integral from 0 to 2 of (6x-9)(8x^2+4)

解答

\int_{0}^{2} (6 x - 9) (8 x^{2} + 4) d x

- 24

求解步骤

\int_{0}^{2} (6 x - 9) (8 x^{2} + 4) d x

使用分布积分法

= [(6 x - 9) (\frac{8 x ^{3}}{3} + 4 x) - \int 6 (\frac{8 x ^{3}}{3} + 4 x) d x]_{0}^{2}

\int 6 (\frac{8 x ^{3}}{3} + 4 x) d x = 6 (\frac{2 x ^{4}}{3} + 2 x^{2})

= [(6 x - 9) (\frac{8 x ^{3}}{3} + 4 x) - 6 (\frac{2 x ^{4}}{3} + 2 x^{2})]_{0}^{2}

乘开

(6 x - 9) (\frac{8 x ^{3}}{3} + 4 x) - 6 (\frac{2 x ^{4}}{3} + 2 x^{2}) : 12 x^{4} - 24 x^{3} + 12 x^{2} - 36 x

= [12 x^{4} - 24 x^{3} + 12 x^{2} - 36 x]_{0}^{2}

带入上下限计算后得:

- 24

= - 24

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