integral from 0 to 1 of (\sqrt[3]{y})^2

解答

\int_{0}^{1} (3 y)^{2} d y

\frac{3}{5}

十进制

0.6

求解步骤

\int_{0}^{1} (3 y)^{2} d y

使用分布积分法

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

\int \frac{2}{3} y^{\frac{2}{3}} d y = \frac{2}{5} y^{\frac{5}{3}}

= [y (3 y)^{2} - \frac{2}{5} y^{\frac{5}{3}}]_{0}^{1}

化简

[y (3 y)^{2} - \frac{2}{5} y^{\frac{5}{3}}]_{0}^{1} : [\frac{3}{5} y^{\frac{5}{3}}]_{0}^{1}

= [\frac{3}{5} y^{\frac{5}{3}}]_{0}^{1}

带入上下限计算后得:

\frac{3}{5}

= \frac{3}{5}

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