integral from 0 to 6 of (4x-1)^2

解答

\int_{0}^{6} (4 x - 1)^{2} d x

1014

求解步骤

\int_{0}^{6} (4 x - 1)^{2} d x

乘开

(4 x - 1)^{2} : 16 x^{2} - 8 x + 1

= \int_{0}^{6} 16 x^{2} - 8 x + 1 d x

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{6} 16 x^{2} d x - \int_{0}^{6} 8 x d x + \int_{0}^{6} 1 d x

\int_{0}^{6} 16 x^{2} d x = 1152

\int_{0}^{6} 8 x d x = 144

\int_{0}^{6} 1 d x = 6

= 1152 - 144 + 6

化简

= 1014

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