What is the integral from-1 to 1 of (x^4-x^8) ?

The integral from-1 to 1 of (x^4-x^8) is 8/45

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integral from-1 to 1 of (x^4-x^8)

\int_{- 1}^{1} (x^{4} - x^{8}) d x

\frac{8}{45}

Decimal

0.17777 \dots

Solution steps

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

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

\int_{- 1}^{1} x^{8} d x = \frac{2}{9}

= \frac{2}{5} - \frac{2}{9}

Simplify

\frac{2}{5} - \frac{2}{9} : \frac{8}{45}

= \frac{8}{45}

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What is the integral from-1 to 1 of (x^4-x^8) ?
The integral from-1 to 1 of (x^4-x^8) is 8/45