Business Calc Homework w answers_Part_68

X2 b lim sec x x2 1 8tan c b b1 1 8u2 16u du 2 2 tan

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Unformatted text preview: e improper integral. Note that the integrand is infinite at x 0. (1 d b 1 b→ u b→ 0 3 b 1 1 1 1 b→ 2 tan 2) d 2) lim ln x 2 du 1 u2 x 3) ( 3)( ( 1 lim 2 dx ( lim b 1 x, so du u by letting 4 2 6 2 ln 2 1 x, so 337 338 Section 8.3 23. Integrate e d by parts. x e 25. 0 e x dx dx x e dx 0 u du d 0 ed dv v x e ed 0 e e b→− lim b→ 0 ed lim b→− dx b b lim (1 e b→ 1) b→ 0 b lim 0 lim ( e ed b→ x e ex lim b b dx 0 C e x dx lim e e ed 0 e x dx b→− x eb) b 0 1 0 e lim b→ e be b lim ( 1 b→ Note that lim be b lim b→ lim c→ 1 ec u eb) ce c→ 0 and lim e b dx 1 1 sin 2 sin c 26. Integrate x ln x dx by parts. c ec lim c→ c u 0. c→ e dv 1 dx x v x ln x dx d 2 cos d v e sin d 2e Integrate 2e cos sin 2e cos d lim 2 cos du 2 sin d e v 12 x ln x 2 b→0 1 4 lim d cos d cos 2e sin b→0 d lim b→0 Thus, 2e sin b2 2 27. 2e cos 2e sin b tan 0 sin d lim sin d 2e 2e sin d e sin sin 2e sin d lim b→ 2e e lim b→ 2e cos lim cos C1 The integral diverges. d b lim ( e b→ sin sin e sin b e cos 0 b b cos b sin 1) lim b→0 d ln cos lim [ ln cos b b→ /2 C 0 e 12 b 4 b d b 0 b ln b 1/b 2 lim b→0 b→ /2 0 cos b→ /2 2 2e 1 0. /2 d 2e 12 x 4 1 4 e 2e 12 x 4 12 b ln b 2 Note that lim b 2 ln b 2e 12 x ln x 2 b lim d by parts. dv 1 x dx 2 x ln x dx b→0 b→0 u 12 x 2 1 x ln x 0 2e x dx 12 x ln x 2 du d by parts. dv ln x 1 du 2 1 lim e b→ 24. Integrate 2e x e b 1 0 0] 1/b 2/b 3 C 1 Section 8.3 sin 28. On [0, ], 0 1 , so 1 32. 0 1 x sin 1 d 0 4 0 b 1 d lim b→ 0 1 b lim b→ x 4 b lim 2 x b→ x lim [2 b b→ 4 4] Since this integral diverges, the given integral diverges. d 0 on [4, ) 1 dx x dx d 339 b lim 2 b→ 1 33. 0 x3 0 lim ( 2 b b→ 2 ) 1 1 on [1, x3 1 b dx x3 lim b→ 1 lim 2 12 x 2 2 b→ Since this integral converges, the given integral converges. 2xe x2 dx x2 2xe 0 dx 2xe x2 dx 0 x2 2xe b dx lim b→ 0 x2 2xe e lim [ e dx 2xe x2 lim e x 0 x lim b→1 b→1 b→0 b2 1 dx 2 lim [ 2e 2 35. 1 ln 1 2 x 1 ln 1 2 1 1 ln 2 1 x x 1 1 ln 2 1...
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