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

解答

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

\frac{1024}{15}

十进制

68.26666 \dots

求解步骤

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

乘开

x^{2} (4 - x)^{3} : 64 x^{2} - 48 x^{3} + 12 x^{4} - x^{5}

= \int_{0}^{4} 64 x^{2} - 48 x^{3} + 12 x^{4} - x^{5} 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}^{4} 64 x^{2} d x - \int_{0}^{4} 48 x^{3} d x + \int_{0}^{4} 12 x^{4} d x - \int_{0}^{4} x^{5} d x

\int_{0}^{4} 64 x^{2} d x = \frac{4096}{3}

\int_{0}^{4} 48 x^{3} d x = 3072

\int_{0}^{4} 12 x^{4} d x = \frac{12288}{5}

\int_{0}^{4} x^{5} d x = \frac{2048}{3}

= \frac{4096}{3} - 3072 + \frac{12288}{5} - \frac{2048}{3}

化简

\frac{4096}{3} - 3072 + \frac{12288}{5} - \frac{2048}{3} : \frac{1024}{15}

= \frac{1024}{15}

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