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人気のある微分積分 >

integral from 0 to 2 of xsqrt(1+x)

解

\int_{0}^{2} x 1 + x d x

解

\frac{2}{3} + \frac{18 3 - 2}{5} - 23

+1

十進法表記

3.03794 \dots

解答ステップ

\int_{0}^{2} x 1 + x d x

u置換積分法を適用する

= \int_{1}^{3} (u - 1) u d u

拡張

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

= \int_{1}^{3} 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

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

\int_{1}^{3} u^{\frac{3}{2}} d u = \frac{18 3 - 2}{5}

\int_{1}^{3} u d u = 23 - \frac{2}{3}

= \frac{18 3 - 2}{5} - (23 - \frac{2}{3})

簡素化

= \frac{2}{3} + \frac{18 3 - 2}{5} - 23

グラフ

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