18 / 38 Notes Notes Notes 6 Section 5.5 : Integration by SubstitutionV63.0121.006/016, Calculus I April 27, 2010
7. 7. Can you do it another way? Example Find tan x dx. (Hint: tan x = sin x cos x ) Solution Let u = sin x. Then du = cos x dx and so dx = du cos x . tan x dx = sin x cos x dx = u cos x du cos x = u du cos2 x = u du 1 − sin2 x = u du 1 − u2 At this point, although it’s possible to proceed, we should probably back up and see if the other way works quicker (it does). V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 19 / 38 For those who really must know all Solution (Continued, with algebra help) tan x dx = u du 1 − u2 = 1 2 1 1 − u − 1 1 + u du = − 1 2 ln |1 − u| − 1 2 ln |1 + u| + C = ln 1 (1 − u)(1 + u) + C = ln 1 √ 1 − u2 + C = ln 1 |cos x| + C = ln |sec x| + C There are other ways to do it, too. V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 20 / 38 Outline Last Time: The Fundamental Theorem(s) of Calculus Substitution for Indeﬁnite Integrals Theory Examples Substitution for Deﬁnite Integrals Theory Examples V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 21 / 38 Notes Notes Notes 7 Section 5.5 : Integration by SubstitutionV63.0121.006/016, Calculus I April 27, 2010
8. 8. Substitution for Deﬁnite Integrals Theorem (The Substitution Rule for Deﬁnite Integrals) If g is continuous and f is continuous on the range of u = g(x), then b a f (g(x))g (x) dx = g(b) g(a) f (u) du. 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 V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 22 / 38 Example Compute π 0 cos2 x sin x dx. 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 = − u2 du = −1 3u3 + C = −1 3 cos3 x + C. Therefore π 0 cos2 x sin x dx = − 1 3 cos3 x π 0 = − 1 3 (−1)3 − 13 = 2 3 . V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 23 / 38 Deﬁnite-ly Quicker 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 0 cos2 x sin x dx = −1 1 −u2 du = 1 −1 u2 du = 1 3 u3 1 −1 = 1 3 1 − (−1) = 2 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. V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 24 / 38 Notes Notes Notes 8 Section 5.5 : Integration by SubstitutionV63.0121.006/016, Calculus I April 27, 2010
9. 9. An exponential example Example Find ln √ 8 ln √ 3 e2x e2x + 1 dx Solution Let u = e2x , so du = 2e2x dx. We have ln √ 8 ln √ 3 e2x e2x + 1 dx = 1 2 8 3 √ u + 1 du Now let y = u + 1, dy = du. So 1 2 8 3 √ u + 1 du = 1 2 9 4 √ y dy = 1 2 9 4 y1/2 dy = 1 2 · 2 3 y3/2 9 4 = 1 3 (27 − 8) = 19 3 V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 25 / 38 About those limits Since e2(ln √ 3) = eln √ 3 2 = eln 3 = 3 we have ln √ 8 ln √ 3 e2x e2x + 1 dx = 1 2 8 3 √ u + 1 du V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 26 / 38 An exponential example Example Find ln √ 8 ln √ 3 e2x e2x + 1 dx Solution Let u = e2x , so du = 2e2x dx. We have ln √ 8 ln √ 3 e2x e2x + 1 dx = 1 2 8 3 √ u + 1 du Now let y = u + 1, dy = du. So 1 2 8 3 √ u + 1 du = 1 2 9 4 √ y dy = 1 2 9 4 y1/2 dy = 1 2 · 2 3 y3/2 9 4 = 1 3 (27 − 8) = 19 3 V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 27 / 38 Notes Notes Notes 9 Section 5.5 : Integration by SubstitutionV63.0121.006/016, Calculus I April 27, 2010
10. 10. About those fractional powers We have 93/2 = (91/2 )3 = 33 = 27 43/2 = (41/2 )3 = 23 = 8 so 1 2 9 4 y1/2 dy = 1 2 · 2 3 y3/2 9 4 = 1 3 (27 − 8) = 19 3 V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 28 / 38 An exponential example Example Find ln √ 8 ln √ 3 e2x e2x + 1 dx Solution Let u = e2x , so du = 2e2x dx. We have ln √ 8 ln √ 3 e2x e2x + 1 dx = 1 2 8 3 √ u + 1 du Now let y = u + 1, dy = du. So 1 2 8 3 √ u + 1 du = 1 2 9 4 √ y dy = 1 2 9 4 y1/2 dy = 1 2 · 2 3 y3/2 9 4 = 1 3 (27 − 8) = 19 3 V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 29 / 38 Another way to skin that cat Example Find ln √ 8 ln √ 3 e2x e2x + 1 dx Solution Let u = e2x + 1,so that du = 2e2x dx. Then ln √ 8 ln √ 3 e2x e2x + 1 dx = 1 2 9 4 √ u du = 1 3 u3/2 9 4 = 1 3 (27 − 8) = 19 3 V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 30 / 38 Notes Notes Notes 10 Section 5.5 : Integration by SubstitutionV63.0121.006/016, Calculus I April 27, 2010
11. 11. A third skinned cat Example Find ln √ 8 ln √ 3 e2x e2x + 1 dx Solution Let u = e2x + 1, so that u2 = e2x + 1 =⇒ 2u du = 2e2x dx Thus ln √ 8 ln √ 3 = 3 2 u · u du = 1 3 u3 3 2 = 19 3 V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 31 / 38 A Trigonometric Example Example Find 3π/2 π cot5 θ 6 sec2 θ 6 dθ. Before we dive in, think about: What “easy” substitutions might help? Which of the trig functions suggests a substitution? V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 32 / 38 Solution Let ϕ = θ 6 . Then dϕ = 1 6 dθ. 3π/2 π cot5 θ 6 sec2 θ 6 dθ = 6 π/4 π/6 cot5 ϕ sec2 ϕ dϕ = 6 π/4 π/6 sec2 ϕ dϕ tan5 ϕ Now let u = tan ϕ. So du = sec2 ϕ dϕ, and 6 π/4 π/6 sec2 ϕ dϕ tan5 ϕ = 6 1 1/ √ 3 u−5 du = 6 − 1 4 u−4 1 1/ √ 3 = 3 2 [9 − 1] = 12. V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 33 / 38 Notes Notes Notes 11 Section 5.5 : Integration by SubstitutionV63.0121.006/016, Calculus I April 27, 2010
12. 12. The limits explained tan π 4 = sin π/4 cos π/4 = √ 2/2 √ 2/2 = 1 tan π 6 = sin π/6 cos π/6 = 1/2 √ 3/2 = 1 √ 3 6 − 1 4 u−4 1 1/ √ 3 = 3 2 −u−4 1 1/ √ 3 = 3 2 u−4 1/ √ 3 1 = 3 2 (3−1/2 )−4 − (1−1/2 )−4 = 3 2 [32 − 12 ] = 3 2 (9 − 1) = 12 V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 34 / 38 Graphs θ y π 3π 2 3π/2 π cot5 θ 6 sec2 θ 6 dθ ϕ y π 6 π 4 π/4 π/6 6 cot5 ϕ sec2 ϕ dϕ The areas of these two regions are the same. V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 35 / 38 Graphs ϕ y π 6 π 4 π/4 π/6 6 cot5 ϕ sec2 ϕ dϕ u y 1 1/ √ 3 6u−5 du 1 √ 3 1 The areas of these two regions are the same. V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 36 / 38 Notes Notes Notes 12 Section 5.5 : Integration by SubstitutionV63.0121.006/016, Calculus I April 27, 2010
13. 13. Final Thoughts Antidiﬀerentiation is a “nonlinear” problem that needs practice, intuition, and perserverance Worksheet in recitation (also to be posted) The whole antidiﬀerentiation story is in Chapter 6 V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 37 / 38 Summary If F is an antiderivative for f , then: f (g(x))g (x) dx = F(g(x)) b a f (g(x))g (x) dx = g(b) g(a) f (u) du = F(g(b)) − F(g(a)) V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 38 / 38 Notes Notes Notes 13 Section 5.5 : Integration by SubstitutionV63.0121.006/016, Calculus I April 27, 2010