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

The integral from 2 to B of (2x-3)^4 is (16B^5-120B^4+360B^3-540B^2+405B-122)/5

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

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

\frac{16 B ^{5} - 120 B ^{4} + 360 B ^{3} - 540 B ^{2} + 405 B - 122}{5}

Solution steps

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

Apply u-substitution

: \int_{1}^{2 B - 3} \frac{u ^{4}}{2} d u

= \int_{1}^{2 B - 3} \frac{u ^{4}}{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^{4} d u

Apply the Power Rule

: [\frac{u ^{5}}{5}]_{1}^{2 B - 3}

= \frac{1}{2} [\frac{u ^{5}}{5}]_{1}^{2 B - 3}

Compute the boundaries:

\frac{( 2 B - 3 ) ^{5}}{5} - \frac{1}{5}

= \frac{1}{2} (\frac{( 2 B - 3 ) ^{5}}{5} - \frac{1}{5})

Simplify

= \frac{16 B ^{5} - 120 B ^{4} + 360 B ^{3} - 540 B ^{2} + 405 B - 122}{5}

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What is the integral from 2 to B of (2x-3)^4 ?
The integral from 2 to B of (2x-3)^4 is (16B^5-120B^4+360B^3-540B^2+405B-122)/5