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人気のある微分積分 >

integral from 1 to e of 2xln(x)

解

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

解

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

+1

十進法表記

4.19452 \dots

解答ステップ

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

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

部分積分を適用する

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

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

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

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

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

限度を計算する:

\frac{e ^{2}}{4} + \frac{1}{4}

= 2 (\frac{e ^{2}}{4} + \frac{1}{4})

簡素化

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

グラフ

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