33 true theorem 782 34 false consider f x 1 35 false

This preview shows page 38 - 41 out of 50 pages.

33. True, Theorem 7.8.2. 34. False; consider f ( x ) = 1. 35. False, neither 0 nor 3 lies in [1 , 2], so the integrand is continuous. 36. False, the integral is divergent. 37. Z + 0 e - x x dx = 2 Z + 0 e - u du = 2 lim + ( - e - u ) 0 = 2 lim + ( 1 - e - ) = 2. 38. Z + 12 dx x ( x + 4) = 2 Z + 2 3 du u 2 + 4 = 2 lim + 1 2 tan - 1 u 2 2 3 = lim + tan - 1 2 - tan - 1 3 = π 6 . 39. Z + 0 e - x 1 - e - x dx = Z 1 0 du u = lim 0 + 2 u 1 = lim 0 + 2(1 - ) = 2. 40. Z + 0 e - x 1 - e - 2 x dx = - Z 0 1 du 1 - u 2 = Z 1 0 du 1 - u 2 = lim 1 sin - 1 u 0 = lim 1 sin - 1 = π 2 . 41. lim + Z 0 e - x cos x dx = lim + 1 2 e - x (sin x - cos x ) 0 = 1 / 2. 42. A = Z + 0 xe - 3 x dx = lim + - 1 9 (3 x + 1) e - 3 x 0 = 1 / 9. 43. (a) 2 . 726585 (b) 2 . 804364 (c) 0 . 219384 (d) 0 . 504067 45. 1 + dy dx 2 = 1 + 4 - x 2 / 3 x 2 / 3 = 4 x 2 / 3 ; the arc length is Z 8 0 2 x 1 / 3 dx = 3 x 2 / 3 # 8 0 = 12.
Exercise Set 7.8 401 46. 1 + dy dx 2 = 1 + x 2 4 - x 2 = 4 4 - x 2 ; the arc length is Z 2 0 r 4 4 - x 2 dx = lim 2 - Z 0 2 4 - x 2 dx = lim 2 - 2 sin - 1 x 2 0 = 2 sin - 1 1 = π . 47. Z ln x dx = x ln x - x + C , Z 1 0 ln x dx = lim 0 + Z 1 ln x dx = lim 0 + ( x ln x - x ) 1 = lim 0 + ( - 1 - ln + ), but lim 0 + ln = lim 0 + ln 1 /‘ = lim 0 + ( - ) = 0, so Z 1 0 ln x dx = - 1. 48. Z ln x x 2 dx = - ln x x - 1 x + C , Z + 1 ln x x 2 dx = lim + Z 1 ln x x 2 dx = lim + - ln x x - 1 x 1 = lim + - ln - 1 + 1 , but lim + ln = lim + 1 = 0, so Z + 1 ln x x 2 = 1. 49. Z 0 e - 3 x dx = lim + Z 0 e - 3 x dx = lim + - 1 3 e - 3 x 0 = 1 3 . 50. Z 4 8 x 2 - 4 dx = lim + Z 4 8 x 2 - 4 dx = lim + 2 ln x - 2 x + 2 4 = 2 ln 3. 51. (a) V = π Z + 0 e - 2 x dx = - π 2 lim + e - 2 x 0 = π/ 2. (b) S = π + 2 π Z + 0 e - x p 1 + e - 2 x dx , let u = e - x to get S = π - 2 π Z 0 1 p 1 + u 2 du = π + 2 π u 2 p 1 + u 2 + 1 2 ln u + p 1 + u 2 1 0 = π + π h 2 + ln(1 + 2) i . 53. (a) For x 1 , x 2 x, e - x 2 e - x . (b) Z + 1 e - x dx = lim + Z 1 e - x dx = lim + - e - x i 1 = lim + ( e - 1 - e - ) = 1 /e . (c) By parts (a) and (b) and Exercise 52(b), Z + 1 e - x 2 dx is convergent and is 1 /e . 54. (a) If x 0 then e x 1 , 1 2 x + 1 e x 2 x + 1 . (b) lim + Z 0 dx 2 x + 1 = lim + 1 2 ln(2 x + 1) 0 = + . (c) By parts (a) and (b) and Exercise 52(a), Z + 0 e x 2 x + 1 dx is divergent. 55. V = lim + Z 1 ( π/x 2 ) dx = lim + - ( π/x ) i 1 = lim + ( π - π/‘ ) = π , A = π + lim + Z 1 2 π (1 /x ) p 1 + 1 /x 4 dx ; use Exercise 52(a) with f ( x ) = 2 π/x , g ( x ) = (2 π/x ) p 1 + 1 /x 4 and a = 1 to see that the area is infinite. 56. (a) 1 x 3 + 1 x for x 2, Z + 2 1 dx = + .
402 Chapter 7 (b) Z + 2 x x 5 + 1 dx Z + 2 dx x 4 = lim + - 1 3 x 3 2 = 1 / 24. (c) Z 0 xe x 2 x + 1 dx Z + 1 xe x 2 x + 1 Z + 1 dx 2 x + 1 = + . 57. The area under the curve y = 1 1 + x 2 , above the x -axis, and to the right of the y -axis is given by Z 0 1 1 + x 2 . Solving for x = r 1 - y y , the area is also given by the improper integral Z 1 0 r 1 - y y dy. 1 2 3 4 5 0.5 1 x y 58. (b) u = x, Z + 0 cos x x dx = 2 Z + 0 cos u du ; Z + 0 cos u du diverges by part (a). 59. Let x = r tan θ to get Z dx ( r 2 + x 2 ) 3 / 2 = 1 r 2 Z cos θ dθ = 1 r 2 sin θ + C = x r 2 r 2 + x 2 + C , so u = 2 πNIr k lim + x r 2 r 2 + x 2 a = 2 πNI kr lim + ( ‘/ p r 2 + 2 - a/ p r 2 + a 2 )] = 2 πNI kr (1 - a/ p r 2 + a 2 ). 60. Let a 2 = M 2 RT to get (a) ¯ v = 4 π M 2 RT 3 / 2 1 2 M 2 RT - 2 = 2 π r 2 RT M = r 8 RT πM . (b) v 2 rms = 4 π M 2 RT 3 / 2 3 π 8 M 2 RT - 5 / 2 = 3 RT M so v rms = r 3 RT M .

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture