# B 39 1 o lim xk 1 ln x lim xk ln x by lhpitals rule

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Unformatted text preview: 22. 3/2 0 1/2 5 = − a3/2 3 cos x dx + (a) = 3/2 − sec(1) 0 √ = π 2 /9 + 2 3 π /2 28. (ln x)/2 1 0 (b) 1 π/6 2 a1/2 x − x3/2 3 (a) 20. = −55/3 2ex + csc x ln x + =1 0 12 x − sec x 2 4 π /2 12 x − 2 cot x 2 tan θ 1 = 5e3 − 5(2) = 5e3 − 10 √ = 179/2 1 π /4 −π/2 −π/4 = 31/160 1 8 4 x 3x5/3 + 14. =0 24. 27. 1 π /2 π /4 23. 12. 2 1 x dx = − 5 5x −6 4 17. 21. = 2/3 1 = 844/5 − cos θ 5ex 2 9 4 5/2 x 5 15. 19. 3 1 x dx = − x −2 11. 13. 230 − 1 − 0 12 x 2 1 x −1 3 = −11/6 0 4 = 17/12 1 3 = 2/3 1 π /2 sin x dx = − cos x =1 0 1 +(ex − x) 0 = −1 − (−1 − e−1 )+ e − 1 − 1 = e +1/e − 2 231 Chapter 7 3 33. 1.098242635; 1 1 dx = ln x x 3 35. 0 2 36. A= = ln 3 ≈ 1.098612289 1 = 12 0 1 3 − x3 + x2 − 2x 3 2 (−x2 + 3x − 2)dx = 1 3 sin x dx = −3 cos x A= = 9/2 0 A1 = = 1/6 1 A=− 38. 0 −2 39. 2 2π/3 2π/3 37. 3 13 x +x 3 (x2 + 1)dx = A= 3 −3 5 A2 = − −2 8 A3 = 13 32 x − x − 10x 3 2 (x2 − 3x − 10)dx = −1 −2 −1 1 x3 dx = − x4 4 −2 = 15/4 −2 y = 23/6, −3 (x2 − 3x − 10)dx = 343/6, 10 A1 (x2 − 3x − 10)dx = 243/6, A = A1 + A2 + A3 = 203/2 5 A3 -3 A2 5 8 x -2 40. (a) the area is positive 5 (b) 41. (a) 1 13 1 1 x − x2 − x + 100 20 25 5 −2 −1 x3 dx = π /2 −π/2 (c) 5 = −2 343 1200 14 x 4 1 = −1 14 (1 − (−1)4 ) = 0; 4 π /2 sin xdx = − cos x = − cos(π/2) + cos(−π/2) = 0 + 0 = 0 −π/2 The area on the left side of the y -axis is equal to the area on the right side, so a a −a f (x)dx = 2 −1 f (x)dx 0 1 (d) 14 1 1 1 x − x3 − x2 + x 400 60 50 5 the area between the curve and the x-axis breaks into equal parts, one above and one below the x-axis, so the integral is zero 1 (b) dx = x2 dx = 13 x 3 1 = −1 1 x2 dx; 0 π /2 π /2 −π/2 13 2 (1 − (−1)3 ) = = 2 3 3 π /2 = sin(π/2) − sin(−π/2) = 1 + 1 = 2 = 2 cos xdx = sin x cos xdx 0 −π/2 42. The numerator is an odd function and the denominator is an even function, so the integrand is an odd function and the integral is zero. 43. (a) 44. 45. (a) (a) x3 + 1 cos 2x sin √ x (b) (b) (b) F (x) = F (x) = 2 ex 14 t +t 4 1 sin 2t 2 x = 1 x = π/4 14 5 x + x − ; F (x) = x3 + 1 4 4 1 1 sin 2x − , F (x) = cos 2x 2 2 Exercise Set 7.6 1 √ 1+ x 46. (a) 47. − 49. F (x) = (b) ln x x cos x (a) 50. 232 48. √ 3x 3x2 + 1, F (x) = √ 3x2 + 1 0 F (x) = |u| √ 13 (b) (c) √ 6/ 13 (c) 0 cos x −(x2 + 3) sin x − 2x cos x , F (x) = x2 + 3 (x2 + 3)2 (a) (a) F (x) = (b) increasing on [3, +∞), decreasing on (−∞, 3] (c) 51. 0 (b) 1/3 7 + 6x − x2 (7 − x)(1 + x) = ; concave up on (−1, 7), concave down on (−∞, −1) and (x2 + 7)2 (x2 + 7)2 on (7, +∞) x−3 = 0 when x = 3, which is a relative minimum, and hence the absolute minimum, x2 + 7 by the ﬁrst derivative test. F (x) = 52. F 3 2 -20 53. -10 10 t 20 (0, +∞) because f is continuous there and 1 is in (0, +∞) (a) (b) at x = 1 because F (1) = 0 54. (−3, 3) because f is continuous there and 1 is in (−3, 3) (a) (b) at x = 1 because F (1) = 0 55. (a) fave = (b) fave = 1 9 57. x1/2 dx = 2; (a) fave = 1 2 √ 2≤ e 1 π 1 2π fave = √ x∗ = 2, x∗ = 4 0 1 e−1 (b) 56. 9 −π 3 1 1 1 dx = ln x x e−1 e = 1 1 1 1 ; , x∗ = e − 1 = e − 1 x∗ e−1 sin x dx = 0; sin x∗ = 0, x∗ = −π, 0, π √ 1 1 1 1 dx = ; ∗ 2 = , x∗ = 3 2 x 3 (x ) 3 √ √ √ x3 + 2 ≤ 29, so 3 2 ≤ 3 √ x3 + 2dx ≤ 3 29 0 58. Let f (x) = x sin x, f (0) = f (1) = 0, f (x) = sin x + x cos x = 0 when x = − tan x, x ≈ 2.0288, so f has an absolute maximum at x ≈ 2.0288; f (2.0288) ≈ 1.8197, so 0 ≤ x sin x ≤ 1.82 and π 0≤ 0 x sin xdx ≤ 1.82π = 5.72 233 59. Chapter 7 5 0 ≤ ln x ≤ ln 5 for x in [1, 5], so 0 ≤ ln xdx ≤ 4 ln 5 1 60. b a = cF (b) − cF (a) = c[F (b) − F (a)] = c F (x) (a) cF (x) (b) F (x) + G(x) b a b a = [F (b) + G(b)] − [F (a) + G(a)] = [F (b) − F (a)] + [G(b) − G(a)] = F (x) b a F (x) − G(x) (c) b a + G(x) b a b a − G(x) b a = [F (b) − G(b)] − [F (a) − G(a)] = [F (b) − F (a)] − [G(b) − G(a)] = F (x) EXERCISE SET 7.7 1. (a) the increase in height in inches, during the ﬁrst ten years (b) the change in the radius in centimeters, during the time interval t = 1 to t = 2 seconds the change in the speed of sound in ft/s, during an increase in temperature from t = 32◦ F to t = 100◦ F (d) the displacement of the particle in cm, during the time interval t = t1 to t = t2 seconds (c) 1 2. V (t)dt gal (a) 0 (b) 3. the change f (x1 ) − f (x2 ) in the values of f over the interval (a) displ = s(3) − s(0) 3 2 v (t)dt = = 0 3 dist = 3 (1 − t)dt + 0 2 (t − 3)dt = (t − t2 /2) 2 0 1 2 |v (t)|dt = (t − t2 /2) + (t2 /2 − t) 0 3 + (t2 /2 − 3t) 0 3 1 v (t)dt = = 0 1 1 1 = t2 /2 5/2 dt + 1 2 = 1/2 2 0 1 2 (5 − 2t)dt = t2 /2 2 (5 − 2t)dt + 2 3 (2t − 5)dt 5/2 3 + (t2 − 5t) 2 = 5/2 5/2 v 1 2 4 6 8 10 t -1 t v (t) = 20 + a(u)du; add areas of the small blocks to get 0 1 v (5) ≈ 20 + (1.5 + 2.7 + 4.6 + 6.2 + 7.6) = 31.3 2 3 3...
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