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

The integral from 0 to 4 of (-2x^2+8x) is 64/3

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

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

\frac{64}{3}

Decimal

21.33333 \dots

Solution steps

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

\int_{0}^{4} 2 x^{2} d x = \frac{128}{3}

\int_{0}^{4} 8 x d x = 64

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

Simplify

- \frac{128}{3} + 64 : \frac{64}{3}

= \frac{64}{3}

What is the integral from 0 to 4 of (-2x^2+8x) ?
The integral from 0 to 4 of (-2x^2+8x) is 64/3