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

解

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

\frac{1}{4}

十進法表記

0.25

解答ステップ

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

部分積分を適用する

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

\int x ln (x) d x = \frac{1}{2} x^{2} ln (x) - \frac{x ^{2}}{4}

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

簡素化

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

限度を計算する:

\frac{1}{4}

= \frac{1}{4}