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人気のある微分積分 >

integral from 1 to 2 of 1/(sqrt(2-x))

解

\int_{1}^{2} \frac{1}{2 - x} d x

解

2

解答ステップ

\int_{1}^{2} \frac{1}{2 - x} d x

u置換積分法を適用する

= \int_{1}^{0} - \frac{1}{u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{1} - \frac{1}{u} d u

定数を除く:

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

= - (- \int_{0}^{1} \frac{1}{u} d u)

べき乗則を適用する

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

簡素化

- (- [2 u^{\frac{1}{2}}]_{0}^{1}) : [2 u]_{0}^{1}

= [2 u]_{0}^{1}

限度を計算する:

2

= 2

グラフ

人気の例

integral from 0 to 2 of (1/((9+x^2)^2))

\int_{0}^{2} (\frac{1}{( 9 + x ^{2} ) ^{2}}) d x

integral from-10 to 10 of sqrt(256-x^2)

\int_{- 10}^{10} 256 - x^{2} d x

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\int_{1}^{e} \frac{7}{x} d x