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

The integral from 0 to 2 of pi(2y-y^2)^2 is pi 16/15

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integral from 0 to 2 of pi(2y-y^2)^2

\int_{0}^{2} π (2 y - y^{2})^{2} d y

π \frac{16}{15}

Decimal

3.35103 \dots

Solution steps

\int_{0}^{2} π (2 y - y^{2})^{2} d y

Take the constant out:

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

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

Expand

(2 y - y^{2})^{2} : 4 y^{2} - 4 y^{3} + y^{4}

= π \cdot \int_{0}^{2} 4 y^{2} - 4 y^{3} + y^{4} d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= π (\int_{0}^{2} 4 y^{2} d y - \int_{0}^{2} 4 y^{3} d y + \int_{0}^{2} y^{4} d y)

\int_{0}^{2} 4 y^{2} d y = \frac{32}{3}

\int_{0}^{2} 4 y^{3} d y = 16

\int_{0}^{2} y^{4} d y = \frac{32}{5}

= π (\frac{32}{3} - 16 + \frac{32}{5})

Simplify

π (\frac{32}{3} - 16 + \frac{32}{5}) : π \frac{16}{15}

= π \frac{16}{15}

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What is the integral from 0 to 2 of pi(2y-y^2)^2 ?
The integral from 0 to 2 of pi(2y-y^2)^2 is pi 16/15