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

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

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

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

\frac{64}{3}

Decimal

21.33333 \dots

Solution steps

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

Simplify

= \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

Simplify

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

= \frac{64}{3}

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