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

The integral from 2 to B of (2x-3)^3 is 2B^4-12B^3+27B^2-27B+10

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integral from 2 to B of (2x-3)^3

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

2 B^{4} - 12 B^{3} + 27 B^{2} - 27 B + 10

Solution steps

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

Apply u-substitution

: \int_{1}^{2 B - 3} \frac{u ^{3}}{2} d u

= \int_{1}^{2 B - 3} \frac{u ^{3}}{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}^{2 B - 3} u^{3} d u

Apply the Power Rule

: [\frac{u ^{4}}{4}]_{1}^{2 B - 3}

= \frac{1}{2} [\frac{u ^{4}}{4}]_{1}^{2 B - 3}

Compute the boundaries:

\frac{( 2 B - 3 ) ^{4}}{4} - \frac{1}{4}

= \frac{1}{2} (\frac{( 2 B - 3 ) ^{4}}{4} - \frac{1}{4})

Simplify

= 2 B^{4} - 12 B^{3} + 27 B^{2} - 27 B + 10

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What is the integral from 2 to B of (2x-3)^3 ?
The integral from 2 to B of (2x-3)^3 is 2B^4-12B^3+27B^2-27B+10