integral from 0 to e of xln(2x)

解答

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

\frac{1}{2} (e^{2} ln (2 e) - \frac{e ^{2}}{2})

十进制

4.40811 \dots

求解步骤

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

使用分布积分法

= [\frac{1}{2} (x^{2} ln (2 x) - 2 \cdot \int \frac{x}{2} d x)]_{0}^{e}

\int \frac{x}{2} d x = \frac{x ^{2}}{4}

= [\frac{1}{2} (x^{2} ln (2 x) - 2 \cdot \frac{x ^{2}}{4})]_{0}^{e}

2 \cdot \frac{x ^{2}}{4} = \frac{x ^{2}}{2}

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

带入上下限计算后得:

\frac{1}{2} (e^{2} ln (2 e) - \frac{e ^{2}}{2})

= \frac{1}{2} (e^{2} ln (2 e) - \frac{e ^{2}}{2})

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