January 27, 2005 11:45    L24-ch06       Sheet number 1 Page number 230               black                               ...
January 27, 2005 11:45      L24-ch06           Sheet number 2 Page number 231                  black              Exercise...
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January 27, 2005 11:45       L24-ch06         Sheet number 4 Page number 233                 black              Exercise S...
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January 27, 2005 11:45        L24-ch06                 Sheet number 6 Page number 235                    black            ...
January 27, 2005 11:45        L24-ch06                 Sheet number 7 Page number 236                    black            ...
January 27, 2005 11:45        L24-ch06         Sheet number 8 Page number 237        black              Exercise Set 6.3  ...
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January 27, 2005 11:45          L24-ch06           Sheet number 11 Page number 240               black             240    ...
January 27, 2005 11:45      L24-ch06           Sheet number 12 Page number 241              black              Exercise Se...
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Chapter 06
Chapter 06
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Chapter 06
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Chapter 06
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Chapter 06

  1. 1. January 27, 2005 11:45 L24-ch06 Sheet number 1 Page number 230 black CHAPTER 6 Integration EXERCISE SET 6.1 1 2 n−1 1. Endpoints 0, , ,..., , 1; using right endpoints, n n n 1 2 n−1 1 An = + + ··· + +1 n n n n n 2 5 10 50 100 An 0.853553 0.749739 0.710509 0.676095 0.671463 1 2 n−1 , ,..., 2. Endpoints 0, , 1; using right endpoints, n n n n n n n 1 1 An = + + + ··· + + n+1 n+2 n+3 2n − 1 2 n n 2 5 10 50 100 An 0.583333 0.645635 0.668771 0.688172 0.690653 π 2π (n − 1)π 3. Endpoints 0, , ,..., , π; using right endpoints, n n n π An = [sin(π/n) + sin(2π/n) + · · · + sin(π(n − 1)/n) + sin π] n n 2 5 10 50 100 An 1.57080 1.93376 1.98352 1.99935 1.99984 π 2π (n − 1)π π 4. Endpoints 0, , ,..., , ; using right endpoints, 2n 2n 2n 2 π An = [cos(π/2n) + cos(2π/2n) + · · · + cos((n − 1)π/2n) + cos(π/2)] 2n n 2 5 10 50 100 An 0.555359 0.834683 0.919400 0.984204 0.992120 n+1 n+2 2n − 1 5. Endpoints 1, , ,..., , 2; using right endpoints, n n n n n n 1 1 An = + + ··· + + n+1 n+2 2n − 1 2 n n 2 5 10 50 100 An 0.583333 0.645635 0.668771 0.688172 0.690653 π π π π 2π π (n − 1)π π 6. Endpoints − , − + , − + ,...,− + , ; using right endpoints, 2 2 n 2 n 2 n 2 π π π 2π π (n − 1)π π π An = cos − + + cos − + + · · · + cos − + + cos 2 n 2 n 2 n 2 n n 2 5 10 50 100 An 1.57080 1.93376 1.98352 1.99936 1.99985 230
  2. 2. January 27, 2005 11:45 L24-ch06 Sheet number 2 Page number 231 black Exercise Set 6.1 231 1 2 n−1 7. Endpoints 0, , , . . . , , 1; using right endpoints, n n n   2 2 2 1 2 n−1 1 An =  1 − + 1− + ··· + 1 − + 0 n n n n n 2 5 10 50 100 An 0.433013 0.659262 0.726130 0.774567 0.780106 2 4 2(n − 1) 8. Endpoints −1, −1 + , −1 + , . . . , −1 + , 1; using right endpoints, n n n   2 2 2 n−2 n−4 n−2 2 An =  1 − + 1− + ··· + 1 − + 0 n n n n n 2 5 10 50 100 An 1 1.423837 1.518524 1.566097 1.569136 2 4 2 9. Endpoints −1, −1 + , −1 + , . . . , 1 − , 1; using right endpoints, n n n An = e−1+ n + e−1+ n + e−1+ n + . . . + e1− n + e1 2 4 6 2 2 n n 2 5 10 50 100 An 3.718281 2.851738 2.59327 2.39772 2.35040 1 2 1 10. Endpoints 1, 1 + , 1 + , . . . , 2 − , 2; using right endpoints, n n n 1 2 1 1 An = ln 1 + + ln 1 + + . . . + ln 2 − + ln 2 n n n n n 2 5 10 50 100 An 0.549 0.454 0.421 0.393 0.390 1 2 n−1 11. Endpoints 0, , ,..., , 1; using right endpoints, n n n 1 2 n−1 1 An = sin−1 + sin−1 + . . . + sin−1 + sin−1 (1) n n n n n 2 5 10 50 100 An 1.04729 0.75089 0.65781 0.58730 0.57894 1 2 n−1 12. Endpoints 0, , ,..., , 1; using right endpoints, n n n 1 2 n−1 1 An = tan−1 + tan−1 + . . . + tan−1 + tan−1 (1) n n n n n 2 5 10 50 100 An 0.62452 0.51569 0.47768 0.44666 0.44274 3 13. 3(x − 1) 14. 5(x − 2) 15. x(x + 2) 16. (x − 1)2 2
  3. 3. January 27, 2005 11:45 L24-ch06 Sheet number 3 Page number 232 black 232 Chapter 6 3 17. (x + 3)(x − 1) 18. x(x − 2) 2 19. The area in Exercise 17 is always 3 less than the area in Exercise 15. The regions are identical except that the area in Exercise 15 has the extra trapezoid with vertices at (0, 0), (1, 0), (0, 2), (1, 4) (with area 3). 1 20. (a) The region in question is a trapezoid, and the area of a trapezoid is (h1 + h2 )w. 2 1 1 (b) From Part (a), A (x) = [f (a) + f (x)] + (x − a) f (x) 2 2 1 1 f (x) − f (a) = [f (a) + f (x)] + (x − a) = f (x) 2 2 x−a 21. A(6) represents the area between x = 0 and x = 6; A(3) represents the area between x = 0 and x = 3; their difference A(6) − A(3) represents the area between x = 3 and x = 6, and 1 A(6) − A(3) = (63 − 33 ) = 63. 3 22. A(9) = 93 /3, A(−3) = (−3)3 /3, and the area between x = −3 and x = 9 is given by A(9)−A(−3) = (93 − (−3)3 )/3 = 252. 23. f (x) = A (x) = 2x; 0 = A(a) = a2 − 4, so take a = 2 (or a = −2), A(x) = x2 − 4, A (x) = 2x = f (x), so a = ±2, f (x) = 2x 24. f (x) = A (x) = 2x − 1, 0 = A(a) = a2 − a, so take a = 0 (or a = 1). √ 25. B is also the area between the graph of f (x) = x and the interval [0, 1] on the y−axis, so A + B is the area of the square. 26. If the plane is rotated about the line y = x then A becomes B and vice versa. EXERCISE SET 6.2 x 1. (a) √ dx = 1 + x2 + C (b) (x + 1)ex dx = xex + C 1 + x2 d 2. (a) (sin x − x cos x + C) = cos x − cos x + x sin x = x sin x dx √ √ d x 1 − x2 + x2 / 1 − x2 1 (b) √ +C = = dx 1 − x2 1 − x2 (1 − x2 )3/2 d 3x2 3x2 5. x3 + 5 = √ so √ dx = x3 + 5 + C dx 2 x3 + 5 2 x3 + 5 d x 3 − x2 3 − x2 x 6. = 2 so dx = 2 +C dx x2 + 3 (x + 3)2 (x2 + 3)2 x +3 √ √ d √ cos (2 x) cos (2 x) √ 7. sin 2 x = √ so √ dx = sin 2 x + C dx x x d 8. [sin x − x cos x] = x sin x so x sin x dx = sin x − x cos x + C dx
  4. 4. January 27, 2005 11:45 L24-ch06 Sheet number 4 Page number 233 black Exercise Set 6.2 233 7 12/7 2 9/2 9. (a) x9 /9 + C (b) x +C (c) x +C 12 9 3 5/3 1 1 10. (a) x +C (b) − x−5 + C = − 5 + C (c) 8x1/8 + C 5 5 5x 2 2 1 5 2 −1 1 5 1 11. 5x + dx = 5x dx + dx = x2 + C = x2 − 4 + C 3x5 3 x5 2 3 4 x4 2 6x 1 1 5 1 12. x−1/2 − 3x7/5 + dx = x−1/2 dx − 3 x7/5 dx + dx = 2x1/2 − 3 x12/5 + x + C 9 9 12 9 1 12 8 13. x−3 − 3x1/4 + 8x2 dx = x−3 dx − 3 x1/4 dx + 8 x2 dx = − x−2 − x5/4 + x3 + C 2 5 3 10 √ 4 1 √ 1 14. 3/4 − 3y+ √ dy = 10 dy − 3 y dy + 4 √ dy y y y 3/4 y 3 3 √ = 10(4)y 1/4 − y 4/3 + 4(2)y 1/2 + C = 40y 1/4 − y 4/3 + 8 y + C 4 4 4 1 15. (x + x4 )dx = x2 /2 + x5 /5 + C 16. (4 + 4y 2 + y 4 )dy = 4y + y 3 + y 5 + C 3 5 12 7/3 3 17. x1/3 (4 − 4x + x2 )dx = (4x1/3 − 4x4/3 + x7/3 )dx = 3x4/3 − x + x10/3 + C 7 10 1 2 1 18. (2 − x + 2x2 − x3 )dx = 2x − x2 + x3 − x4 + C 2 3 4 19. (x + 2x−2 − x−4 )dx = x2 /2 − 2/x + 1/(3x3 ) + C 1 2 20. (t−3 − 2)dt = − t−2 − 2t + C 21. + 3ex dx = 2 ln |x| + 3ex + C 2 x 1 −1 √ t 1 √ 22. t − 2e dt = ln |t| − 2et + C 2 2 23. [3 sin x − 2 sec2 x] dx = −3 cos x − 2 tan x + C 24. [csc2 t − sec t tan t] dt = − cot t − sec t + C 25. (sec2 x + sec x tan x)dx = tan x + sec x + C 26. csc x(sin x + cot x) dx = (1 + csc x cot x) dx = x − csc x + C sec θ 27. dθ = sec2 θ dθ = tan θ + C 28. sin y dy = − cos y + C cos θ 29. sec x tan x dx = sec x + C 30. (φ + 2 csc2 φ)dφ = φ2 /2 − 2 cot φ + C
  5. 5. January 27, 2005 11:45 L24-ch06 Sheet number 5 Page number 234 black 234 Chapter 6 1 1 1 1 31. (1 + sin θ)dθ = θ − cos θ + C 32. sec2 x + dx = tan x + x + C 2 2 2 2 1 3 1 33. √ − dx = sin−1 x − 3 tan−1 x + C 2 1−x2 1 + x2 2 4 1 + x + x3 1 1 34. √ + dx = 4 sec−1 x+ x+ dx = 4 sec−1 x+ x2 +tan−1 x+C x x2 − 1 1 + x2 x2 +1 2 1 − sin x 1 − sin x 35. dx = dx = sec2 x − sec x tan x dx = tan x − sec x + C 1 − sin2 x cos2 x 1 1 1 1 36. dx = dx = sec2 x dx = tan x + C 1 + cos 2x 2 cos2 x 2 2 37. y 38. y 5 2 x x c/4 c/2 1 2 –4 –5 39. f (x) = m = − sin x so f (x) = (− sin x)dx = cos x + C; f (0) = 2 = 1 + C so C = 1, f (x) = cos x + 1 1 40. f (x) = m = (x + 1)2 , so f (x) = (x + 1)2 dx = (x + 1)3 + C; 3 1 1 1 25 1 25 f (−2) = 8 = (−2 + 1)3 + C = − + C, = 8 + = , f (x) = (x + 1)3 + 3 3 3 3 3 3 3 4/3 3 5 3 5 41. (a) y(x) = x1/3 dx = x + C, y(1) = + C = 2, C = ; y(x) = x4/3 + 4 4 4 4 4 π 1 π π (b) y(t) = (sin t + 1) dt = − cos t + t + C, y = − + + C = 1/2, C = 1 − ; 3 2 3 3 π y(t) = − cos t + t + 1 − 3 2 3/2 8 8 (c) y(x) = (x1/2 + x−1/2 )dx = x + 2x1/2 + C, y(1) = 0 = + C, C = − , 3 3 3 2 3/2 8 y(x) = x + 2x1/2 − 3 3 1 −3 1 1 1 1 1 42. (a) y(x) = x dx = − x−2 + C, y(1) = 0 = − + C, C = ; y(x) = − x−2 + 8 16 16 16 16 16 √ √ π 2 2 (b) y(t) = (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 (c) y(x) = x7/2 dx = x + C, y(0) = 0 = C, C = 0; y(x) = x9/2 9 9
  6. 6. January 27, 2005 11:45 L24-ch06 Sheet number 6 Page number 235 black Exercise Set 6.2 235 43. (a) y = 4ex dx = 4ex + C, 1 = y(0) = 4 + C, C = −3, y = 4ex − 3 (b) y(t) = t−1 dt = ln |t| + C, y(−1) = C = 5, C = 5; y(t) = ln |t| + 5 √ 3 3 44. (a) y = √ dt = 3 sin−1 t + C, y = 0 = π + C, C = −π, y = 3 sin−1 t − π 1 − t2 2 dy 2 2 (b) =1− 2 ,y = 1− dx = x − 2 tan−1 x + C, dx x +1 x2 + 1 π π y(1) = = 1 − 2 + C, C = π − 1, y = x − 2 tan−1 x + π − 1 2 4 2 3/2 4 5/2 45. f (x) = x + C1 ; f (x) = x + C1 x + C2 3 15 46. 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 47. 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 48. 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 49. dy/dx = 6xdx = 3x2 + C1 . The slope of the tangent line is −3 so dy/dx = −3 when x = 1. Thus 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. 1 2 2 2 50. (a) f (x) = x sin 3x − sin 3x + x cos 3x − 0.251607 3 27 9 4 (b) f (x) = 4 + x2 + √ −6 4 + x2 51. (a) y (b) y (c) f (x) = x2 /2 − 1 2 x 4 –2 2 x –1 1 52. (a) y (b) y (c) y = (ex + 1)/2 5 1 x 6 x 1
  7. 7. January 27, 2005 11:45 L24-ch06 Sheet number 7 Page number 236 black 236 Chapter 6 53. This slope field is zero along the y-axis, 54. This slope field is independent of y, is near zero and so corresponds to (b). for large negative values of x, and is very large y for large positive x. It must correspond to (d). 10 y 10 5 x 5 –3 –1 1 3 x –5 –4 –2 2 4 –5 –10 –10 55. This slope field has a negative value 56. This slope field appears to be constant along the y-axis, and thus corresponds (approximately 2), and thus corresponds to (c). to differential equation (a). y y 9 10 5 3 x x –2 1 3 –3 –1 2 4 –3 –9 –10 57. (a) F (x) = G (x) = 3x + 4 (b) F (0) = 16/6 = 8/3, G(0) = 0, so F (0) − G(0) = 8/3 (c) F (x) = (9x2 + 24x + 16)/6 = 3x2 /2 + 4x + 8/3 = G(x) + 8/3 58. (a) F (x) = G (x) = 10x/(x2 + 5)2 (b) F (0) = 0, G(0) = −1, so F (0) − G(0) = 1 x2 (x2 + 5) − 5 5 (c) F (x) = 2 = 2+5 =1− 2 = G(x) + 1 x +5 x x +5 59. (a) For x = 0, F (x) = G (x) = 1. But if I is an interval containing 0 then neither F nor G has a derivative at 0, so neither F nor G is an antiderivative on I. (b) Suppose G(x) = F (x) + C for some C. Then F (1) = 4 and G(1) = 4 + C, so C = 0, but F (−1) = −2 and G(−1) = −1, a contradiction. (c) No, because neither F nor G is an antiderivative on (−∞, +∞). 60. (a) Neither F nor G is differentiable at x = 0. For x > 0, F (x) = 1/x, G (x) = 1/x, and for x < 0, F (x) = 1/x, G (x) = 1/x. (b) F (1) = 0, G(1) = 2; F (−1) = 0, G(−1) = 1, so F (x) = G(x) + C is impossible. (c) The hypotheses of the Theorem are violated by any interval containing 0. 61. (sec2 x − 1)dx = tan x − x + C 62. (csc2 x − 1)dx = − cot x − x + C 1 1 1 1 63. (a) (1 − cos x)dx = (x − sin x) + C (b) (1 + cos x) dx = (x + sin x) + C 2 2 2 2
  8. 8. January 27, 2005 11:45 L24-ch06 Sheet number 8 Page number 237 black Exercise Set 6.3 237 d 1 64. For x > 0, [sec−1 x] = √ , and for x < 0, dx |x| x2 − 1 d d 1 1 [sec−1 |x|] = [sec−1 (−x)] = (−1) √ = √ dx dx |x| x2−1 x x2 − 1 which yields formula (14) in both cases. 1087 1087 1/2 1087 1/2 65. v = √ T −1/2 dT = √ T + C, v(273) = 1087 = 1087 + C so C = 0, v = √ T ft/s 2 273 273 273 66. 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 EXERCISE SET 6.3 1. (a) u23 du = u24 /24 + C = (x2 + 1)24 /24 + C (b) − u3 du = −u4 /4 + C = −(cos4 x)/4 + C √ (c) 2 sin u du = −2 cos u + C = −2 cos x+C 3 3 1/2 3 (d) u−1/2 du = u +C = 4x2 + 5 + C 8 4 4 1 1 1 2. (a) sec2 u du = tan u + C = tan(4x + 1) + C 4 4 4 1 1 3/2 1 (b) u1/2 du = u + C = (1 + 2y 2 )3/2 + C 4 6 6 1 2 3/2 2 (c) u1/2 du = u +C = sin3/2 (πθ) + C π 3π 3π 5 9/5 5 (d) u4/5 du = u + C = (x2 + 7x + 3)9/5 + C 9 9 1 1 3. (a) − u du = − u2 + C = − cot2 x + C 2 2 1 10 1 (b) u9 du = u +C = (1 + sin t)10 + C 10 10 1 1 1 (c) cos u du = sin u + C = sin 2x + C 2 2 2 1 1 1 (d) sec2 u du = tan u + C = tan x2 + C 2 2 2 2 7/2 4 5/2 2 3/2 4. (a) (u − 1)2 u1/2 du = (u5/2 − 2u3/2 + u1/2 )du = u − u + u +C 7 5 3 2 4 2 = (1 + x)7/2 − (1 + x)5/2 + (1 + x)3/2 + C 7 5 3 (b) csc2 u du = − cot u + C = − cot(sin x) + C
  9. 9. January 27, 2005 11:45 L24-ch06 Sheet number 9 Page number 238 black 238 Chapter 6 (c) sin u du = − cos u + C = − cos(x − π) + C du 1 1 (d) 2 =− +C =− 5 +C u u x +1 1 5. (a) du = ln |u| + C = ln | ln x| + C u 1 1 1 (b) − eu du = − eu + C = − e−5x + C 5 5 5 1 1 1 1 (c) − du = − ln |u| + C = − ln |1 + cos 3θ| + C 3 u 3 3 du (d) = ln u + C = ln(1 + ex ) + C u 1 du 1 6. (a) u = x3 , = tan−1 (x3 ) + C 3 1 + u2 3 1 (b) u = ln x, √ du = sin−1 (ln x) + C 1 − u2 1 (c) u = 3x, √ du = sec−1 (3x) + C u u2 − 1 √ du √ (d) u = x, 2 = 2 tan−1 u + C = 2 tan−1 ( x) + C 1 + u2 1 1 10 1 9. u = 4x − 3, u9 du = u +C = (4x − 3)10 + C 4 40 40 1 √ 1 3/2 1 10. u = 5 + x4 , u du = u + C = (5 + x4 )3/2 + C 4 6 6 1 1 1 11. u = 7x, sin u du = − cos u + C = − cos 7x + C 7 7 7 12. u = x/3, 3 cos u du = 3 sin u + C = 3 sin(x/3) + C 1 1 1 13. u = 4x, du = 4dx; sec u tan u du = sec u + C = sec 4x + C 4 4 4 1 1 1 14. u = 5x, du = 5dx; sec2 u du = tan u + C = tan 5x + C 5 5 5 1 1 u 1 15. u = 2x, du = 2dx; eu du = e + C = e2x + C 2 2 2 1 1 1 1 16. u = 2x, du = 2dx; du = ln |u| + C = ln |2x| + C 2 u 2 2 1 1 1 17. u = 2x, √ du = sin−1 (2x) + C 2 1 − u2 2 1 1 1 18. u = 4x, du = tan−1 (4x) + C 4 1 + u2 4
  10. 10. January 27, 2005 11:45 L24-ch06 Sheet number 10 Page number 239 black Exercise Set 6.3 239 1 1 3/2 1 19. u = 7t2 + 12, du = 14t dt; u1/2 du = u +C = (7t2 + 12)3/2 + C 14 21 21 1 1 1 20. u = 4 − 5x2 , du = −10x dx; − u−1/2 du = − u1/2 + C = − 4 − 5x2 + C 10 5 5 1 1 1 3 1 21. u = 1 − 2x, du = −2dx, −3 du = (−3) − +C = +C u3 2 u2 2 (1 − 2x)2 1 1 2 22. u = x3 + 3x, du = (3x2 + 3) dx, √ du = x3 + 3x + C 3 u 3 1 du 1 1 1 23. u = 5x4 + 2, du = 20x3 dx, du = − +C =− +C 20 u3 40 u2 40(5x4 + 2)2 1 1 1 1 1 1 24. u = , du = − 2 dx, − sin u du = cos u + C = cos +C x x 3 3 3 x 25. u = sin x, du = cos x dx; eu du = eu + C = esin x + C 1 1 u 1 4 26. u = x4 , du = 4x3 dx; eu du = e + C = ex + C 4 4 4 1 1 1 eu du = − eu + C = − e−2x + C 3 27. u = −2x3 , du = −6x2 , − 6 6 6 1 28. u = ex − e−x , du = (ex + e−x )dx, du = ln |u| + C = ln ex − e−x + C u 1 1 1 1 29. u = ex , du = tan−1 (ex ) + C 30. u = t2 , du = tan−1 (t2 ) + C 1 + u2 2 u2 + 1 2 1 1 1 31. u = 5/x, du = −(5/x2 )dx; − sin u du = cos u + C = cos(5/x) + C 5 5 5 √ 1 √ 32. u = x, du = √ dx; 2 sec2 u du = 2 tan u + C = 2 tan x + C 2 x 1 1 5 1 33. u = cos 3t, du = −3 sin 3t dt, − u4 du = − u + C = − cos5 3t + C 3 15 15 1 1 6 1 34. u = sin 2t, du = 2 cos 2t dt; u5 du = u +C = sin6 2t + C 2 12 12 1 1 1 35. u = x2 , du = 2x dx; sec2 u du = tan u + C = tan x2 + C 2 2 2 1 1 1 1 1 1 36. u = 1 + 2 sin 4θ, du = 8 cos 4θ dθ; du = − +C =− +C 8 u4 24 u3 24 (1 + 2 sin 4θ)3 1 1 1 37. u = 2 − sin 4θ, du = −4 cos 4θ dθ; − u1/2 du = − u3/2 + C = − (2 − sin 4θ)3/2 + C 4 6 6 1 1 4 1 38. u = tan 5x, du = 5 sec2 5x dx; u3 du = u +C = tan4 5x + C 5 20 20
  11. 11. January 27, 2005 11:45 L24-ch06 Sheet number 11 Page number 240 black 240 Chapter 6 1 39. u = tan x, √ du = sin−1 (tan x) + C 1 − u2 1 40. u = cos θ, − du = − tan−1 (cos θ) + C u2 +1 1 1 3 1 41. u = sec 2x, du = 2 sec 2x tan 2x dx; u2 du = u + C = sec3 2x + C 2 6 6 42. u = sin θ, du = cos θ dθ; sin u du = − cos u + C = − cos(sin θ) + C 43. e−x dx; u = −x, du = −dx; − eu du = −eu + C = −e−x + C √ 44. ex/2 dx; u = x/2, du = dx/2; 2 eu du = 2eu + C = 2ex/2 + C = 2 ex + C √ 1 1 √ 45. u = 2 x, du = √ dx; , du = −e−u + C = −e−2 x + C x eu 1 √ 46. u = 2y + 1, du = √ dy; eu du = eu + C = e 2y+1 +C 2y + 1 47. u = 2y + 1, du = 2dy; 1 1 1 1√ 1 1 (u − 1) √ du = u3/2 − u + C = (2y + 1)3/2 − 2y + 1 + C 4 u 6 2 6 2 48. u = 4 − x, du = −dx; √ 8 2 2 8 − (4 − u) u du = − u3/2 + u5/2 + C = (4 − x)5/2 − (4 − x)3/2 + C 3 5 5 3 49. sin2 2θ sin 2θ dθ = (1 − cos2 2θ) sin 2θ dθ; u = cos 2θ, du = −2 sin 2θ dθ, 1 1 1 1 1 − (1 − u2 )du = − u + u3 + C = − cos 2θ + cos3 2θ + C 2 2 6 2 6 50. sec2 3θ = tan2 3θ + 1, u = 3θ, du = 3dθ 1 1 1 1 1 sec4 3θ dθ = (tan2 u + 1) sec2 u du = tan3 u + tan u + C = tan3 3θ + tan 3θ + C 3 9 3 9 3 1 51. 1+ dt = t + ln |t| + C t 2 1 3 52. e2 ln x = eln x = x2 , x > 0, so e2 ln x dx = x2 dx = x +C 3 53. ln(ex ) + ln(e−x ) = ln(ex e−x ) = ln 1 = 0 so [ln(ex ) + ln(e−x )]dx = C cos x 1 54. dx; u = sin x, du = cos xdx; du = ln |u| + C = ln | sin x| + C sin x u √ √ √ √ 55. (a) sin−1 (x/3) + C (b) (1/ 5) tan−1 (x/ 5) + C (c) (1/ π) sec−1 (x/ π) + C
  12. 12. January 27, 2005 11:45 L24-ch06 Sheet number 12 Page number 241 black Exercise Set 6.3 241 1 1 56. (a) u = ex , 2 du = tan−1 (ex /2) + C 4+u 2 1 1 1 (b) u = 2x, √ du = sin−1 (2x/3) + C, 2 9−u 2 2 √ 1 1 √ √ (c) u = 5y, √ du = √ sec−1 ( 5y/ 3) + C u u 2−3 3 57. u = a + bx, du = b dx, 1 (a + bx)n+1 (a + bx)n dx = un du = +C b b(n + 1) 1 58. u = a + bx, du = b dx, dx = du b 1 n n u1/n du = u(n+1)/n + C = (a + bx)(n+1)/n + C b b(n + 1) b(n + 1) 59. u = sin(a + bx), du = b cos(a + bx)dx 1 1 1 un du = un+1 + C = sinn+1 (a + bx) + C b b(n + 1) b(n + 1) 1 2 1 61. (a) with u = sin x, du = cos x dx; u du = u + C1 = sin2 x + C1 ; 2 2 1 2 1 with u = cos x, du = − sin x dx; − u du = − u + C2 = − cos2 x + C2 2 2 (b) because they differ by a constant: 1 1 1 sin2 x + C1 − − cos2 x + C2 = (sin2 x + cos2 x) + C1 − C2 = 1/2 + C1 − C2 2 2 2 25 3 62. (a) First method: (25x2 − 10x + 1)dx = x − 5x2 + x + C1 ; 3 1 1 3 1 second method: u2 du = u + C2 = (5x − 1)3 + C2 5 15 15 1 1 25 3 1 (b) (5x − 1)3 + C2 = (125x3 − 75x2 + 15x − 1) + C2 = x − 5x2 + x − + C2 ; 15 15 3 15 the answers differ by a constant. √ 2 2 63. y = 5x + 1 dx = (5x + 1)3/2 + C; −2 = y(3) = 64 + C, 15 15 2 158 2 158 so C = −2 − 64 = − , and y = (5x + 1)3/2 − 15 15 15 15 1 64. y = (2 + sin 3x) dx = 2x − cos 3x + C and 3 π 2π 1 2π + 1 1 2π + 1 0=y = + + C, C = − , y = 2x − cos 3x − 3 3 3 3 3 3 1 1 1 13 65. y = − e2t dt = − e2t + C, 6 = y(0) = − + C, y = − e2t + 2 2 2 2 1 1 3 π 5 1 π 66. y = dt = tan−1 t + C, =y − =− + C, 25 + 9t2 15 5 30 3 15 4 π 1 3 π C= ,y = tan−1 t + 60 15 5 60
  13. 13. January 27, 2005 11:45 L24-ch06 Sheet number 13 Page number 242 black 242 Chapter 6 1 1 √ 67. (a) u = x2 + 1, du = 2x dx; √ du = u + C = x2 + 1 + C 2 u (b) 5 –5 5 0 1 1 1 1 68. (a) u = x2 + 1, du = 2x dx; du = ln u + C = ln(x2 + 1) + C 2 u 2 2 (b) y 4 x –4 4 √ 2 69. f (x) = m = 3x + 1, f (x) = (3x + 1)1/2 dx = (3x + 1)3/2 + C 9 2 7 2 7 f (0) = 1 = + C, C = , so f (x) = (3x + 1)3/2 + 9 9 9 9 10 70. p(t) = (3 + 0.12t)3/2 dt = (3 + 0.12t)5/2 + C; 3 10 5/2 100 = p(0) = 3 + C, C = 100 − 10 · 33/2 ≈ 48.038 so that 3 10 p(5) = (3 + 5 · (0.12))5/2 + 100 − 10 ∗ 33/2 ≈ 130.005 so that the population at the beginning of 3 the year 2010 is approximately 130,005. du u 71. u = a sin θ, du = a cos θ dθ; √ = aθ + C = sin−1 + C a2 − u2 a du 1 1 u 72. If u > 0 then u = a sec θ, du = a sec θ tan θ dθ, √ = θ = sec−1 + C u u2 − a2 a a a EXERCISE SET 6.4 1. (a) 1 + 8 + 27 = 36 (b) 5 + 8 + 11 + 14 + 17 = 55 (c) 20 + 12 + 6 + 2 + 0 + 0 = 40 (d) 1 + 1 + 1 + 1 + 1 + 1 = 6 (e) 1 − 2 + 4 − 8 + 16 = 11 (f ) 0 + 0 + 0 + 0 + 0 + 0 = 0 2. (a) 1 + 0 − 3 + 0 = −2 (b) 1 − 1 + 1 − 1 + 1 − 1 = 0 (c) π + π + · · · + π = 14π 2 2 2 2 (d) 24 + 25 + 26 = 112 (14 terms) √ √ √ √ √ √ (e) 1+ 2+ 3+ 4+ 5+ 6 (f ) 1 − 1 + 1 − 1 + 1 − 1 + 1 − 1 + 1 − 1 + 1 = 1
  14. 14. January 27, 2005 11:45 L24-ch06 Sheet number 14 Page number 243 black Exercise Set 6.4 243 10 20 10 3. k 4. 3k 5. 2k k=1 k=1 k=1 8 6 5 1 6. (2k − 1) 7. (−1)k+1 (2k − 1) 8. (−1)k+1 k k=1 k=1 k=1 50 50 9. (a) 2k (b) (2k − 1) k=1 k=1 5 5 n 5 10. (a) (−1)k+1 ak (b) (−1)k+1 bk (c) ak xk (d) a5−k bk k=1 k=0 k=0 k=0 100 100 1 7 11. (100)(100 + 1) = 5050 12. 7 k+ 1= (100)(101) + 100 = 35,450 2 2 k=1 k=1 20 3 1 13. (20)(21)(41) = 2870 14. k2 − k 2 = 2870 − 14 = 2856 6 k=1 k=1 30 30 30 30 1 1 15. k(k 2 − 4) = (k 3 − 4k) = k3 − 4 k= (30)2 (31)2 − 4 · (30)(31) = 214,365 4 2 k=1 k=1 k=1 k=1 6 6 1 1 16. k− k3 = (6)(7) − (6)2 (7)2 = −420 2 4 k=1 k=1 n n 3k 3 3 1 3 17. = k= · n(n + 1) = (n + 1) n n n 2 2 k=1 k=1 n−1 n−1 k2 1 1 1 1 18. = k2 = · (n − 1)(n)(2n − 1) = (n − 1)(2n − 1) n n n 6 6 k=1 k=1 n−1 n−1 k3 1 1 1 1 19. = 2 k3 = · (n − 1)2 n2 = (n − 1)2 n2 n n2 4 4 k=1 k=1 n n n 5 2k 5 2 5 2 1 20. − = 1− k= (n) − · n(n + 1) = 4 − n n n n n n n 2 k=1 k=1 k=1 n(n + 1) 22. = 465, n2 + n − 930 = 0, (n + 31)(n − 30) = 0, n = 30. 2 n n 1 + 2 + 3 + ··· + n k 1 1 1 n+1 n+1 1 23. = = 2 k= · n(n + 1) = ; lim = n2 n2 n n2 2 2n n→+∞ 2n 2 k=1 k=1 n n 12 + 22 + 32 + · · · + n2 k2 1 1 1 (n + 1)(2n + 1) 24. = = 3 k2 = · n(n + 1)(2n + 1) = ; n3 n 3 n n 3 6 6n2 k=1 k=1 (n + 1)(2n + 1) 1 1 lim 2 = lim (1 + 1/n)(2 + 1/n) = n→+∞ 6n n→+∞ 6 3 n n 5k 5 5 1 5(n + 1) 5(n + 1) 5 25. = 2 k= · n(n + 1) = ; lim = n2 n n2 2 2n n→+∞ 2n 2 k=1 k=1
  15. 15. January 27, 2005 11:45 L24-ch06 Sheet number 15 Page number 244 black 244 Chapter 6 n−1 n−1 2k 2 2 2 1 (n − 1)(2n − 1) 26. 3 = 3 k2 = · (n − 1)(n)(2n − 1) = 3 6 ; n n n 3n2 k=1 k=1 (n − 1)(2n − 1) 1 2 lim = lim (1 − 1/n)(2 − 1/n) = n→+∞ 3n2 n→+∞ 3 3 5 6 7 27. (a) 2j (b) 2j−1 (c) 2j−2 j=0 j=1 j=2 5 13 28. (a) (k + 4)2 k+8 (b) (k − 4)2k k=1 k=9 4 4 4 4 3 3 6 3 9 3 3 3(n − 1) 3 29. (a) 2+ , 2+ , 2+ , (2 + 3)4 ,..., 2 + n n n n n n n n n 3 3 When [2, 5] is subdivided into n equal intervals, the endpoints are 2, 2 + , 2 + 2 · , 2 + 3 · n n 3 3 , . . . , 2 + (n − 1) , 2 + 3 = 5, and the right endpoint approximation to the area under the n n curve y = x4 is given by the summands above. n−1 4 3 3 (b) 2+k· gives the left endpoint approximation. n n k=0 30. n is the number of elements of the partition, x∗ is an arbitrary point in the k-th interval, k k = 0, 1, 2, . . . , n − 1, n, and ∆x is the width of an interval in the partition. In the usual definition of area, the parts above the curve are given a + sign, and the parts below the curve are given a − sign. These numbers are then replaced with their absolute values and summed. In the definition of net signed area, the parts given above are summed without considering absolute values. In this case there could be lots of cancellation of ’positive’ areas with ’negative’ areas. 31. Endpoints 2, 3, 4, 5, 6; ∆x = 1; 4 (a) Left endpoints: f (x∗ )∆x = 7 + 10 + 13 + 16 = 46 k k=1 4 (b) Midpoints: f (x∗ )∆x = 8.5 + 11.5 + 14.5 + 17.5 = 52 k k=1 4 (c) Right endpoints: f (x∗ )∆x = 10 + 13 + 16 + 19 = 58 k k=1 32. Endpoints 1, 3, 5, 7, 9, ∆x = 2; 4 1 1 1 352 (a) Left endpoints: f (x∗ )∆x = k 1+ + + 2= 3 5 7 105 k=1 4 1 1 1 1 25 (b) Midpoints: f (x∗ )∆x = k + + + 2= 2 4 6 8 12 k=1 4 1 1 1 1 496 (c) Right endpoints: f (x∗ )∆x = k + + + 2= 3 5 7 9 315 k=1
  16. 16. January 27, 2005 11:45 L24-ch06 Sheet number 16 Page number 245 black Exercise Set 6.4 245 33. Endpoints: 0, π/4, π/2, 3π/4, π; ∆x = π/4 4 √ √ (a) Left endpoints: f (x∗ )∆x = 1 + k 2/2 + 0 − 2/2 (π/4) = π/4 k=1 4 (b) Midpoints: f (x∗ )∆x = [cos(π/8) + cos(3π/8) + cos(5π/8) + cos(7π/8)] (π/4) k k=1 = [cos(π/8) + cos(3π/8) − cos(3π/8) − cos(π/8)] (π/4) = 0 4 √ √ (c) Right endpoints: f (x∗ )∆x = k 2/2 + 0 − 2/2 − 1 (π/4) = −π/4 k=1 34. Endpoints −1, 0, 1, 2, 3; ∆x = 1 4 4 5 3 3 15 (a) f (x∗ )∆x = −3 + 0 + 1 + 0 = −2 k (b) f (x∗ )∆x = − + + + k =4 4 4 4 4 k=1 k=1 4 (c) f (x∗ )∆x = 0 + 1 + 0 − 3 = −2 k k=1 35. (a) 0.718771403, 0.705803382, 0.698172179 (b) 0.668771403, 0.680803382, 0.688172179 (c) 0.692835360, 0.693069098, 0.693134682 36. (a) 0.761923639, 0.712712753, 0.684701150 (b) 0.584145862, 0.623823864, 0.649145594 (c) 0.663501867, 0.665867079, 0.666538346 37. (a) 4.884074734, 5.115572731, 5.248762738 (b) 5.684074734, 5.515572731, 5.408762738 (c) 5.34707029, 5.338362719, 5.334644416 38. (a) 0.919403170, 0.960215997, 0.984209789 (b) 1.076482803, 1.038755813, 1.015625715 (c) 1.001028824, 1.000257067, 1.000041125 3 ∗ 3 1 1 3 3 3 1 3 39. ∆x = , x = 1 + k; f (x∗ )∆x = x∗ ∆x = 1+ k = + k n k n k 2 k 2 n n 2 n n2 n n n 3 1 3 3 3 1 3 3n+1 f (x∗ )∆x k = + k = 1 + 2 · n(n + 1) = 1+ 2 n n2 2 n 2 2 2 n k=1 k=1 k=1 3 3 1 3 3 15 A = lim 1+ 1+ = 1+ = n→+∞ 2 2 n 2 2 4 5 ∗ 5 5 5 25 25 40. ∆x = , xk = 0 + k ; f (x∗ )∆x = (5 − x∗ )∆x = k k 5− k = − 2k n n n n n n n n n 25 25 25 1 25 n+1 f (x∗ )∆x k = − 2 k = 25 − · n(n + 1) = 25 − n n n2 2 2 n k=1 k=1 k=1 25 1 25 25 A = lim 25 − 1+ = 25 − = n→+∞ 2 n 2 2
  17. 17. January 27, 2005 11:45 L24-ch06 Sheet number 17 Page number 246 black 246 Chapter 6 3 ∗ 3 k2 3 41. ∆x = , xk = 0 + k ; f (x∗ )∆x = k 9−9 n n n2 n n n n n k2 3 27 k2 27 f (x∗ )∆x = k 9−9 = 1− = 27 − k2 n2 n n n2 n3 k=1 k=1 k=1 k=1 n 27 1 A = lim 27 − k 2 = 27 − 27 = 18 n→+∞ n3 3 k=1 3 ∗ 3 42. ∆x = ,x =k n k n 1 1 9k 2 3 12 27k 2 f (x∗ )∆x = 4 − (x∗ )2 ∆x = 4 − k k = − 4 4 n2 n n 4n3 n n n 12 27 f (x∗ )∆x = k − 3 k2 n 4n k=1 k=1 k=1 27 1 9 (n + 1)(2n + 1) = 12 − 3 · n(n + 1)(2n + 1) = 12 − 4n 6 8 n2 9 1 1 9 A = lim 12 − 1+ 2+ = 12 − (1)(2) = 39/4 n→+∞ 8 n n 8 4 ∗ 4 43. ∆x = , xk = 2 + k n n 3 3 4 4 32 2 32 6 12 8 f (x∗ )∆x = (x∗ )3 ∆x = 2 + k k k = 1+ k = 1 + k + 2 k2 + 3 k3 n n n n n n n n n n n n n 32 6 12 8 f (x∗ )∆x = k 1+ k+ k2 + k3 n n n2 n3 k=1 k=1 k=1 k=1 k=1 32 6 1 12 1 8 1 = n + · n(n + 1) + 2 · n(n + 1)(2n + 1) + 3 · n2 (n + 1)2 n n 2 n 6 n 4 n+1 (n + 1)(2n + 1) (n + 1)2 = 32 1 + 3 +2 2 +2 n n n2 2 1 1 1 1 A = lim 32 1 + 3 1 + +2 1+ 2+ +2 1+ n→+∞ n n n n = 32[1 + 3(1) + 2(1)(2) + 2(1)2 ] = 320 3 2 ∗ 2 2 2 44. ∆x = , x = −3 + k ; f (x∗ )∆x = [1 − (x∗ )3 ]∆x = 1 − −3 + k n k n k k n n 2 54 36 8 = 28 − k + 2 k 2 − 3 k 3 n n n n n 2 (n + 1)(2n + 1) (n + 1)2 f (x∗ )∆x = k 28n − 27(n + 1) + 6 −2 n n n k=1 2 1 1 1 1 A = lim 2 28 − 27 1 + +6 1+ 2+ −2 1+ n→+∞ n n n n = 2(28 − 27 + 12 − 2) = 22
  18. 18. January 27, 2005 11:45 L24-ch06 Sheet number 18 Page number 247 black Exercise Set 6.4 247 3 ∗ 3 45. ∆x = , x = 1 + (k − 1) n k n 1 1 3 3 1 3 9 f (x∗ )∆x = x∗ ∆x = k k 1 + (k − 1) = + (k − 1) 2 2 2 n n 2 n n n n n 1 3 9 1 9 1 3 9n−1 f (x∗ )∆x = k + (k − 1) = 3 + 2 · (n − 1)n = + 2 n n2 2 n 2 2 4 n k=1 k=1 k=1 3 9 1 3 9 15 A = lim + 1− = + = n→+∞ 2 4 n 2 4 4 5 ∗ 5 46. ∆x = , x = (k − 1) n k n 5 5 25 25 f (x∗ )∆x = (5 − x∗ )∆x = 5 − k k (k − 1) = − 2 (k − 1) n n n n n n n 25 25 25 n − 1 f (x∗ )∆x k = 1− 2 (k − 1) = 25 − n n 2 n k=1 k=1 k=1 25 1 25 25 A = lim 25 − 1− = 25 − = n→+∞ 2 n 2 2 3 ∗ 3 (k − 1)2 3 47. ∆x = , xk = 0 + (k − 1) ; f (x∗ )∆x = (9 − 9 k ) n n n2 n n n n n n (k − 1)2 3 27 (k − 1)2 27 54 27 f (x∗ )∆x k = 9−9 = 1− = 27 − 3 2 k + 3 k− n2 n n n2 n n n2 k=1 k=1 k=1 k=1 k=1 1 A = lim = 27 − 27 + 0 + 0 = 18 n→+∞ 3 3 ∗ 3 48. ∆x = , x = (k − 1) n k n 1 1 9(k − 1)2 3 12 27k 2 27k 27 f (x∗ )∆x = 4 − (x∗ )2 ∆x = 4 − k k = − + 3− 3 4 4 n2 n n 4n3 2n 4n n n n n n 12 27 27 27 f (x∗ )∆x = k − 3 k2 + k− 1 n 4n 2n3 4n3 k=1 k=1 k=1 k=1 k=1 27 1 27 n(n + 1) 27 = 12 − · n(n + 1)(2n + 1) + 3 3 6 − 2 4n 2n 2 4n 9 (n + 1)(2n + 1) 27 27 27 = 12 − + + − 2 8 n2 4n 4n2 4n 9 1 1 9 A = lim 12 − 1+ 2+ + 0 + 0 − 0 = 12 − (1)(2) = 39/4 n→+∞ 8 n n 8 4 8 4(n − 1) 4n 2 6 10 4n − 6 4n − 2 49. Endpoints 0, , ,..., , = 4, and midpoints , , , . . . , , . Approxi- n n n n n n n n n n 4k − 2 4 16 n(n + 1) mate the area with the sum 2 = 2 2 − n → 16 as n → +∞. n n n 2 k=1
  19. 19. January 27, 2005 11:45 L24-ch06 Sheet number 19 Page number 248 black 248 Chapter 6 4 8 4(n − 1) 50. Endpoints 1, 1 + ,1 + ,...,1 + , 1 + 4 = 5, and midpoints n n n 2 6 10 4(n − 1) − 2 4n − 2 1 + ,1 + ,1 + ,...,1 + , . Approximate the area with the sum n n n n n n n 4k − 2 4 4 16 8 16 n(n + 1) 8 6− 1+ = 5 − k+ 2 = 20 − + = 20 − 8 = 12, n n n n2 n n2 2 n k=1 k=1 which happens to be exact. 1 ∗ 2k − 1 51. ∆x = ,x = n k 2n (2k − 1)2 1 k2 k 1 f (x∗ )∆x = k = 3− 3+ 3 (2n)2 n n n 4n n n n n 1 1 1 f (x∗ )∆x = k k2 − k+ 1 n3 n3 4n3 k=1 k=1 k=1 k=1 Using Theorem 6.4.4, n 1 1 A = lim f (x∗ )∆x = k +0+0= n→+∞ 3 3 k=1 2 ∗ 2k − 1 52. ∆x = , xk = −1 + n n 2 2k − 1 2 8k 2 8k 2 2 f (x∗ )∆x = k −1 + = 3 − 3+ 3− n n n n n n n n n 8 8 2 f (x∗ )∆x = k k2 − k+ −2 n3 n3 n2 k=1 k=1 k=1 n 8 2 A = lim f (x∗ )∆x = k +0+0−2= n→+∞ 3 3 k=1 2 ∗ 2k 53. ∆x = , x = −1 + n k n 2k 2 2 k f (x∗ )∆x = −1 + k =− +4 2 n n n n n n 4 4 n(n + 1) 2 f (x∗ )∆x = −2 + k k = −2 + = −2 + 2 + n2 n 2 2 n k=1 k=1 n A = lim f (x∗ )∆x = 0 k n→+∞ k=1 The area below the x-axis cancels the area above the x-axis. 3 ∗ 3k 54. ∆x = , xk = −1 + n n 3k 3 3 9 f (x∗ )∆x = −1 + k = − + 2k n n n n n 9 n(n + 1) f (x∗ )∆x = −3 + k n2 2 k=1

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