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In Problems 55–62, find the particular antiderivative of each derivative that satisfies the given condition. 55. C'(x) = 9x^2 – 20x; C(10) = 2,500 56. R'(x) = 500 – 0.4x; R(0) = 0 57. dx/dt = 10/?t; x(1) = 25 58. dR/dt = 50/t^3; R(1) = 50 59. f'(x) = 4x^-2 – 3x^-1 + 2; f(1) = 5 60. f'(x) = x^-1 – 2x^-2 + 1; f(1) = 5 61. dy/dt = 6e^t – 7; y(0) = 0 62. dy/dt = 3 – 2e^t; y(0) = 2

          In Problems 55–62, find the particular antiderivative of each derivative that satisfies the given condition.

55. C'(x) = 9x^2 – 20x; C(10) = 2,500
56. R'(x) = 500 – 0.4x; R(0) = 0
57. dx/dt = 10/?t; x(1) = 25
58. dR/dt = 50/t^3; R(1) = 50
59. f'(x) = 4x^-2 – 3x^-1 + 2; f(1) = 5
60. f'(x) = x^-1 – 2x^-2 + 1; f(1) = 5
61. dy/dt = 6e^t – 7; y(0) = 0
62. dy/dt = 3 – 2e^t; y(0) = 2

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In Problems 55–62, find the particular antiderivative of each derivative that satisfies the given condition.

55. C'(x) = 9x^2 – 20x; C(10) = 2,500
56. R'(x) = 500 – 0.4x; R(0) = 0
57. dx/dt = 10/?t; x(1) = 25
58. dR/dt = 50/t^3; R(1) = 50
59. f'(x) = 4x^-2 – 3x^-1 + 2; f(1) = 5
60. f'(x) = x^-1 – 2x^-2 + 1; f(1) = 5
61. dy/dt = 6e^t – 7; y(0) = 0
62. dy/dt = 3 – 2e^t; y(0) = 2

Added by Anthony A.

Calculus: Early Transcendentals

James Stewart 8th Edition

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In Problems 55-62, find the particular antiderivative of each derivative that satisfies the given condition. 55. C'(x) = 9x^2 - 20x; C(10) = 2,500 56. R'(x) = 500 - 0.4x; R(0) = 0 57. dx/dt = 10/∑t; x(1) = 25 58. dR/dt = 50/t^3; R(1) = 50 59. f'(x) = 4x^-2 - 3x^-1 + 2; f(1) = 5 60. f'(x) = x^-1 - 2x^-2 + 1; f(1) = 5 61. dy/dt = 6e^t - 7; y(0) = 0 62. dy/dt = 3 - 2e^t; y(0) = 2

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