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

解

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

\frac{3 e ^{4} + 1}{4}

十進法表記

41.19861 \dots

解答ステップ

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

部分積分を適用する

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

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

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

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

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

限度を計算する:

\frac{3 e ^{4}}{4} + \frac{1}{4}

= \frac{3 e ^{4}}{4} + \frac{1}{4}

簡素化

= \frac{3 e ^{4} + 1}{4}