What is the integral from-1 to 1 of (4x^2-16/3) ?

The integral from-1 to 1 of (4x^2-16/3) is -8

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integral from-1 to 1 of (4x^2-16/3)

\int_{- 1}^{1} (4 x^{2} - \frac{16}{3}) d x

- 8

Solution steps

\int_{- 1}^{1} 4 x^{2} - \frac{16}{3} 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} 4 x^{2} d x - \int_{- 1}^{1} \frac{16}{3} d x

\int_{- 1}^{1} 4 x^{2} d x = \frac{8}{3}

\int_{- 1}^{1} \frac{16}{3} d x = \frac{32}{3}

= \frac{8}{3} - \frac{32}{3}

Simplify

\frac{8}{3} - \frac{32}{3} : - 8

= - 8

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What is the integral from-1 to 1 of (4x^2-16/3) ?
The integral from-1 to 1 of (4x^2-16/3) is -8