What is the integral from 0 to 2 of (2x-1)^9 ?

The integral from 0 to 2 of (2x-1)^9 is 14762/5

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integral from 0 to 2 of (2x-1)^9

\int_{0}^{2} (2 x - 1)^{9} d x

\frac{14762}{5}

Decimal Notation

2952.4

Solution steps

\int_{0}^{2} (2 x - 1)^{9} d x

Apply u-substitution

: \int_{- 1}^{3} \frac{u ^{9}}{2} d u

= \int_{- 1}^{3} \frac{u ^{9}}{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}^{3} u^{9} d u

Apply the Power Rule

: [\frac{u ^{10}}{10}]_{- 1}^{3}

= \frac{1}{2} [\frac{u ^{10}}{10}]_{- 1}^{3}

Compute the boundaries:

\frac{29524}{5}

= \frac{1}{2} \cdot \frac{29524}{5}

Simplify

= \frac{14762}{5}

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What is the integral from 0 to 2 of (2x-1)^9 ?
The integral from 0 to 2 of (2x-1)^9 is 14762/5