What is the integral from 0 to 1 of (2x-1)^{10} ?

The integral from 0 to 1 of (2x-1)^{10} is 1/11

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integral from 0 to 1 of (2x-1)^{10}

\int_{0}^{1} (2 x - 1)^{10} d x

\frac{1}{11}

Decimal

0.09090 \dots

Solution steps

\int_{0}^{1} (2 x - 1)^{10} d x

Apply u-substitution

: \int_{- 1}^{1} \frac{u ^{10}}{2} d u

= \int_{- 1}^{1} \frac{u ^{10}}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{- 1}^{1} u^{10} d u

Apply the Power Rule

: [\frac{u ^{11}}{11}]_{- 1}^{1}

= \frac{1}{2} [\frac{u ^{11}}{11}]_{- 1}^{1}

Compute the boundaries:

\frac{2}{11}

= \frac{1}{2} \cdot \frac{2}{11}

Simplify

= \frac{1}{11}

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What is the integral from 0 to 1 of (2x-1)^{10} ?
The integral from 0 to 1 of (2x-1)^{10} is 1/11