16 dy 1 4 sec2 2x dx 2 tan 2x c 1 2 tan 2

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Unformatted text preview: /3 so e−u du = 2 lim 0 e−x dx = 1 − e−x +∞ →+∞ +∞ dx = 2 dx =2 x(x + 4) 36. A = 37. = lim (π/3 − sec−1 ) = π/3, dx √ = lim sec−1 x →+∞ x x2 − 1 0 32. 2 = 1/2 (d) 0.504067 333 Chapter 9 dy dx 40. 1 + 2 3/2 16x2 9 + 12x2 = ; the arc length is 2 9 − 4x 9 − 4x2 =1+ 9 + 12x2 dx ≈ 3.633168 9 − 4x2 0 ln x dx = x ln x − x + C , 41. 1 ln x dx = lim (x ln x − x) + →0 0 1 1 ln x dx = lim + but lim ln = lim + + →0 = lim (−1 − ln + ), + →0 →0 →0 1 ln = lim (− ) = 0 so 1/ →0+ ln x dx = −1 0 ln x 1 ln x − + C, dx = − 2 x x x 42. +∞ 1 ln x dx = lim →+∞ x2 but lim ln 1 +∞ 1 = lim →+∞ ln x 1 ln x − dx = lim − →+∞ x2 x x = 0 so →+∞ 1 − = lim →+∞ 1 ln − 1 +1 , ln x =1 x2 1 1 xe−3x dx = − xe−3x − e−3x + C , 3 9 43. +∞ xe−3x dx = lim →+∞ 0 = lim →+∞ +∞ x2 3 0 +∞ e−2x dx = − 0 +∞ (b) S = 2π e−x +∞ 1 = 0 so →+∞ 3e3 = lim x−2 8 dx = lim 2 ln →+∞ −4 x+2 45. (a) V = π 0 1 1 1 − e−3 − e−3 + 3 9 9 →+∞ e3 →+∞ 44. A = →+∞ e−3 = lim but lim 1 1 − xe−3x − e−3x 3 9 xe−3x dx = lim π 2 xe−3x dx = 1/9 0 1 −2 − ln = 2 ln 5 +2 5 = lim 2 ln 3 →+∞ lim e−2x →+∞ = π/2 0 1 + e−2x dx, let u = e−x to get 0 0 S = −2π 1 + u2 du = 2π 1 u 2 1 + u2 + 1 ln | u + 2 1 1 + u2 | =π √ √ 2 + ln(1 + 2) 0 47. (a) For x ≥ 1, x2 ≥ x, e−x ≤ e−x 2 +∞ (b) e−x dx = lim →+∞ 1 e−x dx = lim −e−x 1 →+∞ +∞ (c) By parts (a) and (b) and Exercise 46(b), 1 = lim (e−1 − e− ) = 1/e →+∞ e−x dx is convergent and is ≤ 1/e. 2 1 ex 1 ≤ 2x + 1 2x + 1 dx 1 = lim ln(2x + 1) →+∞ 2 2x + 1 48. (a) If x ≥ 0 then ex ≥ 1, (b) lim →+∞ 0 = +∞ 0 +∞ (c) By parts (a) and (b) and Exercise 46(a), 0 ex dx is divergent. 2x + 1 Exercise Set 9.8 334 (π/x2 ) dx = lim −(π/x) 49. V = lim →+∞ →+∞ 1 1 = lim (π − π/ ) = π →+∞ 2π (1/x) 1 + 1/x4 dx; use Exercise 46(a) with f (x) = 2π/x, g (x) = (2π/x) 1 + 1/x4 A = lim →+∞ 1 and a = 1 to see that the area is infinite. √ +∞ x3 + 1 for x ≥ 2, 1dx = +∞ 50. (a) 1 ≤ x 2 +∞ (b) x5 2 (c) x dx ≤ +1 +∞ 2 1 dx = lim − 3 →+∞ x4 3x ex 1 ≤ for x ≥ 0, 2x + 1 2x + 1 2x 2x 1 + t3 dt ≥ 51. t3/2 dt = 0 0 2x 0 2 5/2 t 5 1 1 dx = lim ln(2x + 1) →+∞ 2 2x + 1 2x = 0 2 (2x)5/2 = +∞ so x→+∞ 5 0 = +∞ 0 2 (2x)5/2 , 5 +∞ t3/2 dt = lim lim x→+∞ +∞ = 1/24 2 1 + t3 dt = +∞; by L’Hˆpital’s Rule o 0 2x 1 + t3 dt 0 lim x5/2 x→+∞ 52. (b) u = √ √ 2 1 + (2x)3 2 1/x3 + 8 = 8 2/5 = lim x→+∞ (5/2)x3/2 x→+∞ 5/2 = lim +∞ x, 0 √ cos x √ dx = 2 x +∞ +∞ cos u du; 0 1 dx =2 r (r2 + x2 )3/2 53. Let x = r tan θ to get cos u du diverges by part (a). 0 cos θ dθ = 1 x sin θ + C = √ +C r2 r2 r2 + x2 2πN Ir k 54. Let a2 = lim →+∞ x √ r2 r2 + x2 = 2πN I lim ( / r2 + kr →+∞ = so u = 2πN I (1 − a/ r2 + a2 ). kr a 2 − a/ r2 + a2 ) M to get 2RT 4 (a) v = √ ¯ π M 2RT 4 2 (b) vrms = √ π 3/2 M 2RT 1 2 3/2 −2 2 2RT 8RT M = =√ 2RT M πM π √ −5/2 3RT 3π M so vrms = = 8 2RT M 3RT M 55. (a) Satellite’s weight = w(x) = k/x2 lb when x = distance from center of Earth; w(4000) = 6000 4000+ so k = 9.6 × 1010 and W = 9.6 × 1010 x−2 dx mi·lb. 4000 +∞ (b) 9.6 × 1010 x−2 dx = lim −9.6 × 1010 /x →+∞ 4000 +∞ 56. (a) L{1} = 0 (b) L{e2t } = 0 4000 1 e−st dt = lim − e−st →+∞ s +∞ +∞ e−st e2t dt = 0 = 2.4 × 107 mi·lb = 0 1 s e−(s−2)t dt = lim − →+∞ 1 −(s−2)t e s−2 = 0 1 s−2 335 Chapter 9 +∞ (c) L{sin t} = e−st (−s sin t − cos t) →+∞ s2 + 1 e−st sin t dt = lim 0 +∞ (d) L{cos t} = +∞ 57. (a) L{f (t)} = te−st dt = lim −(t/s + 1/s2 )e−st +∞ 2 −st te +∞ 3 58. 0 = 0 2 →+∞ 100 1000 0.8862269 0.8862269 e−3s = s 10,000 0.8862269 3 s +1 2 s3 = 0 1 e−st dt = lim − e−st →+∞ s 10 s2 1 s2 dt = lim −(t /s + 2t/s + 2/s3 )e−st 2 0 (c) L{f (t)} = = →+∞ 0 (b) L{f (t)} = 0 e−st (−s cos t + sin t) e−st cos t dt = lim 2 →+∞ s + 1 0 1 s2 + 1 = 0.8862269 √ 59. (a) u = √ (b) x = 3 60. (a) 0 ax, du = 2σu, dx = 2 +∞ −x2 3 dx = 0 x6 x6 0 2 4 e 4 0 x6 +∞ +∞ π /a e−u du = 1 2 0 e +∞ xe−x dx = dx < 2 3 1 −9 e < 7 × 10−5 2 +∞ 4 = +∞, 1 0 1 if p = 1, then 0 1 px e p →+∞ 63. If p = 1, then dx = lim ln x x →0+ 4 1 dx so x6 + 1 0 epx dx = lim 0 +∞ 1 1 dx = < 2 × 10−4 6 x 5(4)5 →+∞ if p = 0, then = lim →+∞ 0 −1/p, p < 0 . +∞, p>0 1p (e − 1) = p 1 dx x1−p = lim p x →0+ 1 − p = +∞; 1 = lim [(1 − →0+ 1−p 1/(1 − p), p < 1 . +∞, p>1 )/(1 − p)] = 1 − x, u2 = 1 − x, 2u du = −dx; 0 1 2 − u2 du = 2 1 √ 2 − u2 du = u 2 − u2 + 2 sin−1 (u/ 2) 0 1 0 cos(u2 )du ≈ 1.809 0 +∞ −x2 dx so E = 1 dx + +1 (1)dx = lim x 0 65. 2 2 dx = √ π 3 +∞ 62. If p = 0, then −2 0 /2σ 2 3 1 dx < 6+1 x E= √ 2 0 +∞ −x2 e−x dx + 1 dx = +1 +∞ 64. u = e−x 2 1 dx ≈ 1.047;...
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This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

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