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

The integral from 0 to x of (1-r/x)^{1/n}r is x(-(xn)/(2n+1)+(xn)/(n+1))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

\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})

Solution steps

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

Apply u-substitution

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

= \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

Take the constant out:

\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)

Expand

(- 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)

Apply the Sum Rule:

\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}))

Simplify

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

\int_{5}^{12} x d x

\int_{1}^{2} x^{5} e^{x^{2}} d x

\int_{0}^{1} 4 arctan (x^{2}) d x

\int_{0}^{1} 4 - r^{2} r d r

\int_{0}^{1} (14 ln (x)) d x

What is the integral from 0 to x of (1-r/x)^{1/n}r ?
The integral from 0 to x of (1-r/x)^{1/n}r is x(-(xn)/(2n+1)+(xn)/(n+1))