integral from 1/2 to e/2 of tln(2t)

解答

\int_{\frac{1}{2}}^{\frac{e}{2}} t ln (2 t) d t

\frac{e ^{2} + 1}{16}

十进制

0.52431 \dots

求解步骤

\int_{\frac{1}{2}}^{\frac{e}{2}} t ln (2 t) d t

使用分布积分法

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

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

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

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

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

带入上下限计算后得:

\frac{e ^{2}}{16} + \frac{1}{16}

= \frac{e ^{2}}{16} + \frac{1}{16}

化简

= \frac{e ^{2} + 1}{16}

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