What is the integral from 0 to 5 of (4x-3)^3 ?

The integral from 0 to 5 of (4x-3)^3 is 5215

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integral from 0 to 5 of (4x-3)^3

\int_{0}^{5} (4 x - 3)^{3} d x

5215

Solution steps

\int_{0}^{5} (4 x - 3)^{3} d x

Apply u-substitution

: \int_{- 3}^{17} \frac{u ^{3}}{4} d u

= \int_{- 3}^{17} \frac{u ^{3}}{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_{- 3}^{17} u^{3} d u

Apply the Power Rule

: [\frac{u ^{4}}{4}]_{- 3}^{17}

= \frac{1}{4} [\frac{u ^{4}}{4}]_{- 3}^{17}

Compute the boundaries:

20860

= \frac{1}{4} \cdot 20860

Simplify

= 5215

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What is the integral from 0 to 5 of (4x-3)^3 ?
The integral from 0 to 5 of (4x-3)^3 is 5215