Business Calc Homework w answers_Part_68

Business Calc Homework w answers_Part_68 - 336 Section 8.3...

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Unformatted text preview: 336 Section 8.3 1 0 dx x 2/3 8. 1 0 1 0 b dx x 2/3 1 1 dx x 2/3 1 2 b→ 1 b→0 b lim 1 4 0 4 b 2 b→4 0 dr lim b→0 r b lim b→0 1 11. 0 1 x 2 b 4) r 0.001 1000b0.001) x 1 sin 1 b→1 lim b→ lim b→ lim b→ 4 b tan 4 2 2 tan 3 4 2 b s 4 s 2 ds b b→2 1b b b 1 2b) 3 0 b→2 0 lim 4 b→2 4 2 s2 1 sin 2 lim 4 b→2 sin 1 1 2 1 s 1 s2 4 s lim s lim b 3 2 2 0 2 2 1 2 b→0 16. (1/2) dx (x/2)2 1 1x 2) d 2 2 lim ( 0) 2 2 1 (2 b 1 2 3 ln 2 b→0 x b 3 ln b lim b 0 1 1 3 ln 2 b→0 2 2 b lim 2 3 ln t b t 1 1 2 0 dx 0 1 1 2 3 ln 1 1000 15. lim (sin 12. 1 b→ 1 lim 2 dx x2 4 t lim 3 ln 1 b b→1 2 1 b→ 4 dr lim b b b 1000r b→1 b 3 dt t 1 lim 3 ln t 0.999 b dx b t 2 b→ b→0 2 ln 3 3 b→ r 4 lim (1000 1 (t 1)] dt t(t 1) 2 lim 3 ln 1 0.999 3[t lim 0 b→4 ln 1 lim r 4 lim ( 2 10. ln b→ dr 4 lim 1 3 dt t2 t 14. 0 ln x 3 6 b→4 ln 3 ln 3 3b1/3) lim r dx 1 b b→ 2 dr 9. 3 1 1 lim 1 b 3 x ln x b→ 3x b→0 dx x 2/3 x 1 3 dx x 2/3 lim (3 1)] dx 1) 1 1 ln x b→ 3) 1/3 b→0 1 1 x b lim 1 lim (x 1)(x 2 b→0 b 1) (x lim lim (3b1/3 dx x 2/3 [(x b lim b→0 1 2 2 dx lim x 2 1 b→ b 3x1/3 lim 13. dx x 2/3 lim b→0 2 dx x 2/3 b2 2 sin 2 (s/2)2 1s 2 1b 2 2 4 ds b 0 2 s2 ds Section 8.3 dx 17. First integrate (1 x) 20. 1 x b d 5 2 b→ 6 dx. b→ (1 x) 1 2 tan 1 2 2 ln lim ln 2 3 2 3 ln C x C dx 0 x) x 1 b b b (1 lim 2 tan lim (2 tan 1 b→0 b→0 1 lim (2 tan c→ 2 1 x x2 1 x 1) b 16 tan 1 x dx 1 x2 0 1 x2 1 b→ lim 1 b→1 b x x2 1 8 4 x 1 22. 1 2 1 sec b) 0 2 sec b 2 1 lim (sec dx x 3 b→0 3 1 2 x 1 lim ( 2 x 4 b sec x x b b→0 b 1 2) 0 4 dx lim b→0 x 2) b 4 dx b x 2 b→0 6 6 4 b ds s s2 sec 1 lim (sec 1 lim b→1 1 1 2 s b b→1 1 2 2 sec sec 1 1 1 b) 3 dx x 2 x b 4 b→0 2 2 lim sec 1 lim lim 3 b→1 dx lim x dx 1 2 4 2 0 dx 0 b lim ds b)2 b 1 2 1 4 b→0 x x2 b→ 1 1 2 2 dx 1 x dx 1 0 dx 1 1 b 1 b b→ x)2 0 2 1 2 1 sec lim 1 2 x2 x lim sec s s2 0 b→ dx x b→ 1 16 tan 1 x dx 1 x2 lim dx 2 1 dx 2 C lim 8(tan dx sec 19. x)2 b→ lim (sec x x2 1 8(tan C b) b→1 1 8u2 16u du 2 2 tan . 1 1 b→1 x x2 c 1 x2 1 16 tan 1 x dx 1 x2 x c→ b lim 2 x) lim 2 tan 2 tan c (1 dx du x) x lim 8 (tan dx x x2 1 2 tan 1 dx 1 1 x 0 2 18. c→ x 1 lim (1 dx c x) 16 tan 1 x dx by letting u 1 x2 1 2 dx 1 x) x dx lim b→0 dx (1 21. Integrate 3 1 lim ln Now evaluate the 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 b b 0 1 0 b] x dt 0 x t lim b→0 b 2 lim [2 b→0 x lim b→1 b→1 2 0 1 dx b x 0 b 0 0 dx 1 x dx 1 x ln 1 lim ( ln 1 x 0 b 0) Since this integral diverges, the given integral diverges. on (0, ] since sin t 0 on [0, ]. dt lim b→0 x) b t sin t 1 lim The integral converges. t 2 dx b dx 1 0 b→1 1 1 2(1 1 dx 1 2e b→0 1 1 x) b→1 b 2 x) (1 x)] dx x)(1 x) 2(1 b→1 x 2e 2e x2 lim dx x b→0 1/2[(1 (1 0 0 e x b lim dx 1 1 Since this integral diverges, the given integral diverges. e lim 2 lim 1 4 31. 0 0 b→1 4 dx x lim The integral converges. e 1 1 0 2 dx b b b→ 1 2 lim x2 x2 lim [ 1 4 x 2 dx b lim 1 2 b 1] 2xe b→ b→ 30. x 1 2 0 0 dx 0 1 dx 1 b b→ 0 2 34. 0 b2 lim b→ 1 Since this integral converges, the given integral converges. 0 x2 b 12 b 2 b→ 29. dx x3 lim 20 ) t t b 2 b] 2 Since this integral converges, the given integral converges. 340 Section 8.3 1 1 36. ln x dx 2 ln x dx by symmetry of ln x about the 1 0 40. 1 dx 0 y-axis. Integrate ln x dx by parts. 1 dx 0 x dx dx 1 x b lim b→ x x dx 1 x b u ln x dv 1 dx x du ln x dx v lim 2 lim (2 x ln x dx x ln x x 2 lim b→0 0 b→0 1 x ln x x b 2 lim [ 1 b ln b b→0 Note that lim b ln b b] b→0 1/b 1/b 2 b→0 1 42. 0 0. 2 1 on [2, ) x2 b lim b→ 2 2x 1 2b 2 1 dx x lim (ln b b→ dx b 2 2 b lim ln x b→ lim (ln b b→ ln 2) 1 ex e 1 x. ex e x(e2x 1) 1 (e x)2 e x dx Integrate by letting u e x so du 1 (e x)2 dx e x dx x x e e 1 (e x)2 du 1 u2 e x tan 1 u tan 2 1 ex C Since this integral diverges, the given integral diverges. 39. Let f (x) x1 and g(x) x2 1 . Both are continuous on x3/2 [1, ). x1 f (x) lim lim x→ g(x) x→ x→ x b 1 dx lim x 3/2 dx x3/2 1 b→ 1 b→ lim ( 2b b→ 1/2 x dx e ex x 0 x 1 lim b→ tan 1 1 b→ 4 2) 2 Since this integral converges, the given integral converges. 0 dx ex e 0 b x lim b→ x 0 ex 0 4 dx ex e 1 ex lim [tan 1 b b→ tan 1 lim tan 1 e b] tan 1 1] x b 0 b→ 2 dx e ex b b→ b 0 dx e ex b lim 1 1/2 dx e dx ex e C lim [tan 1 lim 2x 1 x 0 ex 0 lim ln ) ) 1 1 ex Since this integral converges, the given integral converges. dx x 2x b→ 43. First rewrite 1 e 1 x b Since this integral converges, the given integral converges. e 1] b→ 38. 0 b lim lim 1 b 2 on [ , x2 b→ b lim [ e sin x x2 b→ e b→ cos x on [ , ) x b dx lim lim ln x x b→ b→ lim 1 1 on [1, ) 1e e b 1 d lim ed e 1 b→ lim 2 2 dx x2 The integral converges. 1 2) Since this integral diverges, the given integral diverges. 2 lim ln b 1/b lim b→0 b 1 x dx x 41. 0 ln x dx b 2 lim 37. 0 1 Since this integral diverges, the given integral diverges. C 1 lim b b→ ln x dx b→0 x b→ x 1 2 dx 4 e 4 Thus, the given integral converges. x e x dx. ...
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This document was uploaded on 10/31/2011 for the course MAC 2311 at University of Florida.

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