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

The integral from-1 to 1 of (4-4x^2)^2 is 256/15

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integral from-1 to 1 of (4-4x^2)^2

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

\frac{256}{15}

Decimal Notation

17.06666 \dots

Solution steps

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

Expand

(4 - 4 x^{2})^{2} : 16 - 32 x^{2} + 16 x^{4}

= \int_{- 1}^{1} 16 - 32 x^{2} + 16 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} 16 d x - \int_{- 1}^{1} 32 x^{2} d x + \int_{- 1}^{1} 16 x^{4} d x

\int_{- 1}^{1} 16 d x = 32

\int_{- 1}^{1} 32 x^{2} d x = \frac{64}{3}

\int_{- 1}^{1} 16 x^{4} d x = \frac{32}{5}

= 32 - \frac{64}{3} + \frac{32}{5}

Simplify

32 - \frac{64}{3} + \frac{32}{5} : \frac{256}{15}

= \frac{256}{15}

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What is the integral from-1 to 1 of (4-4x^2)^2 ?
The integral from-1 to 1 of (4-4x^2)^2 is 256/15