# EVNREV07 - 358 Chapter 7 1 e 32 f x dx 50 0.4 Integration...

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Unformatted text preview: 358 Chapter 7 1 e 32 f x dx 50 0.4 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals 84. (a) f x 90 x 70 2 18 (b) P 72 ≤ x < (c) 0.5 0.2525 0.5 0.2475 0.2525 P 70 ≤ x ≤ 72 1.0 These are the same answers because by symmetry, P 70 ≤ x < and P 70 ≤ x < P 70 ≤ x ≤ 72 P 72 ≤ x < . 0.5 50 90 0.5 − 0.2 86. False. This is equivalent to Exercise 85. 88. True Review Exercises for Chapter 7 2. xe x 2 1 dx 1 x2 1 e 2x dx 2 1 x21 e 2 C 4. x 1 x2 dx 1 2 1 x2 12 12 2x dx C 1 1 x2 2 12 1 u5 5 1 x4 x4 x4 2x2 x 1 2x2 1 2x2 x x2 1 2 1 x2 C x 6. 2x 2x 3 dx u4 2 2x 5 u2 2 3 3u2 du 3 32 u3 C C 8. 1 dx x2 dx x 1 2 x u du 1 2 1 2 x2 2x x2 1 1 C 2 dx u 2x 3, x , dx 10. x2 (1) dv u (2) dv u 1 ex dx ex dx x2 x2 ⇒ v 1 ex ex 2 xex dx x2 1 ex 2xex 2 ex dx ex x2 2x 1 1 1 ⇒ du v ex ⇒ du 2x dx ex dx ⇒ x dx 12. u arctan 2x, du arctan 2x dx 2 1 4x2 dx, dv dx,v 2x dx 4x2 4x2 x 14. ln x2 1 dx 1 2 ln x2 1 dx 1 1 1 x x2 x2 dx 1 dx 1 x2 1x ln 2x 1 1 1 dx C x arctan 2x x arctan 2x 1 1 x ln x2 2 C 1 x ln x2 2 1 x ln x2 2 dv u dx ln x2 ⇒ v x 1 ln 1 4 1 ⇒ du 2x x2 1 dx Review Exercises for Chapter 7 e2x dx 1 e2x 1 ln 1 2 e2x C 359 16. ex arctan ex dx ex arctan ex ex arctan ex dv u ex dx ⇒ v ex ex 1 e2x dx arctan ex ⇒ du 18. sin2 x dx 2 1 1 2 cos x d x 1 x 2 1 sin x C 1 2 x sin x C 20. tan sec4 or tan sec4 d tan3 tan sec2 d 14 tan 4 12 tan 2 C1 d sec3 sec tan d 1 sec4 4 C2 22. cos 2 sin cos 2 d cos 2 sin sin2 cos 3 sin cos cos sin 2 d 1 sin 4 cos 4 d C 24. x2 x 9 dx 3 tan 3 sec 3 tan2 3 sec2 3 sec tan d 1d C 9 3 arcsec x2 x 3 d x x2 − 9 θ 3 3 tan x2 x 3 sec , dx C 9 3 tan 3 sec tan d , 26. 9 4x2 dx 1 2 1 2 9 2x 2 2 dx 2x 9 9 4x2 4x2 C C 28. 1 sin 2 cos2 d 1 2 1 1 2 cos2 2 cos 2 sin C d 1 2x 9 arcsin 2 3 x 2 1 arctan 2 u 2 cos , du 2 sin d 9 2x arcsin 4 3 360 Chapter 7 Integration Techniques, L’ H ô pital ’s Rule, and Improper Integrals 30. (a) x4 x dx 64 tan3 64 sec4 sec3 d sec tan d 5 8 d, C C (b) x4 x dx 2 u4 4u2 du 20 32 sec2 3 sec3 2u3 2 3u 15 24 u2 4 x, dx x 15 C 8 C 64 sec3 15 24 x 4 (c) x4 4 tan2 , dx x 2 sec x dx u3 2 3x x 15 32 3x 2u du 8 tan sec2 4u1 2 du C 8 C (d) x4 x dx 2x 4 3 2x 4 3 24 x 15 v x x 32 2 3 4 x x C 3 2 dx 2u3 2 3u 15 24 u 4 x, du dx x 15 32 20 3x 32 4 4 15 8 x 32 52 C 32 3x 2 4 3 dx dv u x 4 x dx ⇒ ⇒ du 32. 2x3 2x3 5x2 x2 5x2 x2 4x x 4 4 2x 2x 3 3 4 x 4 x 3 x x 1 3 1 dx x2 3x 4 ln x 3 ln x 1 C 4x x A x 3A x 1 dx 34. 4x 3x 4x Let x Let x 2 12 2 B x 1 1 3B 2 3 4 3 dx 2 3 1 x 1 1 u C 1 2 2 1: 2 2: 6 4x 3x 3B ⇒ B 3A 3B ⇒ A 4 3 1 x 1 1 uu ln u 1 1 2 dx 12 dx 4 ln x 3 1 du u tan tan 1 1 2 3x 1 C 2 2 ln x 3 1 1 x 1 C 36. sec2 tan tan 1 d du ln u du ln C ln 1 cot C u uu tan , du 1 1 1 A u Au sec2 B u 1 d 1 Bu 1 Let u Let u 0: 1 1: 1 B A⇒ A R eview Exercises for Chapter 7 x 2 3x 2 4 3x 27 6x 8 27 2 x 1 ex 2 361 38. dx 2 3x 3x C C (Formula 21) 40. dx 1 1 dx 2 1 eu 1 u 2 12 x 2 ln 1 ln 1 eu ex 2 u C C x2 (Formula 84) 42. 3 2x 9x2 1 dx 3 2 3x 1 3x 2 1 C 3 dx u 3x 3 arcsec 3x 2 1 dx tan x 1 1 11 2 x 1 tan x (Formula 33) 44. 1 dx sin x C u x ln cos x (Formula 71) 46. tann x dx tann tann 1 n 1 2x sec2 x 1 dx tann tann 2 2 48. x dx u csc 2x dx x 2 csc 2x 1 dx 2x cot 2x C 2 x sec2 x dx 1 2 ln csc 2x 2x, du 1 dx 2x tann x x dx 50. u 3x3 x2 3x3 1 1 4x 12 4x x dx x, x Ax x2 Ax Ax3 B 1 u 4u3 u4 4u du 2u2 1, dx D 12 1 A Cx Cx 1, D B 4u4 4u3 4u2 du 4u du 4u5 5 4u3 3 C 4 1 15 x 32 3x 2 C 52. Cx x2 B x2 Bx2 C 0 x x2 1 D A B 3, B D 3x3 x2 0, A 0⇒D 4⇒C 4x dx 12 3 dx 1 x x2 1 2 x2 1 1 2 dx C 3 ln x2 2 54. sin cos 2 d sin2 1 2 sin cos sin 2 d cos2 1 cos 2 2 d C 1 2 2 cos 2 C 362 Chapter 7 4 x2 dx 2x Integration Techniques, L’ H ô pital ’s Rule, and Improper Integrals 2 cos 2 cos 4 sin sin d 56. y csc ln csc ln 2 d cos cos 4 2 x2 2 cos x2 C C 4 x 4 sin 1 cos x2 x 2 sin , dx 2 cos d, 58. y u 1 1 1 cos d d 1 cos 12 sin d 21 cos C cos , du x sin d 1 2 60. 0 x 2x 4 dx 2 ln x 2 ln 3 ln 9 8 x 4 2 ln 4 0.118 ln x ln 2 2 0 62. 0 xe3x dx e3x 3x 9 2 1 0 16 5e 9 1 224.238 3 64. 0 x 1 x dx 22 3 3 1 x 0 4 3 4 3 8 3 4 66. A 0 1 25 x2 1 x ln 10 x dx 5 5 4 0 1 1 ln 10 9 1 ln 9 10 0.220 68. By symmetry, y A x 1 17 5 x, y 3.4, 0 sin x sin 2 x lim x 4 0. 5 44 4 3.4 3 2 1 y 70. s 0 1 sin2 2x dx 3.82 (3.4, 0) x 1 3 4 5 −1 −2 −3 72. lim x →0 x →0 lim cos x 2 cos 2 x ln x 2 1 2 74. lim xe x→ x2 x→ lim x 2 ex x→ lim 1 2 2 xe x 0 76. y ln y x →1 1 x →1 lim ln x ln x 1 1 x1 1 1 x ln2 x 2 x →1 x →1 lim ln x 1 ln x 1 x →1 lim x →1 lim ln2 x x1 x lim 1 ln x x 1 x2 x →1 lim 2x ln x 0, y 1. 0 Since ln y P roblem Solving for Chapter 7 78. lim x →1 363 2 ln x 2 x 1 x →1 lim 2x 2 2 ln x ln x x 1 2 2x 1 1x ln x 2x 1 b x →1 lim x x x→1 lim 2 x ln x x→1 lim 1 2 1 ln x 1 1 80. 0 6 x 1 dx b→1 lim 6 ln x 1 0 82. 0 e 1x b x 2 dx a→0 b→ lim e 1x a 1 0 1 Diverges 84. V 0 xe x2e 0 x 2dx 2x dx b→ lim e 2x 2 2x 4 1 x5 1 x10 1 9x9 b 2x 1 0 4 1 1 x5 1 x10 1 4x4 2 dx x15 1 9x9 1 7x14 b 2 86. 2 1 dx < x15 1 14x14 b 2 x5 1 x5 1 x5 1 1 1 dx < 2 b→ lim 1 4x4 < 2 2 dx < lim b→ 0.015846 < 2 dx < 0.015851 Problem Solving for Chapter 7 1 2. (a) 0 1 b→0 ln x dx lim 1 x ln b→0 x b lim b ln b ln b b1 2 b→0 b lim 1 1b 1 b2 1 Note: lim b ln b b→0 1 b→0 lim 0 ln x 2 dx 0 b→0 lim x ln x b→0 2x ln x 2 2x b 2 b→0 1 lim b ln b n 2b ln b 2b 2 (b) Note first that lim b ln b Also, Assume 0 1 0 (Mathematical induction). n 1 ln x 1 n dx x ln x n 1 ln x n dx. ln x n dx ln x n 1 dx 1 n n!. 1 b→0 1 Then, 0 lim n x ln x 1 n 1 b n 1 0 ln x n dx n 1 0 1 n n! 1 n 1 !. 364 Chapter 7 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals x x x x c c c c c 1 c x 1 x2 c ln 4 x 4. x→ lim 1 4 ln 1 4 ln 4 x→ lim x ln c x→ lim ln x ln x 1x 1 x→ lim x x→ lim x 2c cx x→ c x2 x2 2cx2 c2 2c 2x c ln 4 ln 4 ln 4 2 ln 2 ln 2 lim 6. sin Area BD, cos DAB OD 1 1 2 2 1 21 12 cos 1 sin 2 cos sin sin lim →0 1 DA BD 2 2 1 1 BD 2 sin Shaded area R lim R →0 DAB Shaded area lim →0 1 1 sin cos sin sin cos sin 4 cos sin 1 cos cos 1 cos sin2 lim →0 cos sin sin sin lim 4 cos 1 3 1 →0 2 sin cos 1 lim →0 3 8. u dx 2 0 x tan , cos x 2 1 2 2 du u2 1 dx cos x 1 1 u2 ,2 u2 cos x 2 1 1 u2 u2 u2 u2 1 0 1 0 1 3 2 3 u2 u2 u2 du 2 1 u2 du 2 1 arctan 3 1 3 3 9 u 3 1 0 2 arctan 3 2 36 0.6046 P roblem Solving for Chapter 7 10. Let u b 365 cx, du c2 x2 c dx. cb e 0 dx 0 e u2 du c 1 c cb e 0 u2 du dx 1 c e 0 x2 As b → x , cb → . Hence, 0 e c2 x2 dx. 0 by symmetry. Mx m 2 0 e c2x2 y 2 e 0 c2x2 dx dx dx dx 2 1 2 e 0 2c2x2 e 0 c2x2 1 2 1 1 2 e x dx 2c 0 1 2 e x dx c0 2 4 0, 2 . 4 x, dx f y dy. 22 Thus, x, y 12. (a) Let y f 1 f x dx 1 x ,f y y f y dy yf y xf 1 f y dy x y, f x f y dy sin x sin y dy cos y 1 C x2 C u y, du dy dv f y dy, v fy (b) f 1 x arcsin x arcsin x dx x arcsin x x arcsin x x arcsin x 1 x y 1 − x2 (c) f x e ex, f 1 x x ln x ln x e 1 1 y 1 x 1⇔y 0; x e⇔y 1 ln x dx 1 e y dy 0 e e ey 0 e 1 1 366 Chapter 7 Integration Techniques, L’ H ô pital ’s Rule, and Improper Integrals 14. (a) Let x 2 2 u, dx du. sin 2 I 0 sin x dx cos x sin x 0 2 u du sin 2 u cos 2 u 2 0 cos u du sin u cos u Hence, 2 2I 0 2 sin x dx cos x sin x 1 dx 2 ⇒I u 2 0 cos x dx sin x cos x 0 4 . 0 sinn 2 (b) I cosn 2 u 2 du sinn 2 u 2 0 cosn u du sinn u cosn u 2 Thus, 2I 0 1 dx 2 ⇒I 4 Pn . 16. Nx Dx Nx Let x P1 x P1 x c1 x P2 c2 c2 x ... x cn c3 . . . x cn P2 x c3 . . . c1 cn cn c1 x cn c3 . . . x cn ... Pn x c1 x c2 . . . x cn 1 c1: N c1 P1 P1 c1 c1 c2 c1 c2 c1 N c1 c3 . . . c1 Let x c2: N c2 P2 P2 c2 c2 c1 c2 c3 . . . c2 cn c1 c2 N c2 c3 . . . c2 Let x cn: N cn Pn Pn cn cn c1 cn c2 . . . cn cn cn 1 c1 cn c2 x N cn c2 . . . cn c3 . . . x cn 1 If D x Dx and D c1 D c2 x x c1 x c2 x cn , then by the Product Rule x c1 x c3 . . . x cn ... x c1 x c2 x c3 . . . x cn 1 c3 . . . x c1 c2 c2 c1 c1 c2 c3 . . . c1 c3 . . . c2 cn cn D cn Thus, Pk cn c1 cn c2 . . . cn cn 1 . N ck D ck for k 1, 2, . . ., n. Problem Solving for Chapter 7 50,000 dt 50,000 400t ln 50,000 400 t dt 400 t 1 400t dt 50,000 400 t 367 18. st 32 t 16 t 2 16 t 2 16 t 2 16 t 2 12,000 ln 12,000 ln 50,000 12,000 t ln 50,000 12,000 t ln 12,000 t ln 12,000 t ln 50,000 12,000 t 12,000 t 50,000 50,000 400 t 50,000 50,000 400t C 0 50,000 dt 50,000 400 t 400 t C 1,500,000 ln 50,000 s0 C st When t 20. Let u b 1,500,000 ln 50,000 1,500,000 ln 50,000 16 t 2 12,000 t 1 ln 50,000 50,000 400 t 1,500,000 ln 50,000 400 t 50,000 100, s 100 x x ax ax 557,168.626 feet b , du x x b a ax 2x a a x b dx, dv b b f x dx, v x a x f x. b f x dx b dx bf x a a a b f x dx b b a u 2x a b dv f x dx 2 f x dx 2x b bfx a a 2 a f x dx ...
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## This note was uploaded on 05/18/2011 for the course MAC 2311 taught by Professor All during the Fall '08 term at University of Florida.

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