integral from 0 to 1 of k(1-x)x^n

解

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

k (\frac{1}{n + 1} - \frac{1}{n + 2})

解答ステップ

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

定数を除く:

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

= k \cdot \int_{0}^{1} (1 - x) x^{n} d x

拡張

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

= k \cdot \int_{0}^{1} x^{n} - x^{1 + n} d x

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

= k (\int_{0}^{1} x^{n} d x - \int_{0}^{1} x^{1 + n} d x)

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

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

= k (\frac{1}{n + 1} - \frac{1}{n + 2})