sol5 - Integration by Parts Solution Math 125 In this work sheet well study the technique of integration by parts Recall that the basic formula looks

# sol5 - Integration by Parts Solution Math 125 In this work...

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Unformatted text preview: Integration by Parts - Solution Math 125 In this work sheet we’ll study the technique of integration by parts. Recall that the basic formula looks like this: u dv = u · v − 1 First a warm-up problem. Consider the integral v du x sin(3x) dx. Let u = x and let dv = sin(3x) dx. Compute du by diﬀerentiating and v by integrating, and use the basic formula to compute the original integral. Don’t forget the arbitrary constant! du = dx 1 v = − 3 cos(x) 1 x sin(3x) dx = − x cos(3x) + 3 2 Compute 1 1 1 cos(3x) dx = − x cos(3x) + sin(3x) + C 3 3 9 ln x dx. (The proper technique is, indeed, integration by parts. What should you take to be u and dv? The choices are pretty limited. Try one and see what happens.) Let u = ln(x) and dv = dx. Then du = ln(x) dx = x ln(x) − x· 1 dx and v = x. x 1 dx = x ln(x) − x dx = x ln(x) − x + C The volume is (2πyey − 2πey )|1 = 2π ≈ 6.283 0 2πyey dy = 2πyey − du = 2π dy u = 2πy 2πey dy = 2πyey − 2πey + C v = ey dv = ey dy 0 2πyey dy 1 (c) Use the shell method to ﬁnd the volume of the solid swept out by region B. The volume is πx ln(x)2 − 2πx ln(x) + 2πx (∗) = πx ln(x)2 − 2πx ln(x) + dU = 1 dx x U = ln(x) 1 e = π(e − 2) ≈ 2.2565 2π dx = πx ln(x)2 − 2πx ln(x) + 2πx + C V = 2πx dV = 2π dx π ln(x)2 dx = πx ln(x)2 − du = 2 ln(x) dx x u = ln(x)2 2π ln(x) dx = (∗) v = πx dv = π dx 1 e π ln(x)2 dx (b) Use the disk method to ﬁnd the volume of the solid swept out by region A. 1 2 ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡£¡£¡£¡£¡£¡ ¤ ¤ ¤ ¤ ¤ ¤ ¤ ¤ ¤ ¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¤¡¤¡¤¡¤¡¤¡ £¡£¡£¡£¡£¡£¡£¡£¡£¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡¢£¡£¡£¡£¡£¡¤ ¤¡¢¤¡¢¤¡¢¤¡¢¤¡¢¤¡¢¤¡¢¤¡¢¤¡¤ ¤ ¤ ¤ ¤ ¢¡¡¡¡¡¡¡¡¡¢ ¡¡¡¡¡¤£ £¡£¡£¡£¡£¡£¡£¡£¡£¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ £¡£¡£¡£¡£¡£ ¤¡¤¡¤¡¤¡¤¡¤¡¤¡¤¡¤¡ ¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¤¡¤¡¤¡¤¡¤¡ £¡£¡£¡£¡£¡£¡£¡£¡£¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡¢£¡£¡£¡£¡£¡¤ ¤¡¢¤¡¢¤¡¢¤¡¢¤¡¢¤¡¢¤¡¢¤¡¢¤¡¤ ¤ ¤ ¤ ¤ ¢¡¡¡¡¡¡¡¡¡¢ ¡¡¡¡¡¤£ £¡£¡£¡£¡£¡£¡£¡£¡£¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ £¡£¡£¡£¡£¡£ ¤¡¤¡¤¡¤¡¤¡¤¡¤¡¤¡¤¡ ¢¡¡¡¡¡¡¡¡¡¡¡¡¡¡ £¡£¡£¡£¡£¡£¡£¡£¡£¡ ¢¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¢¡¡¡¡¡£¤ ¤¡¢¤¡¢¤¡¢¤¡¢¤¡¢¤¡¢¤¡¢¤¡¢¤¡¤ ¤ ¤ ¤ ¤ £¡£¡£¡£¡£¡£¡£¡£¡£¡ ¢¡¡¡¡¡¡¡¡¡¢ ¢¡¡¡¡¡¤£¤ ¤£¡¤£¡¤£¡¤£¡¤£¡¤£¡¤£¡¤£¡¤£¡¤ ¤ ¤ ¤ ¤ ¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡£¡£¡£¡£¡£¡ ¡¢ ¡¢ ¡¢ ¡¢ ¡¢ ¡¢ ¡¢ ¡¢ ¡¤¤¡¤¤¡¤¤¡¤¤¡¤¤¡ ¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡ ¤¡¤¡¤¡¤¡¤¡£ ¤¡¤¡¤¡¤¡¤¡¤¡¤¡¤¡¤¡£ £ £ £ £ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡£¡£¡£¡£¡£¡ ¡¡¡¡¡¡¡¡¡ ¢¡¡¡¡¡£¤£¤ £¡£¡£¡£¡£¡£¡£¡£¡£¡ ¤¡¤¡¤¡¤¡¤¡¤¡¤¡¤¡¤¡¡¡¡¡¡ £¡£¡£¡£¡£¡£¡£¡£¡£¡ ¡¡¡¡¡¡¡¡¡£ £ £ £ £ Region B looks larger. e3 4 5 A (a) Before computing the integrals, which solid do you think has a larger volume? Why? B 1 3 The regions A and B in the ﬁgure are revolved around the x-axis to form two solids of revolution. y=ln(x) 4 Suppose we try to integrate 1/x by parts, taking u = 1/x and dv = dx. We have du = (−1/x2 ) dx and v = x, so 1 −1 1 dx = · x − x · 2 dx x x x 1 = 1+ dx. x Canceling the integral from both sides, we get the disconcerting result that 0 = 1. What went wrong? What happens if we replace the indeﬁnite integrals by deﬁnite integrals, that is, if we try b 1 dx by this method? to calculate a x This issue here is that indeﬁnite integrals are deﬁned only up to an arbitrary constant. The correct equation should look like 0 = 1 + C b For the second question, a 1 dx = 1|b + a x b a 1 dx. Here the term 1|b = 1 − 1 = 0. a x ...
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