integral from 0 to 1 of x(sqrt(x)-x^3)

解答

\int_{0}^{1} x (x - x^{3}) d x

\frac{1}{5}

十进制

0.2

求解步骤

\int_{0}^{1} x (x - x^{3}) d x

使用分布积分法

= [x (\frac{2}{3} x^{\frac{3}{2}} - \frac{x ^{4}}{4}) - \int \frac{2}{3} x^{\frac{3}{2}} - \frac{x ^{4}}{4} d x]_{0}^{1}

\int \frac{2}{3} x^{\frac{3}{2}} - \frac{x ^{4}}{4} d x = \frac{4}{15} x^{\frac{5}{2}} - \frac{x ^{5}}{20}

= [x (\frac{2}{3} x^{\frac{3}{2}} - \frac{x ^{4}}{4}) - (\frac{4}{15} x^{\frac{5}{2}} - \frac{x ^{5}}{20})]_{0}^{1}

化简

[x (\frac{2}{3} x^{\frac{3}{2}} - \frac{x ^{4}}{4}) - (\frac{4}{15} x^{\frac{5}{2}} - \frac{x ^{5}}{20})]_{0}^{1} : [\frac{1}{5} (- x^{5} + 2 x^{\frac{5}{2}})]_{0}^{1}

= [\frac{1}{5} (- x^{5} + 2 x^{\frac{5}{2}})]_{0}^{1}

带入上下限计算后得:

\frac{1}{5}

= \frac{1}{5}

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