What is the integral from 0 to 8 of-2x^2+16x ?

The integral from 0 to 8 of-2x^2+16x is 512/3

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integral from 0 to 8 of-2x^2+16x

\int_{0}^{8} - 2 x^{2} + 16 x d x

\frac{512}{3}

Decimal

170.66666 \dots

Solution steps

\int_{0}^{8} - 2 x^{2} + 16 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}^{8} 2 x^{2} d x + \int_{0}^{8} 16 x d x

\int_{0}^{8} 2 x^{2} d x = \frac{1024}{3}

\int_{0}^{8} 16 x d x = 512

= - \frac{1024}{3} + 512

Simplify

- \frac{1024}{3} + 512 : \frac{512}{3}

= \frac{512}{3}

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What is the integral from 0 to 8 of-2x^2+16x ?
The integral from 0 to 8 of-2x^2+16x is 512/3