What is the integral from-1 to 1 of (8-8x^2)^2 ?

The integral from-1 to 1 of (8-8x^2)^2 is 1024/15

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integral from-1 to 1 of (8-8x^2)^2

\int_{- 1}^{1} (8 - 8 x^{2})^{2} d x

\frac{1024}{15}

Decimal Notation

68.26666 \dots

Solution steps

\int_{- 1}^{1} (8 - 8 x^{2})^{2} d x

Expand

(8 - 8 x^{2})^{2} : 64 - 128 x^{2} + 64 x^{4}

= \int_{- 1}^{1} 64 - 128 x^{2} + 64 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} 64 d x - \int_{- 1}^{1} 128 x^{2} d x + \int_{- 1}^{1} 64 x^{4} d x

\int_{- 1}^{1} 64 d x = 128

\int_{- 1}^{1} 128 x^{2} d x = \frac{256}{3}

\int_{- 1}^{1} 64 x^{4} d x = \frac{128}{5}

= 128 - \frac{256}{3} + \frac{128}{5}

Simplify

128 - \frac{256}{3} + \frac{128}{5} : \frac{1024}{15}

= \frac{1024}{15}

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What is the integral from-1 to 1 of (8-8x^2)^2 ?
The integral from-1 to 1 of (8-8x^2)^2 is 1024/15