integral from 1/2 to e/2 of xln(2x)

解

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

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

十進法表記

0.52431 \dots

解答ステップ

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

部分積分を適用する

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

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

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

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

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

限度を計算する:

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

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

簡素化

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