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

The integral from-1 to 1 of (x^2-2)^2 is 86/15

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

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

\frac{86}{15}

Decimal

5.73333 \dots

Solution steps

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

Expand

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

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

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

\int_{- 1}^{1} 4 x^{2} d x = \frac{8}{3}

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

= \frac{2}{5} - \frac{8}{3} + 8

Simplify

\frac{2}{5} - \frac{8}{3} + 8 : \frac{86}{15}

= \frac{86}{15}

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What is the integral from-1 to 1 of (x^2-2)^2 ?
The integral from-1 to 1 of (x^2-2)^2 is 86/15