integral from 0 to 1 of x^2(1-x)^5

解答

\int_{0}^{1} x^{2} (1 - x)^{5} d x

\frac{1}{168}

十进制

0.00595 \dots

求解步骤

\int_{0}^{1} x^{2} (1 - x)^{5} d x

乘开

x^{2} (1 - x)^{5} : x^{2} - 5 x^{3} + 10 x^{4} - 10 x^{5} + 5 x^{6} - x^{7}

= \int_{0}^{1} x^{2} - 5 x^{3} + 10 x^{4} - 10 x^{5} + 5 x^{6} - x^{7} 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}^{1} x^{2} d x - \int_{0}^{1} 5 x^{3} d x + \int_{0}^{1} 10 x^{4} d x - \int_{0}^{1} 10 x^{5} d x + \int_{0}^{1} 5 x^{6} d x - \int_{0}^{1} x^{7} d x

\int_{0}^{1} x^{2} d x = \frac{1}{3}

\int_{0}^{1} 5 x^{3} d x = \frac{5}{4}

\int_{0}^{1} 10 x^{4} d x = 2

\int_{0}^{1} 10 x^{5} d x = \frac{5}{3}

\int_{0}^{1} 5 x^{6} d x = \frac{5}{7}

\int_{0}^{1} x^{7} d x = \frac{1}{8}

= \frac{1}{3} - \frac{5}{4} + 2 - \frac{5}{3} + \frac{5}{7} - \frac{1}{8}

化简

\frac{1}{3} - \frac{5}{4} + 2 - \frac{5}{3} + \frac{5}{7} - \frac{1}{8} : \frac{1}{168}

= \frac{1}{168}

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