integral from 0 to 1 of 1/((1-x)^{1/2)}

解

\int_{0}^{1} \frac{1}{( 1 - x ) ^{\frac{1}{2}}} d x

2

解答ステップ

\int_{0}^{1} \frac{1}{( 1 - x ) ^{\frac{1}{2}}} d x

部分積分を適用する

= [\frac{x}{( 1 - x ) ^{\frac{1}{2}}} - \int \frac{x}{2 ( 1 - x ) ^{\frac{3}{2}}} d x]_{0}^{1}

\int \frac{x}{2 ( 1 - x ) ^{\frac{3}{2}}} d x = \frac{- x + 2}{( 1 - x ) ^{\frac{1}{2}}}

= [\frac{x}{( 1 - x ) ^{\frac{1}{2}}} - \frac{- x + 2}{( 1 - x ) ^{\frac{1}{2}}}]_{0}^{1}

結合

\frac{x}{( 1 - x ) ^{\frac{1}{2}}} - \frac{- x + 2}{( 1 - x ) ^{\frac{1}{2}}} : - 2 (- x + 1)^{\frac{1}{2}}

= [- 2 (- x + 1)^{\frac{1}{2}}]_{0}^{1}

限度を計算する:

2

= 2