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

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

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

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

- \frac{64}{3}

Decimal

- 21.33333 \dots

Solution steps

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 2}^{2} - 2 x^{2} + 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_{- 2}^{2} 2 x^{2} d x + \int_{- 2}^{2} 8 d x)

\int_{- 2}^{2} 2 x^{2} d x = \frac{32}{3}

\int_{- 2}^{2} 8 d x = 32

= - (- \frac{32}{3} + 32)

Join

- \frac{32}{3} + 32 : \frac{64}{3}

= - \frac{64}{3}

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