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

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

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

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

1

Solution steps

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

Apply u-substitution

: \int_{1}^{0} - u^{n - 1} d u

= n \cdot \int_{1}^{0} - u^{n - 1} d u

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

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

Take the constant out:

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

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

Apply the Power Rule

: [\frac{u ^{n}}{n}]_{0}^{1}

= n (- (- [\frac{u ^{n}}{n}]_{0}^{1}))

Simplify

= n [\frac{u ^{n}}{n}]_{0}^{1}

Compute the boundaries:

\frac{1}{n}

= n \frac{1}{n}

Simplify

= 1

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