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人気のある微分積分 >

integral from 0 to 1 of-ln(1-x)

解

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

解

1

解答ステップ

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

定数を除く:

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

= - \int_{0}^{1} ln (1 - x) d x

u置換積分法を適用する

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

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

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

部分積分を適用する

= - (- (- [u ln (u) - \int 1 d u]_{0}^{1}))

\int 1 d u = u

= - (- (- [u ln (u) - u]_{0}^{1}))

簡素化

= - [u ln (u) - u]_{0}^{1}

限度を計算する:

- 1

= - (- 1)

簡素化

= 1

グラフ

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