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人気のある微分積分 >

integral from 0 to 1 of xsqrt(x/(1-x))

解

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

解

\frac{3 π}{8}

+1

十進法表記

1.17809 \dots

解答ステップ

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

u置換積分法を適用する

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

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

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

拡張

\frac{- u + 1}{u} (- u + 1) : - \frac{- u + 1}{u} u + \frac{- u + 1}{u}

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

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int_{0}^{1} \frac{- u + 1}{u} u d u = \frac{π}{8}

\int_{0}^{1} \frac{- u + 1}{u} d u = \frac{π}{2}

= - (- (- \frac{π}{8} + \frac{π}{2}))

簡素化

- (- (- \frac{π}{8} + \frac{π}{2})) : \frac{3 π}{8}

= \frac{3 π}{8}

グラフ

人気の例

integral from-3 to 2 of 2x

\int_{- 3}^{2} 2 x d x

integral from 0 to 8 of sqrt(1+x^{-2/3)}

\int_{0}^{8} 1 + x^{- \frac{2}{3}} d x

integral from 0 to 1 of (9-8x^3+16x^7)

\int_{0}^{1} (9 - 8 x^{3} + 16 x^{7}) d x

integral from 0 to pi of 36sin(x)

\int_{0}^{π} 36 sin (x) d x

integral from-8 to 4 of-x^2-3x

\int_{- 8}^{4} - x^{2} - 3 x d x