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人気のある微分積分 >

integral from 0 to 1 of 2pi(2-y)(4y^2)

解

\int_{0}^{1} 2 π (2 - y) (4 y^{2}) d y

解

\frac{10 π}{3}

+1

十進法表記

10.47197 \dots

解答ステップ

\int_{0}^{1} 2 π (2 - y) \cdot 4 y^{2} d y

定数を除く:

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

= 2 π 4 \cdot \int_{0}^{1} (2 - y) y^{2} d y

簡素化

= 8 π \cdot \int_{0}^{1} (2 - y) y^{2} d y

拡張

(2 - y) y^{2} : 2 y^{2} - y^{3}

= 8 π \cdot \int_{0}^{1} 2 y^{2} - y^{3} d y

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 8 π (\int_{0}^{1} 2 y^{2} d y - \int_{0}^{1} y^{3} d y)

\int_{0}^{1} 2 y^{2} d y = \frac{2}{3}

\int_{0}^{1} y^{3} d y = \frac{1}{4}

= 8 π (\frac{2}{3} - \frac{1}{4})

簡素化

8 π (\frac{2}{3} - \frac{1}{4}) : \frac{10 π}{3}

= \frac{10 π}{3}

グラフ

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