integral from 0 to 1 of (1-y)y^n

解

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

\frac{1}{n + 1} - \frac{1}{n + 2}

解答ステップ

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

拡張

(1 - y) y^{n} : y^{n} - y^{1 + n}

= \int_{0}^{1} y^{n} - y^{1 + n} d y

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} y^{n} d y - \int_{0}^{1} y^{1 + n} d y

\int_{0}^{1} y^{n} d y = \frac{1}{n + 1}

\int_{0}^{1} y^{1 + n} d y = \frac{1}{n + 2}

= \frac{1}{n + 1} - \frac{1}{n + 2}