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人気のある微分積分 >

integral from 0 to e of 2piy(1-ln(y))

解

\int_{0}^{e} 2 π y (1 - ln (y)) d y

解

\frac{e ^{2} π}{2}

+1

十進法表記

11.60670 \dots

解答ステップ

\int_{0}^{e} 2 π y (1 - ln (y)) d y

定数を除く:

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

= 2 π \cdot \int_{0}^{e} y (1 - ln (y)) d y

部分積分を適用する

= 2 π [\frac{1}{2} (y^{2} (1 - ln (y)) - 2 \cdot \int - \frac{y}{2} d y)]_{0}^{e}

\int - \frac{y}{2} d y = - \frac{y ^{2}}{4}

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

簡素化

2 π [\frac{1}{2} (y^{2} (1 - ln (y)) - 2 (- \frac{y ^{2}}{4}))]_{0}^{e} : 2 π [\frac{1}{4} (3 y^{2} - 2 y^{2} ln (y))]_{0}^{e}

= 2 π [\frac{1}{4} (3 y^{2} - 2 y^{2} ln (y))]_{0}^{e}

限度を計算する:

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

= 2 π \frac{e ^{2}}{4}

簡素化

= \frac{e ^{2} π}{2}

グラフ

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