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人気のある微分積分 >

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

解

\int_{0}^{2} x^{3} x^{2} + 4 d x

解

\frac{64 2 + 64}{15}

+1

十進法表記

10.30064 \dots

解答ステップ

\int_{0}^{2} x^{3} x^{2} + 4 d x

u置換積分法を適用する

= \int_{2}^{22} (u^{2} - 4) u^{2} d u

拡張

(u^{2} - 4) u^{2} : u^{4} - 4 u^{2}

= \int_{2}^{22} u^{4} - 4 u^{2} 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}^{22} u^{4} d u - \int_{2}^{22} 4 u^{2} d u

\int_{2}^{22} u^{4} d u = \frac{128 2 - 32}{5}

\int_{2}^{22} 4 u^{2} d u = \frac{4 ( 16 2 - 8 )}{3}

= \frac{128 2 - 32}{5} - \frac{4 ( 16 2 - 8 )}{3}

簡素化

\frac{128 2 - 32}{5} - \frac{4 ( 16 2 - 8 )}{3} : \frac{64 2 + 64}{15}

= \frac{64 2 + 64}{15}

グラフ

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