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f.6 math integration question3
發問:
求∫(0&-3) [x / 開方(1 - 5x)] dx 求∫(0&-3) [x / 開方(1 - 5x)] dx ∫(b&a)..means F(b) - F(a) ∫(b&a)..means F(b) - F(a) 更新: 求過程,ans. is -1.44
最佳解答:
Sub u = √(1 - 5x), then du = - 5dx/[2√(1 - 5x)], = - 5dx/(2u) So dx = - 2udu/5 When x = 0, u = 1 and when x = - 3, u = 4. Therefore: ∫(x = -3 → 0) x √(1 - 5x) dx = ∫(u = 4 → 1) (1 - u2)/(5u) (-2udu/5) = (2/25)∫(u = 4 → 1) (1 - u2) du = (2/25) [u - u3/3] (u = 4 → 1) = 36/25 2009-09-28 00:30:20 補充： Just simply reverse the upper and lower limits of the integral can give the ans. 2009-09-28 22:34:47 補充： ∫(x = 0 → -3) x √(1 - 5x) dx Sub u = √(1 - 5x), it becomes: (2/25)∫(u = 1 → 4) (1 - u^2) du = - 36/25 or - 1.44
其他解答:
f.6 math integration question3
發問:
求∫(0&-3) [x / 開方(1 - 5x)] dx 求∫(0&-3) [x / 開方(1 - 5x)] dx ∫(b&a)..means F(b) - F(a) ∫(b&a)..means F(b) - F(a) 更新: 求過程,ans. is -1.44
最佳解答:
Sub u = √(1 - 5x), then du = - 5dx/[2√(1 - 5x)], = - 5dx/(2u) So dx = - 2udu/5 When x = 0, u = 1 and when x = - 3, u = 4. Therefore: ∫(x = -3 → 0) x √(1 - 5x) dx = ∫(u = 4 → 1) (1 - u2)/(5u) (-2udu/5) = (2/25)∫(u = 4 → 1) (1 - u2) du = (2/25) [u - u3/3] (u = 4 → 1) = 36/25 2009-09-28 00:30:20 補充： Just simply reverse the upper and lower limits of the integral can give the ans. 2009-09-28 22:34:47 補充： ∫(x = 0 → -3) x √(1 - 5x) dx Sub u = √(1 - 5x), it becomes: (2/25)∫(u = 1 → 4) (1 - u^2) du = - 36/25 or - 1.44
其他解答:
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ans. is -1.44 2009-09-28 17:43:26 補充： how??please show it to me
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