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人気のある微分積分 >

integral from 0 to 2 of r^3sqrt(1+4r^2)

解

\int_{0}^{2} r^{3} 1 + 4 r^{2} d r

解

\frac{391 17 + 1}{120}

+1

十進法表記

13.44278 \dots

解答ステップ

\int_{0}^{2} r^{3} 1 + 4 r^{2} d r

u置換積分法を適用する

= \int_{1}^{17} \frac{u ( u - 1 )}{32} d u

定数を除く:

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

= \frac{1}{32} \cdot \int_{1}^{17} u (u - 1) d u

拡張

u (u - 1) : u^{\frac{3}{2}} - u

= \frac{1}{32} \cdot \int_{1}^{17} u^{\frac{3}{2}} - u d u

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

= \frac{1}{32} (\int_{1}^{17} u^{\frac{3}{2}} d u - \int_{1}^{17} u d u)

\int_{1}^{17} u^{\frac{3}{2}} d u = \frac{578 17 - 2}{5}

\int_{1}^{17} u d u = \frac{34 17 - 2}{3}

= \frac{1}{32} (\frac{578 17 - 2}{5} - \frac{34 17 - 2}{3})

簡素化

\frac{1}{32} (\frac{578 17 - 2}{5} - \frac{34 17 - 2}{3}) : \frac{391 17 + 1}{120}

= \frac{391 17 + 1}{120}

グラフ

人気の例

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