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

The integral from-2 to 2 of (4-y^2)^2 is 512/15

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

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

\frac{512}{15}

Decimal

34.13333 \dots

Solution steps

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

Expand

(4 - y^{2})^{2} : 16 - 8 y^{2} + y^{4}

= \int_{- 2}^{2} 16 - 8 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

= \int_{- 2}^{2} 16 d y - \int_{- 2}^{2} 8 y^{2} d y + \int_{- 2}^{2} y^{4} d y

\int_{- 2}^{2} 16 d y = 64

\int_{- 2}^{2} 8 y^{2} d y = \frac{128}{3}

\int_{- 2}^{2} y^{4} d y = \frac{64}{5}

= 64 - \frac{128}{3} + \frac{64}{5}

Simplify

64 - \frac{128}{3} + \frac{64}{5} : \frac{512}{15}

= \frac{512}{15}

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What is the integral from-2 to 2 of (4-y^2)^2 ?
The integral from-2 to 2 of (4-y^2)^2 is 512/15