What is the integral from 0 to 2 of y^2(1/2 (2-y)) ?

The integral from 0 to 2 of y^2(1/2 (2-y)) is 2/3

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integral from 0 to 2 of y^2(1/2 (2-y))

\int_{0}^{2} y^{2} (\frac{1}{2} (2 - y)) d y

\frac{2}{3}

Decimal

0.66666 \dots

Solution steps

\int_{0}^{2} y^{2} \frac{1}{2} (2 - y) d y

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{2} y^{2} (2 - y) d y

Expand

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

= \frac{1}{2} \cdot \int_{0}^{2} 2 y^{2} - y^{3} 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

= \frac{1}{2} (\int_{0}^{2} 2 y^{2} d y - \int_{0}^{2} y^{3} d y)

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

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

= \frac{1}{2} (\frac{16}{3} - 4)

Simplify

\frac{1}{2} (\frac{16}{3} - 4) : \frac{2}{3}

= \frac{2}{3}

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What is the integral from 0 to 2 of y^2(1/2 (2-y)) ?
The integral from 0 to 2 of y^2(1/2 (2-y)) is 2/3