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人気のある微分積分 >

integral from 0 to 2 of 2pi(x+1)(4-2x)

解

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

解

\frac{40 π}{3}

+1

十進法表記

41.88790 \dots

解答ステップ

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

定数を除く:

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

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

(x + 1) (4 - 2 x) = 2 (x + 1) (2 - x)

= 2 π \cdot \int_{0}^{2} 2 (x + 1) (2 - x) d x

定数を除く:

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

= 2 π 2 \cdot \int_{0}^{2} (x + 1) (2 - x) d x

拡張

(x + 1) (2 - x) : x - x^{2} + 2

= 2 π 2 \cdot \int_{0}^{2} x - x^{2} + 2 d x

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

= 2 π 2 (\int_{0}^{2} x d x - \int_{0}^{2} x^{2} d x + \int_{0}^{2} 2 d x)

\int_{0}^{2} x d x = 2

\int_{0}^{2} x^{2} d x = \frac{8}{3}

\int_{0}^{2} 2 d x = 4

= 2 π 2 (2 - \frac{8}{3} + 4)

簡素化

2 π 2 (2 - \frac{8}{3} + 4) : \frac{40 π}{3}

= \frac{40 π}{3}

グラフ

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