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

The integral from 2 to 5 of (2x-3)^4 is 8403/5

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integral from 2 to 5 of (2x-3)^4

\int_{2}^{5} (2 x - 3)^{4} d x

\frac{8403}{5}

Solution steps

\int_{2}^{5} (2 x - 3)^{4} d x

Apply u-substitution

: \int_{1}^{7} \frac{u ^{4}}{2} d u

= \int_{1}^{7} \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}^{7} u^{4} d u

Apply the Power Rule

: [\frac{u ^{5}}{5}]_{1}^{7}

= \frac{1}{2} [\frac{u ^{5}}{5}]_{1}^{7}

Compute the boundaries:

\frac{16806}{5}

= \frac{1}{2} \cdot \frac{16806}{5}

Simplify

= \frac{8403}{5}

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What is the integral from 2 to 5 of (2x-3)^4 ?
The integral from 2 to 5 of (2x-3)^4 is 8403/5