What is the integral from 0 to 2 of pi(4-2x)^2 ?

The integral from 0 to 2 of pi(4-2x)^2 is (32pi)/3

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integral from 0 to 2 of pi(4-2x)^2

\int_{0}^{2} π (4 - 2 x)^{2} d x

\frac{32 π}{3}

Decimal

33.51032 \dots

Solution steps

\int_{0}^{2} π (4 - 2 x)^{2} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= π \cdot \int_{0}^{2} (4 - 2 x)^{2} d x

Apply u-substitution

: \int_{4}^{0} - \frac{u ^{2}}{2} d u

= π \cdot \int_{4}^{0} - \frac{u ^{2}}{2} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= π (- \int_{0}^{4} - \frac{u ^{2}}{2} d u)

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= π (- (- \frac{1}{2} \cdot \int_{0}^{4} u^{2} d u))

Apply the Power Rule

: [\frac{u ^{3}}{3}]_{0}^{4}

= π (- (- \frac{1}{2} [\frac{u ^{3}}{3}]_{0}^{4}))

Simplify

π (- (- \frac{1}{2} [\frac{u ^{3}}{3}]_{0}^{4})) : \frac{π}{2} [\frac{u ^{3}}{3}]_{0}^{4}

= \frac{π}{2} [\frac{u ^{3}}{3}]_{0}^{4}

Compute the boundaries:

\frac{64}{3}

= \frac{π}{2} \cdot \frac{64}{3}

Simplify

= \frac{32 π}{3}

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What is the integral from 0 to 2 of pi(4-2x)^2 ?
The integral from 0 to 2 of pi(4-2x)^2 is (32pi)/3