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

The integral from 3/2 to 5/2 of (2x-4)^3 is 0

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

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

0

Solution steps

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

Apply u-substitution

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

= \int_{- 1}^{1} \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}^{1} u^{3} d u

Apply the Power Rule

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

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

Compute the boundaries:

0

= \frac{1}{2} \cdot 0

Simplify

= 0

\int_{0}^{2} x^{3} - 2 x^{2} + 10 x d x

\int_{1}^{2} ln (x^{2} + 1) d x

\int_{0}^{4} 3 x^{2} ln (2 x) d x

\int_{0}^{x} y e^{- y} d y

\int_{0}^{1} t t^{2} + 1 d t

What is the integral from 3/2 to 5/2 of (2x-4)^3 ?
The integral from 3/2 to 5/2 of (2x-4)^3 is 0