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

The integral from 0 to 4/3 of pi(4-3x)^2 is (64pi)/9

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

\int_{0}^{\frac{4}{3}} π (4 - 3 x)^{2} d x

\frac{64 π}{9}

Decimal

22.34021 \dots

Solution steps

\int_{0}^{\frac{4}{3}} π (4 - 3 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}^{\frac{4}{3}} (4 - 3 x)^{2} d x

Apply u-substitution

: \int_{4}^{0} - \frac{u ^{2}}{3} d u

= π \cdot \int_{4}^{0} - \frac{u ^{2}}{3} 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}}{3} d u)

Take the constant out:

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

= π (- (- \frac{1}{3} \cdot \int_{0}^{4} u^{2} d u))

Apply the Power Rule

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

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

Simplify

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

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

Compute the boundaries:

\frac{64}{3}

= \frac{π}{3} \cdot \frac{64}{3}

Simplify

= \frac{64 π}{9}

z \to \frac{1}{2} lim (\frac{3 z + 1}{2 z - 5})

\int (2 - 2 y) d y

\frac{\partial}{\partial y} (8 x e^{x^{2} + y^{2}})

f (x) = 4 x

\frac{d}{d x} (3 ln (x))

What is the integral from 0 to 4/3 of pi(4-3x)^2 ?
The integral from 0 to 4/3 of pi(4-3x)^2 is (64pi)/9