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

The integral from-1 to 1 of pi(1-y^2)^2 is pi 16/15

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integral from-1 to 1 of pi(1-y^2)^2

\int_{- 1}^{1} π (1 - y^{2})^{2} d y

π \frac{16}{15}

Decimal

3.35103 \dots

Solution steps

\int_{- 1}^{1} π (1 - 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_{- 1}^{1} (1 - y^{2})^{2} d y

Expand

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

= π \cdot \int_{- 1}^{1} 1 - 2 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

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

\int_{- 1}^{1} 1 d y = 2

\int_{- 1}^{1} 2 y^{2} d y = \frac{4}{3}

\int_{- 1}^{1} y^{4} d y = \frac{2}{5}

= π (2 - \frac{4}{3} + \frac{2}{5})

Join

2 - \frac{4}{3} + \frac{2}{5} : \frac{16}{15}

= π \frac{16}{15}

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What is the integral from-1 to 1 of pi(1-y^2)^2 ?
The integral from-1 to 1 of pi(1-y^2)^2 is pi 16/15