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人気のある微分積分 >

integral from 0 to 4 of 1/(2+sqrt(x))

解

\int_{0}^{4} \frac{1}{2 + x} d x

解

2 (2 - 2 ln (2))

+1

十進法表記

1.22741 \dots

解答ステップ

\int_{0}^{4} \frac{1}{2 + x} d x

u置換積分法を適用する

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

定数を除く:

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

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

拡張

\frac{u - 2}{u} : 1 - \frac{2}{u}

= 2 \cdot \int_{2}^{4} 1 - \frac{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_{2}^{4} 1 d u - \int_{2}^{4} \frac{2}{u} d u)

\int_{2}^{4} 1 d u = 2

\int_{2}^{4} \frac{2}{u} d u = 2 ln (2)

= 2 (2 - 2 ln (2))

グラフ

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