integral from 0 to 2 of (x^2ye^{xy})

解答

\int_{0}^{2} (x^{2} y e^{x y}) d x

4 e^{2 y} + \frac{- 2 ( 2 y e ^{2 y} - e ^{2 y} ) - 2}{y ^{2}}

求解步骤

\int_{0}^{2} x^{2} y e^{x y} d x

提出常数:

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

= y \cdot \int_{0}^{2} e^{y x} x^{2} d x

使用分布积分法

= y [e^{y x} \frac{1}{y} x^{2} - \int e^{y x} \frac{2}{y} x d x]_{0}^{2}

\int e^{y x} \frac{2}{y} x d x = \frac{2}{y} (e^{y x} \frac{1}{y} x - e^{y x} \frac{1}{y ^{2}})

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

带入上下限计算后得:

e^{2 y} \frac{4}{y} - \frac{2}{y} (e^{2 y} \frac{2}{y} - e^{2 y} \frac{1}{y ^{2}}) - \frac{2}{y ^{3}}

= y (e^{2 y} \frac{4}{y} - \frac{2}{y} (e^{2 y} \frac{2}{y} - e^{2 y} \frac{1}{y ^{2}}) - \frac{2}{y ^{3}})

化简

= 4 e^{2 y} + \frac{- 2 ( 2 y e ^{2 y} - e ^{2 y} ) - 2}{y ^{2}}

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