What is the integral from 0 to 1 of (1-x)^9 ?

The integral from 0 to 1 of (1-x)^9 is 1/10

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integral from 0 to 1 of (1-x)^9

\int_{0}^{1} (1 - x)^{9} d x

\frac{1}{10}

Decimal

0.1

Solution steps

\int_{0}^{1} (1 - x)^{9} d x

Apply u-substitution

= \int_{1}^{0} - u^{9} d u

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

= - \int_{0}^{1} - u^{9} d u

Take the constant out:

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

= - (- \int_{0}^{1} u^{9} d u)

Apply the Power Rule

= - (- [\frac{u ^{10}}{10}]_{0}^{1})

Simplify

= [\frac{u ^{10}}{10}]_{0}^{1}

Compute the boundaries:

\frac{1}{10}

= \frac{1}{10}

What is the integral from 0 to 1 of (1-x)^9 ?
The integral from 0 to 1 of (1-x)^9 is 1/10