integral from 0 to x of (1-r/x)^{1/n}r

解

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

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

解答ステップ

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

u置換積分法を適用する

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

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

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

定数を除く:

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

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

拡張

(- xu + x) u^{\frac{1}{n}} : - x u^{1 + \frac{1}{n}} + x u^{\frac{1}{n}}

= - (- x \cdot \int_{0}^{1} - x u^{1 + \frac{1}{n}} + x u^{\frac{1}{n}} 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

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

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

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

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

簡素化

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