# ODDREV07 - 108 Chapter 7 1, x 0 Integration Techniques,...

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Unformatted text preview: 108 Chapter 7 1, x 0 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals 83. For n I1 x2 1 4 dx b→ lim 1 2 b x2 0 1 4 2x dx b→ lim 1 1 6 x2 1 b 3 0 1 . 6 For n > 1, In 0 x2n x 2 1 1 n 3 dx b→ lim 2n dv 1 6 x2 x x2n 2 2 x2 1 x2 b b n 2 0 n n 2n 1 2 x2n 0 3 x 2 1 1 n n 2 dx 0 n n 1 I 2n 1 u (a) x 2n 2, du x 2n dx dx 2 5 2 x 2n lim 3 dx, x 1 1 6 11 46 n 3 dx, v 1 2 x2 2 0 x2 x 3 1 1 x5 1 4 b→ 1 4 3 0 (b) 0 x2 x 2 5 1 4 0 x2 x3 2 1 1 5 dx 1 24 1 60 1 0 b (c) 0 6 0 x dx 21 5 24 85. False. f x Diverges 87. True 1x 1 is continuous on 0, , lim 1 x x→ 1 0, but x 1 dx b→ lim ln x 1 0 . Review Exercises for Chapter 7 1. x x2 1 dx 1 2 x2 1 12 2x dx 3. x x2 1 dx 1 2x dx 2 x2 1 1 ln x2 2 1 C 1 x2 1 2 32 12 x 3 ln 2x dx x ln 2x 2 2 32 C 32 1 C 5. C 7. 16 dx 16 x2 16 arcsin x 4 C 9. e2x sin 3x dx 1 2x e cos 3x 3 1 2x e cos 3x 3 2 3 e2x cos 3x dx 2 3 e2x sin 3x dx 2 1 2x e sin 3x 33 2 2x e sin 3x 9 3 cos 3x C 13 9 e2x sin 3x dx e2x sin 3x dx 1 2x e cos 3x 3 e2x 2 sin 3x 13 v (1) dv u sin 3x dx ⇒ e2x 1 cos 3x 3 2e2x dx (2) dv u cos 3x dx ⇒ e2x v 1 sin 3x 3 2e2x dx ⇒ du ⇒ du Review Exercises for Chapter 7 2 x 3 5 5 12 x cos 2x 2 12 x cos 2x 2 C C (1) dv u (2) dv 5 32 109 11. u x, du xx dx, dv 5 dx 2 xx 3 2 xx 3 x x 2 x 15 x 5 5 5 5 5 1 2 dx, v 5 3 2 dx 32 13. x2 sin 2x dx x cos 2x dx 1 x sin 2x 2 x sin 2x 2 1 cos 2x 2 2x dx 1 sin 2x 2 dx 1 2 sin 2x dx C 32 2 x 3 4 x 15 4 x 15 4 3 10 32 52 12 x cos 2x 2 sin 2x dx ⇒ x2 v 1 cos 2x 4 32 2 x 3 6 x 15 3x 5 C C 32 ⇒ du v cos 2x dx ⇒ x u x2 1 1 8 11 82 4x2 ⇒ du 15. x arcsin 2x dx x2 arcsin 2x 2 x2 arcsin 2x 2 x2 arcsin 2x 2 1 8x2 16 dx 2 2x 2 dx 1 2x 2 2x 1 4x2 4x2 arcsin 2x C C (by Formula 43 of Integration Tables) 1 arcsin 2x 2x 1 dv u x dx ⇒ v x2 2 2 1 4x2 dx arcsin 2x ⇒ du 17. cos3 x 1 dx 1 1 sin sin2 x x x x x 1 13 13 12 1 cos 13 sin 3 sin2 1 cos 2 x x x cos 2 x 1 dx 1 1 x 1 C C 1 C C 1 sin 3 1 sin 3 1 sin 3 x dx 2 x 2 19. sec4 tan2 tan2 1 sec2 x dx 2 sec2 C x dx 2 x 2 tan3 3 2 3 tan x 2 C x x sec2 dx 2 2 2 tan x 2 2 3x tan 3 2 1 sin 1 sin cos2 21. 1 d d sec2 sec tan d tan sec C 110 Chapter 7 Integration Techniques, L’ H ô pital ’s Rule, and Improper Integrals 23. 12 dx x2 4 x2 24 cos d 4 sin2 2 cos 3 3 cot 34 x csc2 d C x2 d, 4 C x2 2 cos θ 2 x 4 − x2 x 25. 2 sin , dx x dx 2 tan 2 sec2 4 sec2 2 cos x2 + 4 d θ 2 x 4 x2 x3 dx 4 x2 8 tan3 2 sec 8 tan3 8 8 8 sec2 sec3 3 x2 4 24 4 2 sec 2 d sec d 1 tan sec d sec 32 C x2 2 4 8 3 x2 8 4 C C C x2 12 2 xx 3 12 x 3 12 x 3 4 4 4 C 4 12 x2 27. 4 x2 dx 2 2 2 2 cos 1 2 cos cos 2 d d 2 x θ 4 − x2 1 sin 2 2 sin cos x 2 x 2 C C 4 2 x4 4 x2 x2 2 cos x2 C C 2 arcsin 1 x 4 arcsin 2 2 x 2 sin , dx 2 cos d, Review Exercises for Chapter 7 x3 dx 4 x2 sin3 cos4 cos 4 111 29. (a) 8 8 d cos 3 8 2 (b) sin d C C u2 2x 4 2 4 3 4 2x dx x2 x2 x2 dx 32 x3 dx 4 x2 u2 13 u 3 u2 u 3 4 3 4 du 4u 12 x2 2x dx x2 C C 8 C 8 sec 3 4 3 x (c) 2 tan , dx x3 dx 4 x2 2 sec2 x2 4 x2 4 dv u x2 x 4 x2 dx ⇒ sec2 x2 x2 d x2 x2 v 4 x2, 2u du C 4 3 x2 x2 8 C ⇒ du A B 3 2 B A5 x 5 3 x 2 Bx 31. x x2 x x x x2 3 x x 28 6 28 2⇒ ⇒ 28 6 6 x Ax 30 25 dx 3 6 5 2 dx 5 ln x 3 6 ln x 2 C 5 ⇒B ⇒A 6 x 33. x2 2x x 1 x2 1 x2 Let x Let x Let x 1: 3 0: 0 2: 8 x2 x2 2x A x A x2 2A ⇒ A A 5A 2x x 1 1 Bx x2 C 1 Bx Cx 1 3 2 3 2 1 2 1 1 x 1 dx dx 1 2 1 4 x x2 2x x2 1 1 1 3 dx 1 dx 3 2 1 x2 1 dx C⇒C 2B dx C⇒B 3 2 3 2 1 x x 3 1 3 ln x 2 1 6 ln x 4 1 1 1 ln x2 4 ln x2 3 arctan x 2 6 arctan x C C 112 Chapter 7 Integration Techniques, L’ H ô pital ’s Rule, and Improper Integrals 35. x2 x x2 2x 15 1 A x Ax 15 x2 3 5 2x B x 2x 15 5 Bx 9 8 25 8 9 8 9 ln x 8 1 x 3 3 dx 25 8 1 x 5 5 dx C 3 15 2x 3x 5 15 2x 9 5: 25 x2 2x Let x Let x 3: 8A ⇒ A 8B ⇒ B dx x dx x2 15 25 ln x 8 37. x 2 3x 2 dx 1 2 9 2 3x ln 2 3x C 39. 1 x dx sin x2 1 21 1 tan u 2 1 du sin u sec u sec x2 C C u x2 (Formula 4) (Formula 56) 1 tan x2 2 41. x2 x 4x 8 dx 1 ln x2 2 1 ln x2 2 1 ln x2 2 4x 4x 4x 8 8 8 4 2 x2 1 4x 8 dx 2x 32 4 16 C (Formula 15) (Formula 14) 2 arctan 32 16 x 2 C arctan 1 43. 1 dx sin x cos x 1 1 1 sin x cos x ln tan x C dx u x 45. dv u dx ln x ln x n dx n ⇒ v x n ln x n n 1 (Formula 58) ⇒ du x ln x 1 dx x n 1 dx n ln x 47. sin cos d 1 2 sin 2 d 1 cos 2 4 1 4 cos 2 d 1 cos 2 4 1 sin 2 8 C 1 sin 2 8 2 cos 2 C dv u sin 2 d ⇒ v d 1 cos 2 2 ⇒ du R eview Exercises for Chapter 7 x1 1 4 113 49. x1 2 dx 4 4 4 u u3 du 1 u2 u2 13 u 3 4 51. 1 u2 1 du C u 1 cos x dx 1 sin x dx 1 cos x cos x cos x 12 1 u 3x1 4u3 du sin x dx arctan u 4 21 1 cos x, du C sin x dx 43 x 3 y 4 3 arctan x1 4 C x, x u4, dx 53. cos x ln sin x dx sin x ln sin x sin x ln sin x cos x dx sin x C 55. y 3x3 9 dx ln C x2 9 2x3 (by Formula 24 of Integration Tables) dv u cos x dx ⇒ v sin x cos x dx sin x 2x2 x2 2x x 2dx 2x ln x x dx x 1 dx 1 1 x 1 1 dx C 5 ln sin x ⇒ du 57. y ln x2 x dx x ln x2 x ln x 2 x ln x2 x ln x2 x x x x 59. 2 x x2 4 3 2 dx 12 x 5 5 4 52 2 1 5 dv u dx ln x2 ⇒ v x 2x x2 1 dx x x ⇒ du 4 61. 1 ln x dx x 4 1 ln x 2 4 2 1 1 ln 4 2 0 2 2 ln 2 2 0.961 63. 0 x sin x dx x cos x sin x 0 65. A 0 0 x4 2 u4 2 x dx 2 4 u2 u 2u du 67. By symmetry, x y 21 2 0, 4 3 1 0, A 1 1 1 . 2 x2 2 dx 1 x 13 x 3 1 1 4u2 du 4u3 3 0 2 4 3 2 u 4 u5 5 128 15 u2, dx 2u du x, y x, x 4 69. s 0 1 cos2 x dx 3.82 71. lim x →1 ln x 2 x1 x →1 lim 2 1 x ln x 1 0 73. lim x→ e2x x2 x→ lim 2e2x 2x x→ lim 4e2x 2 75. y ln y x→ lim ln x x→ 2x lim 2 ln ln x x 0, y 1. x→ lim 2 x ln x 1 0 Since ln y 114 Chapter 7 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals 0.09 n 1 n 77. lim 1000 1 n→ 1000 lim n→ 1 0.09 n n Let y n→ lim 0.09 n . n 0.09 n ln 1 n→ ln y n→ lim n ln 1 lim 0.09 n 1 n 0.09 n n→ lim 1 0.09 n2 0.09 n 1 n2 n→ lim 0.09 0.09 1 n 0.09 Thus, ln y 0.09 ⇒ y n e0.09 and lim 1000 1 n→ 1000e0.09 1094.17. 16 79. 0 1 4 x dx b→0 lim 43 x 3 16 4 b 32 3 81. 1 x2 ln x dx Diverges b→ lim x3 9 b 1 3 ln x 1 Converges t0 83. 0 500,000e 0.05t dt 500,000 e 0.05 0.05t t0 0 500,000 e 0.05 10,000,000 1 (a) t0 (b) t0 → 20: \$6,321,205.59 : \$10,000,000 0.05t0 1 0.05t0 e 85. (a) P 13 ≤ x < (b) P 15 ≤ x < 20 1 0.95 2 1 0.95 2 e 13 x 12.9 2 2 0.95 2 dx dx 0.4581 0.0135 e 15 x 12.9 2 2 0.95 2 Problem Solving for Chapter 7 1 1. (a) 1 1 1 1 1 x2 dx x2 2 dx x 1 x3 3 1 1 1 21 1 1 3 x 4 3 2x3 3 u x5 5 cos2 1 2x2 x4 dx x2 1 21 1 2 3 1 5 16 15 (b) Let x 1 sin u, dx x2 n dx cos u du, 1 2 sin2 u. 1 1 cos2 u n cos u du 2 2 cos2n 2 1 u du (Wallis’s Formula) 2 2 2 3 2 4 5 6. . . 2n 7 2n 1 1 22 42 62 . . . 2n 2 3 4 5 . . . 2n 2n 22n 2n 1 2 22n n! 2 2n 1 ! n! 2 1! P roblem Solving for Chapter 7 x x x x c c c c c 1 c x 1 x2 2c cx x→ x 115 3. x→ lim 9 ln 9 ln 9 x→ lim x ln c x→ lim ln x ln x 1x 1 x→ lim x c ln 9 x→ lim x c x2 x2 ln 9 ln 9 ln 9 2 ln 3 ln 3 lim 2cx2 c2 2c 2c c 5. sin AQ BR PB OP AP OR PB, cos OB OB OR cos BPR are similar: 1 OR sin OR cos sin cos cos sin y The triangles AR AQ AQR and OR BR ⇒ BP sin OR sin OR lim OR →0 lim →0 cos sin sin sin cos cos 1 cos R O θ Q P lim →0 B A (1, 0) x lim →0 sin cos sin sin cos cos cos sin 1 cos lim →0 lim 2 →0 116 7. (a) Chapter 7 Integration Techniques, L’ H ô pital ’s Rule, and Improper Integrals Area 0.2986 0.2 0 0 4 (b) Let x 3 tan , dx x2 3 sec2 x2 9 dx d , x2 9 9 sec2 . 32 9 tan2 9 sec2 tan2 sec sin2 cos 1 d d 3 sec2 d x2 + 9 x θ 3 cos2 cos d tan tan 9 4 5 x 3 ln 3 9 sin sin 0 ln sec 4 C tan 1 Area 0 x2 x2 9 32 43 dx ln sec ln 5 ln 3 x2 3 4 3 9 x x2 4 5 9 4 0 (c) x A 3 sinh u, dx 4 0 3 cosh u du, x2 sin 32 1 9 sinh2 u 2 9 cosh2 u x 2 x2 9 43 dx 0 sinh 0 sinh 0 1 1 9 sinh2 u 9 cosh2 u 3 tanh2 u du 3 cosh u du 43 43 1 u sinh ln ln 4 3 4 3 5 3 tanh u 0 1 sech2 u du sinh 1 43 4 3 tanh sinh 16 9 1 4 3 1 4 3 4 3 16 9 1 tanh ln 5 3 tanh ln ln 3 ln 3 ln 3 tanh ln 3 3 3 4 5 13 13 P roblem Solving for Chapter 7 9. y 1 ln 1 y 2 117 x2 , y 1 12 1 1 2x x2 1 2 4x2 x2 1 2x2 1 x4 4x2 x2 2 1 1 x2 x2 2 Arc length 0 12 0 12 y 2 dx 1 1 1 x2 dx x2 2 1 1 x x ln 1 2 ln 2 1 1 ln 1 x2 dx 1 x x 0 0 12 1 0 dx 12 x 1 2 1 2 ln 3 ln 1 ln 3 2 ln 3 1 2 ln 2 0.5986 11. Consider Let u If 1 dx. ln x 1 dx, x dx eu. Then 1 dx ln x 1u e du u eu du. u ln x, du 1 dx were elementary, then ln x 1 dx is not elementary. ln x x2 x4 ax a d 1 1 dx 0 eu du would be too, which is false. u Hence, 13. x4 1 b x2 c x3 cx ac 2 d b d x2 ad bc x bd a c, b x4 1 1, a x2 1 2x Ax x2 1 2 x2 1 x2 2x 1 0 1 Cx D dx 2x 1 2 x 4 dx 2x 1 1 1 x4 0 B dx 2x 1 2 x 4 dx 2x 1 2x 2 1 1 x2 1 1 0 1 2 x2 0 2 arctan 4 2 arctan 4 0.5554 0.8670 arctan arctan 2 2x 1 1 0 2 ln x2 8 2 ln 2 8 2 1 2x ln 2 1 2 ln x2 2x 2 44 1 0 4 2 0 8 0.3116 118 Chapter 7 Integration Techniques, L’ H ô pital ’s Rule, and Improper Integrals 15. Using a graphing utility, (a) lim cot x x→0 1 x 1 x 1 x 0 cot x 1 x 2 . 3 (b) lim cot x x→0 (c) lim cot x x→0 Analytically, (a) lim cot x x→0 1 x 1 x x→0 (b) lim cot x x→0 lim lim x cot x x 1 x→0 lim x cos x sin x x sin x x→0 x→0 cos x x sin x cos x sin x x cos x cos x sin x x cos x cos x x sin x 1 x2 1 lim sin x x sin x x cos x x→0 lim 1 x 0. (c) cot x 1 x cot x cot2 x x2 cot2 x x2 x→0 lim x2 cot2 x x2 1 x→0 lim lim lim 2x cot2 x cot2 x 2x2 cot x csc2 x 2x x cot x csc2 x 1 x→0 x→0 cos2 x sin x x cos x sin3 x 1 sin x sin2 x sin x sin3 x x cos x sin3 x 1 x cos x x→0 lim lim x→0 Now, lim x→0 sin x x cos x sin3 x x→0 lim lim cos x cos x x sin x 3 sin2 x cos x x→0 x 3 sin x cos x x 1 sin x 3 cos x 1 3 1 1 . 3 2 . 3 x→0 lim 1 x Thus, lim cot x x→0 1 x cot x The form 0 is indeterminant. Problem Solving for Chapter 7 x3 x 4 119 17. 3x2 13x2 x3 4x 3 1 12x 3x2 26x 1 12 P1 x 1 12 P2 x 1 x P3 4 x P4 3 ⇒ c1 0, c2 1, c3 4, c4 3 Nx Dx P1 P2 P3 P4 Thus, N0 D0 N1 D1 N D N3 D3 x3 x 4 1 10 4 4 1 42 1 12x 111 140 1 10 111 140 3x2 13x2 1 12 x 1 10 x1 111 140 x4 1 42 . x3 19. By parts, b b f x g x dx a f xg x a b f x g x dx f x g x dx a b b f xgx a b a g x f x dx f x g x dx. a ...
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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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