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

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

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

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

\frac{512}{15}

Decimal

34.13333 \dots

Solution steps

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

Expand

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

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

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

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

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

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

Simplify

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

= \frac{512}{15}

\int (e^{2 x} \cdot, x) d x

d er i v a t i v e \frac{8 \cdot x}{x ^{2} + p}

\int e^{x} \cdot e^{x} d x

n = 1 \sum \infty \frac{4}{5 ^{n}}

d er i v a t i v e 6 x + \frac{8}{3}

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