What is the integral from 0 to 1 of 4y^2(1-y^2) ?

The integral from 0 to 1 of 4y^2(1-y^2) is 8/15

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integral from 0 to 1 of 4y^2(1-y^2)

\int_{0}^{1} 4 y^{2} (1 - y^{2}) d y

\frac{8}{15}

Decimal

0.53333 \dots

Solution steps

\int_{0}^{1} 4 y^{2} (1 - y^{2}) d y

Take the constant out:

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

= 4 \cdot \int_{0}^{1} y^{2} (1 - y^{2}) d y

Expand

y^{2} (1 - y^{2}) : y^{2} - y^{4}

= 4 \cdot \int_{0}^{1} y^{2} - 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

= 4 (\int_{0}^{1} y^{2} d y - \int_{0}^{1} y^{4} d y)

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

\int_{0}^{1} y^{4} d y = \frac{1}{5}

= 4 (\frac{1}{3} - \frac{1}{5})

Simplify

4 (\frac{1}{3} - \frac{1}{5}) : \frac{8}{15}

= \frac{8}{15}

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What is the integral from 0 to 1 of 4y^2(1-y^2) ?
The integral from 0 to 1 of 4y^2(1-y^2) is 8/15