integral from 1 to e of zln(z)

解答

\int_{1}^{e} z ln (z) d z

\frac{e ^{2} + 1}{4}

十进制

2.09726 \dots

求解步骤

\int_{1}^{e} z ln (z) d z

使用分布积分法

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

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

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

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

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

带入上下限计算后得:

\frac{e ^{2}}{4} + \frac{1}{4}

= \frac{e ^{2}}{4} + \frac{1}{4}

化简

= \frac{e ^{2} + 1}{4}

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