integral from 0 to 2 of (6x-6)(8x^2+2)

解答

\int_{0}^{2} (6 x - 6) (8 x^{2} + 2) d x

64

求解步骤

\int_{0}^{2} (6 x - 6) (8 x^{2} + 2) d x

乘开

(6 x - 6) (8 x^{2} + 2) : 48 x^{3} + 12 x - 48 x^{2} - 12

= \int_{0}^{2} 48 x^{3} + 12 x - 48 x^{2} - 12 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}^{2} 48 x^{3} d x + \int_{0}^{2} 12 x d x - \int_{0}^{2} 48 x^{2} d x - \int_{0}^{2} 12 d x

\int_{0}^{2} 48 x^{3} d x = 192

\int_{0}^{2} 12 x d x = 24

\int_{0}^{2} 48 x^{2} d x = 128

\int_{0}^{2} 12 d x = 24

= 192 + 24 - 128 - 24

化简

= 64

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