What is the integral from 2 to 5 of (4x-3)^9 ?

The integral from 2 to 5 of (4x-3)^9 is (17^{10}-9765625)/(40)

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integral from 2 to 5 of (4x-3)^9

\int_{2}^{5} (4 x - 3)^{9} d x

\frac{1 7 ^{10} - 9765625}{40}

Decimal

50399603370.6

Solution steps

\int_{2}^{5} (4 x - 3)^{9} d x

Apply u-substitution

: \int_{5}^{17} \frac{u ^{9}}{4} d u

= \int_{5}^{17} \frac{u ^{9}}{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_{5}^{17} u^{9} d u

Apply the Power Rule

: [\frac{u ^{10}}{10}]_{5}^{17}

= \frac{1}{4} [\frac{u ^{10}}{10}]_{5}^{17}

Compute the boundaries:

\frac{1 7 ^{10}}{10} - \frac{1953125}{2}

= \frac{1}{4} (\frac{1 7 ^{10}}{10} - \frac{1953125}{2})

Simplify

= \frac{1 7 ^{10} - 9765625}{40}

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What is the integral from 2 to 5 of (4x-3)^9 ?
The integral from 2 to 5 of (4x-3)^9 is (17^{10}-9765625)/(40)