# Ch 5 solutions

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Unformatted text preview: /2 2 + 6 − (12 2 − 3) = 21/2 3 3 (c) x + 3)dx = 0 ( 2 0 / + 3x) 1 9 4/ + 3 − (1/ / + 1) 3= 6. ( 3 ￿ 12 ￿2 1￿2 ￿ −1 − 2 ( 3)dx dx = ( 22 / 3x (c) (c) (x +x + 3)= (x2 /x +2 +)3x) = 4=24/2 + 6 − 2 −2 −= 21/21/2 / + 6 − (1/ (1/ 3) 3) = 2 3. (a) (b) −1 (c) y y y =x+1 −y −1 1 −1 3. 3. (a) (a) 2 y y y =2−x 1 2 2 1 0 1 0 0 2 1 12 y 3 3 5. 4. 4. (a)x y = =xx /4 (a) 2 5 dx yy 7. 5 2.5 4 5√ 2 x y = xy = x −1 (b) = 81/4 − 16/4 = 65/4. (b)(b) y 5 4 (x*, 3 x dx =(x*, f(x*))f(x*)) 2x3/2 x = 1 2.5 2.5 1 0 2.5 5 16 − 2 = 14. 5 5 x 9. ￿ xln 2 ln 2 1x x 0 20 2.5 2.5 5 25 e 1 3 = (4 − 1) = . 2 2 dx = e 2 0 ￿3 3 5. x3 dx = x4 /4 = 81/4 − 16/4 = 65/4 ￿3 ￿3 ￿ 23 ￿ 3 2 5. x3 dx3= x=/x4 /4 81/81/4 − 16=465/65/4 x dx 4 4 = = 4 − 16/4 / = 4 5. 0 −1 y 1 x x (c) (c) f(x*) = 5 y=5 f(x*) f(x*) = 5 =5 y=5 y=5 9 3 9 x 3 2 1 0 1 0 0 (x*, f(x*)) (x*, f(x*)) 1 2 x x 2 3 1 2 -1 3 2 1 1 0.5 (x*, f(x*)) (x*, f(x*)) 2 x x ￿1 0 -1 -10.5 0.5 2 2 0 4 5 6. x dx = x /5 = 1/5 − (−1)/5 = 2/5 ￿1 1￿1 ￿ −1￿ 1 − 1 6. x4 dx4= x=/x5 /5 = 1=51/5 −1)/5 =52=52/5 x dx 5 5 6. / − ( − (−1)/ / ￿ 1 x y y = xy+=3x + 3 3y (x*, f(x*)) (c) (c) 0 3 1 2 y =2 x 3 3 3 + y 1 x 9 y = xy+=1x + 1 y x 3 3 (x*, f(x*)) y 1 2 x 1 1 y (b) 2 (x*, f(x*)) (c) (c) 3 2 3 1 1 −1 x 3 4 y f(x*) f(x*) = 2 =2 y 1 (x*, f(x*)) (x*, f(x*)) x 1 f(x*) = 2 y y=2 =2 (x*, f(x*)) y= y = 2 − x2 − x 3. (a) 4. (a) y=2 (b)(b) x Exercise Set 5.6 3 √ 11. (a) 141 3 2 3/2 x 3 x dx = 0 0 √ 2 4 = 2 3 = f (x∗ )(3 − 0), so f (x∗ ) = √ , x∗ = . 3 3 0 0 13 12 x+ x = 504, so f (x∗ )(0 − (−12)) = 504, (x∗ )2 + x∗ = 42, x∗ = 6, −7 but only 3 2 −12 −12 −7 lies in the interval. f (−7) = 49 − 7 = 42, so the area is that of a rectangle 12 wide and 42 high. (x2 + x) dx = (b) 1 13. −2 4 15. 1 17. 13 x − 3x2 + 12x 3 (x2 − 6x + 12) dx = 4 dx = −4x−1 x2 1 = −2 8 1 − 3 + 12 − − − 12 − 24 3 3 = 48. 4 = −1 + 4 = 3. 1 9 4 5/2 x 5 = 844/5. 4 π/2 19. − cos θ]−π/2 = 0. π/4 21. sin x]−π/4 = √ 2. 3 23. 5ex ]ln 2 = 5e3 − 5(2) = 5e3 − 10. √ 1/ 2 −1 25. sin x 0 √ = sin−1 (1/ 2) − sin−1 0 = π/4. 2 27. sec−1 x 29. √ 2 = sec−1 2 − sec−1 √ 2 t − 2t3/2 4 1 −1 1/2 |2x − 1| dx = −1 π /2 (b) 0 π/2 0 −1 2 (b) 1 1/2 (2x − 1) dx = (x − x2 ) π /2 (− cos x)dx = sin x 2−x dx + x 35. (a) 17/6 0 4 2 0 (ex − 1)dx = (x − ex ) x−2 dx = 2 ln x x −1 =2− π/2 √ = 1/2 5 . 2 2/2. 0 = −1 − (−1 − e−1 ) + e − 1 − 1 = e + 1/e − 2. 4 − 1 + 2 − 2 ln x 1 x2 , x≤1 2 (b) F (x) = 1 x3 + 1 , x > 1 3 6 2 = 2 ln 2 + 1 − 2 ln 4 + 2 ln 2 = 1. 37. False; consider F (x) = x2 /2 if x ≥ 0 and F (x) = −x2 /2 if x ≤ 0. 39. True. + ( x2 − x) 1 + ( ex − x) 2 1 −1 3π/4 − sin x 0 1 (1 − ex )dx + 1 1/2 1 (1 − 2x) dx + 3π/4 cos x dx + 33. (a) 2 = π/3 − π/4 = π/12. = −12. 1 31. (a) √ 4815450 142 Chapter 5 3 41. 0.665867079; 1 3 1 1 dx = − 2 x x = 2/3. 1 1 1 2 43. 3.106017890; sec x dx = tan x −1 −1 3 2 45. A = (x + 1)dx = 0 Page number 256 black1 49. Area = − (x2 − x) dx + 0 y 6=1 = 12. 0 2π/3 3 sin x dx = −3 cos x 0 3 13 x +x 3 2π/3 47. A = = 2 tan 1 ≈ 3.114815450. = 9/2. 0 2 1 (x2 − x) dx = 5/6 + 1/6 = 1. Chapter 5 2 y 1 A1 A1 A2 i xx 2 6 A2 –1 0 51. Area = − −1 1 (ex − 1) dx + 0 (ex − 1) dx = 1/e + e − 2. y 2 2 1 A2 –1 A1 x 1 –1 + 1 /2 = 1 y 0.8 53. (a) 1 A = 0 1 √ dx = sin−1 x 1 − x2 A2 x 0.8 = sin−1 (0.8). 0 2 A (b) The1 calculator was in degree mode instead of radian mode; the correct answer is 0.93. –1 55. (a) The increase in height in inches, during the ﬁrst ten years. –3 (b) The change in the radius in centimeters, during the time interval t = 1 to t = 2 seconds. ) (c) 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 hours. adian mode; the correct answer is 0.93. ten years the time interval t = 1 to t = 2 seconds Exercise Set 5.7 143 x x 57. (a) F (x) = 3x2 − 3. 59. (a) sin x2 61. − (b) e (3t2 − 3) dt = (t3 − 3t) (b) 1 1 = x3 − 3x + 2, and d3 (x − 3x + 2) = 3x2 − 3. dx √ x x cos x 63. F (x) = √ x2 + 9 , F (x) = √ x . x2 + 9 (a) 0 (b) 5 (c) 4 5 x−3 = 0 when x = 3, which is a relative minimum, and hence the absolute minimum, by the ﬁrst x2 + 7 derivative test. 65. (a) F (x) = (b) Increasing on [3, +∞), decreasing on (−∞, 3]. (c) F (x) = (7 − x)(1 + x) 7 + 6 x − x2 = ; concave up on (−1, 7), concave down on (−∞, −1) and on (7, +∞). 2 + 7)2 (x (x2 + 7)2 67. (a) (0, +∞) because f is continuous there and 1 is in (0, +∞). (b) At x = 1 because F (1) = 0. 69. (a) Amount of water = (rate of ﬂow)(time) = 4t gal, total amount = 4(30) = 120 gal. 60 (b) Amount of water = (4 + t/10)dt = 420 gal. 0 120 (c) Amount of water = (10 + √ 0 n 71. k=1 π sec2 4n n lim n→+∞ k=1 πk 4n π sec2 4n √ t)dt = 1200 + 160 30 ≈ 2076.36 gal. n f (x∗ )∆x where f (x) = sec2 x, x∗...
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## This document was uploaded on 02/02/2014.

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