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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

- \frac{( n - y ) ^{n}}{n ^{n}} + 1

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

= n \cdot \int_{1}^{1 - \frac{y}{n}} - u^{- 1 + n} d u

Take the constant out:

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

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

Apply the Power Rule

= n (- [\frac{u ^{n}}{n}]_{1}^{1 - \frac{y}{n}})

Simplify

= - n [\frac{u ^{n}}{n}]_{1}^{1 - \frac{y}{n}}

Compute the boundaries:

\frac{( n - y ) ^{n}}{n ^{n + 1}} - \frac{1}{n}

= - n (\frac{( n - y ) ^{n}}{n ^{n + 1}} - \frac{1}{n})

Simplify

= - \frac{( n - y ) ^{n}}{n ^{n}} + 1

\int_{- x}^{x^{2}} 8 x - 10 y + 2 d y

\int_{0}^{1} \frac{e ^{x} + 1}{e ^{x}} d x

\int_{1}^{2} 4 - 2 x d x

\int_{0}^{1} x (3 x^{2} - 1)^{5} d x

\int_{1}^{2} (\frac{x}{2} - \frac{9}{x}) d x

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