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

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

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

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

\frac{1}{n + 1}

Solution steps

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

Apply u-substitution

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

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

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

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

Take the constant out:

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

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

Apply the Power Rule

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

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

Simplify

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

Compute the boundaries:

\frac{1}{n + 1}

= \frac{1}{n + 1}

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