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人気のある微分積分 >

integral from 0 to 2 of x^2sqrt(4-x)

解

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

解

\frac{2048 - 1136 2}{105}

+1

十進法表記

4.20431 \dots

解答ステップ

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

u置換積分法を適用する

= \int_{4}^{2} - (- u + 4)^{2} u d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{2}^{4} - (- u + 4)^{2} u d u

定数を除く:

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

= - (- \int_{2}^{4} (- u + 4)^{2} u d u)

拡張

(- u + 4)^{2} u : u^{\frac{5}{2}} - 8 u^{\frac{3}{2}} + 16 u

= - (- \int_{2}^{4} u^{\frac{5}{2}} - 8 u^{\frac{3}{2}} + 16 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

= - (- (\int_{2}^{4} u^{\frac{5}{2}} d u - \int_{2}^{4} 8 u^{\frac{3}{2}} d u + \int_{2}^{4} 16 u d u))

\int_{2}^{4} u^{\frac{5}{2}} d u = \frac{256 - 16 2}{7}

\int_{2}^{4} 8 u^{\frac{3}{2}} d u = \frac{8 ( 64 - 8 2 )}{5}

\int_{2}^{4} 16 u d u = \frac{16 ( 16 - 4 2 )}{3}

= - (- (\frac{256 - 16 2}{7} - \frac{8 ( 64 - 8 2 )}{5} + \frac{16 ( 16 - 4 2 )}{3}))

簡素化

- (- (\frac{256 - 16 2}{7} - \frac{8 ( 64 - 8 2 )}{5} + \frac{16 ( 16 - 4 2 )}{3})) : \frac{2048 - 1136 2}{105}

= \frac{2048 - 1136 2}{105}

グラフ

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