integral from 0 to 2 of (2x-3)(8x^2+4)

解答

\int_{0}^{2} (2 x - 3) (8 x^{2} + 4) d x

- 8

求解步骤

\int_{0}^{2} (2 x - 3) (8 x^{2} + 4) d x

乘开

(2 x - 3) (8 x^{2} + 4) : 16 x^{3} + 8 x - 24 x^{2} - 12

= \int_{0}^{2} 16 x^{3} + 8 x - 24 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} 16 x^{3} d x + \int_{0}^{2} 8 x d x - \int_{0}^{2} 24 x^{2} d x - \int_{0}^{2} 12 d x

\int_{0}^{2} 16 x^{3} d x = 64

\int_{0}^{2} 8 x d x = 16

\int_{0}^{2} 24 x^{2} d x = 64

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

= 64 + 16 - 64 - 24

化简

= - 8

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