integral from 0 to 1/2 of xln(2x)

解

\int_{0}^{\frac{1}{2}} x ln (2 x) d x

- \frac{1}{16}

十進法表記

- 0.0625

解答ステップ

\int_{0}^{\frac{1}{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)]_{0}^{\frac{1}{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})]_{0}^{\frac{1}{2}}

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

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

限度を計算する:

- \frac{1}{16}

= - \frac{1}{16}