Business Calc Homework w answers_Part_68

3 sin 28 on 0 0 1 so 1 32 0 1 x sin 1 d 0 4 0 b 1 d

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Unformatted text preview: 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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