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人気のある微分積分 >

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

解

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

解

23 - 2

+1

十進法表記

1.46410 \dots

解答ステップ

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

u置換積分法を適用する

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

定数の積分:

\int a d x = a x

= [2 u]_{1}^{3}

限度を計算する:

23 - 2

= 23 - 2

グラフ

人気の例

integral from 1 to 4 of (12(ln(x))^3)/x

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\int_{1}^{2} (8 x^{7} - 3 x^{2} + 2 x) d x

integral from 0 to x of sin(u)

\int_{0}^{x} sin (u) d u