integral from 0.25 to 0.6 of 4(1-x)^3

解答

\int_{0.25}^{0.6} 4 (1 - x)^{3} d x

0.29080625

求解步骤

\int_{0.25}^{0.6} 4 (1 - x)^{3} d x

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int_{0.25}^{0.6} (1 - x)^{3} d x

乘开

(1 - x)^{3} : 1 - 3 x + 3 x^{2} - x^{3}

= 4 \cdot \int_{0.25}^{0.6} 1 - 3 x + 3 x^{2} - x^{3} 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

= 4 (\int_{0.25}^{0.6} 1 d x - \int_{0.25}^{0.6} 3 x d x + \int_{0.25}^{0.6} 3 x^{2} d x - \int_{0.25}^{0.6} x^{3} d x)

\int_{0.25}^{0.6} 1 d x = 0.35

\int_{0.25}^{0.6} 3 x d x = 0.44625

\int_{0.25}^{0.6} 3 x^{2} d x = 0.200375

\int_{0.25}^{0.6} x^{3} d x = 0.0314234375

= 4 (0.35 - 0.44625 + 0.200375 - 0.03142 \dots)

化简

= 0.29080625

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