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人気のある微分積分 >

integral from 1 to e^2 of 2piyln(y)

解

\int_{1}^{e^{2}} 2 π y ln (y) d y

解

\frac{π ( 3 e ^{4} + 1 )}{2}

+1

十進法表記

258.85851 \dots

解答ステップ

\int_{1}^{e^{2}} 2 π y 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_{1}^{e^{2}} y ln (y) d y

部分積分を適用する

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

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

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

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

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

限度を計算する:

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

= 2 π (\frac{3 e ^{4}}{4} + \frac{1}{4})

簡素化

= \frac{π ( 3 e ^{4} + 1 )}{2}

グラフ

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