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人気のある微分積分 >

integral from 0 to 1 of sqrt(1+\sqrt{x)}

解

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

解

2 \cdot \frac{4 2 + 4}{15}

+1

十進法表記

1.28758 \dots

解答ステップ

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

u置換積分法を適用する

= \int_{1}^{2} u \cdot 2 (u - 1) d u

定数を除く:

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

= 2 \cdot \int_{1}^{2} u (u - 1) d u

拡張

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

= 2 \cdot \int_{1}^{2} 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

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

\int_{1}^{2} u^{\frac{3}{2}} d u = \frac{8 2 - 2}{5}

\int_{1}^{2} u d u = \frac{4 2 - 2}{3}

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

結合

\frac{8 2 - 2}{5} - \frac{4 2 - 2}{3} : \frac{4 2 + 4}{15}

= 2 \cdot \frac{4 2 + 4}{15}

グラフ

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