Nn 1 2 n k 13 k 1 n 1n 2 2 a f is continuous

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Unformatted text preview: ation is increasing most rapidly when t = 8.4 years. 23. Solve φ − 0.0167 sin φ = 2π (90)/365 to get φ = 1.565978 so r = 150 × 106 (1 − 0.0167 cos φ) = 149.988 × 106 km. 24. Solve φ − 0.0934 sin φ = 2π (1)/1.88 to get φ = 3.325078 so r = 228 × 106 (1 − 0.0934 cos φ) = 248.938 × 106 km. CHAPTER 7 Integration EXERCISE SET 7.1 1. A = 1(1)/2 = 1/2; ∆x = (b − a)/n = 1/n, x∗ = k/n, f (x∗ ) = k/n, An = k k n An 2. 1 2 3 4 5 6 7 8 9 10 1.0000 0.7500 0.6666 0.6250 0.6000 0.5833 0.5714 0.5625 0.5556 0.5500 A = 4(2)/2 = 4; ∆x = (b − a)/n = 2/n, x∗ = 2k/n, f (x∗ ) = 4 − 2(2k/n) = 4 − 4k/n = 4(1 − k/n), k k 2 n 2 1 An = 4 1 − + 1− + ··· + 1 − n n n n n An 3. 2 n1 1 + + ··· + nn nn 1 2 3 4 5 6 7 8 9 10 0.0000 2.0000 2.6667 3.0000 3.2000 3.3333 3.4286 3.50000 3.5556 3.6000 A = 2(2 + 14)/2 = 16; ∆x = (b − a)/n = 2/n, x∗ = 2k/n, f (x∗ ) = 2 + 12k/n = 2(1 + 6k/n), k k 2 12 6n 6 An = 2 1 + + 1+ + ··· + 1 + n n n n n An 1 2 3 4 5 6 7 8 9 10 28.0000 22.0000 20.0000 19.0000 18.4000 18.0000 17.7143 17.5000 17.3333 17.2000 4. A = π (1)2 /4 = π/4; ∆x = (b − a)/n = 1/n, x∗ = k/n, f (x∗ ) = 1 − (k/n)2 , k k 1 1 + (1/n)2 + 1 + (2/n)2 + · · · + 1 + (n/n)2 An = n n 1 2 3 4 5 6 7 8 9 10 An 0.0000 0.4330 0.5627 0.6239 0.6593 0.6822 0.6982 0.7100 0.7190 0.7261 5. A(1) − A(0) = 1/2 6. A(2) − A(0) = 4 7. A(2) − A(0) = 16 8. A(1) − A(0) = 9. A(x) = ex , area = A(1) − A(0) = e − 1 10. 1π = π/4 22 A(x) = − cos x, area = A(π ) − A(0) = 2 EXERCISE SET 7.2 1. (a) 2. (a) (b) √ x dx = 1 + x2 1 + x2 + C (b) (x + 1)ex dx = xex + C d (sin x − x cos x + C ) = cos x − cos x + x sin x = x sin x dx √ √ x 1 d 1 − x2 + x2 / 1 − x2 √ +C = = 2 dx 1 − x2 (1 − x2 )3/2 1−x 3x2 x3 + 5 = √ 2 x3 + 5 3. d dx 4. x d 3 − x2 =2 dx x2 + 3 (x + 3)2 so so 3x2 √ dx = 2 x3 + 5 x3 + 5 + C 3 − x2 x dx = 2 +C (x2 + 3)2 x +3 209 Exercise Set 7.2 210 √ cos (2 x) √ x √ √ cos (2 x) √ dx = sin 2 x + C x 5. √ d sin 2 x dx 6. d [sin x − x cos x] = x sin x dx 7. (a) x9 /9 + C (b) 7 12/7 x +C 12 (c) 2 9/2 x +C 9 8. (a) 3 5/3 x +C 5 (b) 1 1 − x−5 + C = − 5 + C 5 5x (c) 8x1/8 + C 9. (a) 1 2 10. = so so x sin x dx = sin x − x cos x + C 1 x−3 dx = − x−2 + C 4 (b) u4 /4 − u2 + 7u + C 3 5/3 x − 5x4/5 + 4x + C 5 11. 2 12 1 1 (x−3 + x1/2 − 3x1/4 + x2 )dx = − x−2 + x3/2 − x5/4 + x3 + C 2 3 5 3 12. 3 8 (7y −3/4 − y 1/3 + 4y 1/2 )dy = 28y 1/4 − y 4/3 + y 3/2 + C 4 3 13. (x + x4 )dx = x2 /2 + x5 /5 + C 14. 4 1 (4 + 4y 2 + y 4 )dy = 4y + y 3 + y 5 + C 3 5 15. x1/3 (4 − 4x + x2 )dx = 16. 1 2 1 (2 − x + 2x2 − x3 )dx = 2x − x2 + x3 − x4 + C 2 3 4 17. (x + 2x−2 − x−4 )dx = x2 /2 − 2/x + 1/(3x3 ) + C 18. 1 (t−3 − 2)dt = − t−2 − 2t + C 2 20. 22. (4x1/3 − 4x4/3 + x7/3 )dx = 3x4/3 − √ 1 −1 √ t 1 t − 2e dt = ln t − 2et + C 2 2 4 tan x − csc x + C 19. 12 7/3 3 x + x10/3 + C 7 10 2 ln x + 3ex + C 21. −4 cos x + 2 sin x + C 23. (sec2 x + sec x tan x)dx = tan x + sec x + C 24. (sec x tan x + 1)dx = sec x + x + C 25. ln θ − 2eθ + cot θ + C 26. sin y dy = − cos y + C 27. sec x tan x dx = sec x + C 28. (φ + 2 csc2 φ)dφ = φ2 /2 − 2 cot φ + C 29. (1 + sin θ)dθ = θ − cos θ + C 211 Chapter 7 30. 2 sin x cos x dx = 2 cos x 31. 1 − sin x dx = 1 − sin2 x 33. (a) sin x dx = −2 cos x + C 1 − sin x = cos2 x sec2 x − sec x tan x dx = tan x − sec x + C (b) f (x) = x2 /2 + 5 (b) y f (x) = ex /2 + 1/2 6 4 2 -1 34. 1 (a) 2 x y 6 4 2 -1 1 35. 2 x 36. y y 5 4 π/4 π/2 2 x x 1 2 3 4 5 -5 37. f (x) = m = − sin x so f (x) = 38. f (x) = m = (x + 1)2 , so f (x) = C =8+ (− sin x)dx = cos x + C ; f (0) = 2 = 1 + C so C = 1, f (x) = cos x + 1 (x + 1)2 dx = 1 25 1 25 = , f (x) = (x + 1)3 + 3 3 3 3 1 1 1 (x + 1)3 + C ; f (−2) = 8 = (−2 + 1)3 + C = − + C , 3 3 3 3 3 4/3 3 5 x + C , y (1) = + C = 2, C = 5/4; y (x) = x4/3 + 4 4 4 4 (a) y (x) = x1/3 dx = (b) y (t) = t−1 dt = ln |t| + C , y (−1) = C = 5, C = 5; y (t) = ln |t| + 5 (c) 39. y (x) = (x1/2 + x−1/2 )dx = y (x) = 40. (a) y (x) = 8 2 3/2 8 x + 2x1/2 + C , y (1) = 0 = + C , C = − , 3 3 3 2 3/2 8 x + 2x1/2 − 3 3 1 1 1 1 1 1 −3 x ; y (x) = − x−2 + dx = − x−2 + C , y (1) = 0 = − + C , C = 8 16 16 16 16 16 Exercise Set 7.2 (b) (c) 212 √ √ π 2 2 (sec2 t − sin t) dt = tan t + cos t + C , y ( ) = 1 = 1 + + C, C = − ; 4 2 2 √ 2 y (t) = tan t + cos t − 2 2 9/2 2 y (x) = x7/2 dx = x + C , y (0) = 0 = C , C = 0; y (x) = x9/2 9 9 y (t) = 2 3/2 4 5/2 x + C1 ; f (x) = x + C1 x + C2 3 15 41. f (x) = 42. f (x) = x2 /2 + sin x + C1 , use f (0) = 2 to get C1 = 2 so f (x) = x2 /2 + sin x + 2, f (x) = x3 /6 − cos x + 2x + C2 , use f (0) = 1 to get C2 = 2 so f (x) = x3 /6 − cos x + 2x + 2 43. dy/dx = 2x + 1, y = (2x + 1)dx = x2 + x + C ; y = 0 when x = −3 so (−3)2 + (−3) + C = 0, C = −6 thus y = x2 + x − 6 44. dy/dx = x2 , y = x2 dx = x3 /3 + C ; y = 2 when x = −1 so (−1)3 /3 + C = 2, C = 7/3 thus y = x3 /3 + 7/3 45. 6xdx = 3x2 + C1 . The slope of the tangent line is −3 so dy/dx = −3 when x = 1. Thus dy/dx = 3(1)2 + C1 = −3, C1 = −6 so dy/dx = 3x2 − 6, y = (3x2 − 6)dx = x3 − 6x + C2 ; If x = 1, then y = 5 − 3(1) = 2 so (1)2 − 6(1) + C2 = 2, C2 = 7 thus y = x3 − 6x + 7. 46. dT /dx = C1 , T = C1 x + C2 ; T = 25 when x = 0 so C2 = 25, T = C1 x + 25. T = 85 when x = 50 so 50C1 + 25 = 85, C1 = 1.2, T = 1.2x + 25 47. (a) F (x) = G (x) = 3x + 4 (b) F (0) = 16/6 = 8/3...
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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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