What is the integral from 1 to 2 of (2x-3)^4 ?

The integral from 1 to 2 of (2x-3)^4 is 1/5

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integral from 1 to 2 of (2x-3)^4

\int_{1}^{2} (2 x - 3)^{4} d x

\frac{1}{5}

Solution steps

\int_{1}^{2} (2 x - 3)^{4} d x

Apply u-substitution

: \int_{- 1}^{1} \frac{u ^{4}}{2} d u

= \int_{- 1}^{1} \frac{u ^{4}}{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^{4} d u

Apply the Power Rule

: [\frac{u ^{5}}{5}]_{- 1}^{1}

= \frac{1}{2} [\frac{u ^{5}}{5}]_{- 1}^{1}

Compute the boundaries:

\frac{2}{5}

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

Simplify

= \frac{1}{5}

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What is the integral from 1 to 2 of (2x-3)^4 ?
The integral from 1 to 2 of (2x-3)^4 is 1/5