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

解

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

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

解答ステップ

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

定数を除く:

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

= 2 \cdot \int_{0}^{1} (1 - y)^{- 1 + x} y^{2} d y

u置換積分法を適用する

= 2 \cdot \int_{1}^{0} - (- u + 1)^{2} u^{- 1 + x} d u

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

= 2 (- \int_{0}^{1} - (- u + 1)^{2} u^{- 1 + x} d u)

定数を除く:

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

= 2 (- (- \int_{0}^{1} (- u + 1)^{2} u^{- 1 + x} d u))

拡張

(- u + 1)^{2} u^{- 1 + x} : u^{x + 1} - 2 u^{x} + u^{- 1 + x}

= 2 (- (- \int_{0}^{1} u^{x + 1} - 2 u^{x} + u^{- 1 + x} 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

= 2 (- (- (\int_{0}^{1} u^{x + 1} d u - \int_{0}^{1} 2 u^{x} d u + \int_{0}^{1} u^{- 1 + x} d u)))

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

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

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

= 2 (- (- (\frac{1}{x + 2} - \frac{2}{x + 1} + \frac{1}{x})))

簡素化

= 2 (\frac{1}{x + 2} - \frac{2}{x + 1} + \frac{1}{x})