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

The integral from-1 to 1 of (2-2x^2)^2 is 64/15

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integral from-1 to 1 of (2-2x^2)^2

\int_{- 1}^{1} (2 - 2 x^{2})^{2} d x

\frac{64}{15}

Decimal

4.26666 \dots

Solution steps

\int_{- 1}^{1} (2 - 2 x^{2})^{2} d x

Expand

(2 - 2 x^{2})^{2} : 4 - 8 x^{2} + 4 x^{4}

= \int_{- 1}^{1} 4 - 8 x^{2} + 4 x^{4} d x

Apply the Sum Rule:

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

= \int_{- 1}^{1} 4 d x - \int_{- 1}^{1} 8 x^{2} d x + \int_{- 1}^{1} 4 x^{4} d x

\int_{- 1}^{1} 4 d x = 8

\int_{- 1}^{1} 8 x^{2} d x = \frac{16}{3}

\int_{- 1}^{1} 4 x^{4} d x = \frac{8}{5}

= 8 - \frac{16}{3} + \frac{8}{5}

Simplify

8 - \frac{16}{3} + \frac{8}{5} : \frac{64}{15}

= \frac{64}{15}

What is the integral from-1 to 1 of (2-2x^2)^2 ?
The integral from-1 to 1 of (2-2x^2)^2 is 64/15