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

The integral from 1 to 2 of (2x-4)^5 is -16/3

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integral from 1 to 2 of (2x-4)^5

\int_{1}^{2} (2 x - 4)^{5} d x

- \frac{16}{3}

Decimal Notation

- 5.33333 \dots

Solution steps

\int_{1}^{2} (2 x - 4)^{5} d x

Apply u-substitution

: \int_{- 2}^{0} \frac{u ^{5}}{2} d u

= \int_{- 2}^{0} \frac{u ^{5}}{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_{- 2}^{0} u^{5} d u

Apply the Power Rule

: [\frac{u ^{6}}{6}]_{- 2}^{0}

= \frac{1}{2} [\frac{u ^{6}}{6}]_{- 2}^{0}

Compute the boundaries:

- \frac{32}{3}

= \frac{1}{2} (- \frac{32}{3})

Simplify

= - \frac{16}{3}

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What is the integral from 1 to 2 of (2x-4)^5 ?
The integral from 1 to 2 of (2x-4)^5 is -16/3