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人気のある微分積分 >

integral from 1 to 2 of x^3sqrt(9+x^2)

解

\int_{1}^{2} x^{3} 9 + x^{2} d x

解

\frac{169 13}{5} + 1010 - 3913

+1

十進法表記

12.87390 \dots

解答ステップ

\int_{1}^{2} x^{3} 9 + x^{2} d x

u置換積分法を適用する

= \int_{10}^{13} (u^{2} - 9) u^{2} d u

拡張

(u^{2} - 9) u^{2} : u^{4} - 9 u^{2}

= \int_{10}^{13} u^{4} - 9 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_{10}^{13} u^{4} d u - \int_{10}^{13} 9 u^{2} d u

\int_{10}^{13} u^{4} d u = \frac{169 13}{5} - 2010

\int_{10}^{13} 9 u^{2} d u = 3 (1313 - 1010)

= \frac{169 13}{5} - 2010 - 3 (1313 - 1010)

簡素化

\frac{169 13}{5} - 2010 - 3 (1313 - 1010) : \frac{169 13}{5} + 1010 - 3913

= \frac{169 13}{5} + 1010 - 3913

グラフ

人気の例

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