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人気のある微分積分 >

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

解

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

解

2 - 1

+1

十進法表記

0.41421 \dots

解答ステップ

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

u置換積分法を適用する

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

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

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

定数を除く:

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

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

べき乗則を適用する

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

簡素化

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

= \frac{1}{2} [2 u]_{1}^{2}

限度を計算する:

22 - 2

= \frac{1}{2} (22 - 2)

簡素化

= 2 - 1

グラフ

人気の例

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