What is the integral from 0 to 2 of (x^4-8x) ?

The integral from 0 to 2 of (x^4-8x) is -48/5

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integral from 0 to 2 of (x^4-8x)

\int_{0}^{2} (x^{4} - 8 x) d x

- \frac{48}{5}

Decimal

- 9.6

Solution steps

\int_{0}^{2} x^{4} - 8 x 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_{0}^{2} x^{4} d x - \int_{0}^{2} 8 x d x

\int_{0}^{2} x^{4} d x = \frac{32}{5}

\int_{0}^{2} 8 x d x = 16

= \frac{32}{5} - 16

Simplify

\frac{32}{5} - 16 : - \frac{48}{5}

= - \frac{48}{5}

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

\int_{- 2}^{2} \frac{x ^{3}}{1 + x ^{2}} d x

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\int_{0}^{l n (x)} y e^{z} d z

\int_{2}^{10} \frac{1}{12} d x

What is the integral from 0 to 2 of (x^4-8x) ?
The integral from 0 to 2 of (x^4-8x) is -48/5