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

The integral from 2 to B of (4x-3)^3 is 16B^4-48B^3+54B^2-27B-34

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

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

16 B^{4} - 48 B^{3} + 54 B^{2} - 27 B - 34

Solution steps

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

Apply u-substitution

: \int_{5}^{4 B - 3} \frac{u ^{3}}{4} d u

= \int_{5}^{4 B - 3} \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_{5}^{4 B - 3} u^{3} d u

Apply the Power Rule

: [\frac{u ^{4}}{4}]_{5}^{4 B - 3}

= \frac{1}{4} [\frac{u ^{4}}{4}]_{5}^{4 B - 3}

Compute the boundaries:

\frac{( 4 B - 3 ) ^{4}}{4} - \frac{625}{4}

= \frac{1}{4} (\frac{( 4 B - 3 ) ^{4}}{4} - \frac{625}{4})

Simplify

= 16 B^{4} - 48 B^{3} + 54 B^{2} - 27 B - 34

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What is the integral from 2 to B of (4x-3)^3 ?
The integral from 2 to B of (4x-3)^3 is 16B^4-48B^3+54B^2-27B-34