What is the integral from 2 to 8 of (4x-9)^{20} ?

The integral from 2 to 8 of (4x-9)^{20} is (23^{21}+1)/(84)

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integral from 2 to 8 of (4x-9)^{20}

\int_{2}^{8} (4 x - 9)^{20} d x

\frac{2 3 ^{21} + 1}{84}

Decimal

4.699 E 26

Solution steps

\int_{2}^{8} (4 x - 9)^{20} d x

Apply u-substitution

: \int_{- 1}^{23} \frac{u ^{20}}{4} d u

= \int_{- 1}^{23} \frac{u ^{20}}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int_{- 1}^{23} u^{20} d u

Apply the Power Rule

: [\frac{u ^{21}}{21}]_{- 1}^{23}

= \frac{1}{4} [\frac{u ^{21}}{21}]_{- 1}^{23}

Compute the boundaries:

\frac{2 3 ^{21}}{21} + \frac{1}{21}

= \frac{1}{4} (\frac{2 3 ^{21}}{21} + \frac{1}{21})

Simplify

= \frac{2 3 ^{21} + 1}{84}

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What is the integral from 2 to 8 of (4x-9)^{20} ?
The integral from 2 to 8 of (4x-9)^{20} is (23^{21}+1)/(84)