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

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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

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

= 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

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

= 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

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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