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

The integral from-2 to 2 of (x^4-4x^2) is -128/15

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

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

- \frac{128}{15}

Decimal Notation

- 8.53333 \dots

Solution steps

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

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

\int_{- 2}^{2} 4 x^{2} d x = \frac{64}{3}

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

Simplify

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

= - \frac{128}{15}

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What is the integral from-2 to 2 of (x^4-4x^2) ?
The integral from-2 to 2 of (x^4-4x^2) is -128/15