What is the integral from-4 to 4 of [(32-x^2)-(x^2)] ?

The integral from-4 to 4 of [(32-x^2)-(x^2)] is 512/3

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integral from-4 to 4 of [(32-x^2)-(x^2)]

\int_{- 4}^{4} [(32 - x^{2}) - (x^{2})] d x

\frac{512}{3}

Decimal

170.66666 \dots

Solution steps

\int_{- 4}^{4} 32 - x^{2} - x^{2} d x

Simplify

= \int_{- 4}^{4} 32 - 2 x^{2} 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_{- 4}^{4} 32 d x - \int_{- 4}^{4} 2 x^{2} d x

\int_{- 4}^{4} 32 d x = 256

\int_{- 4}^{4} 2 x^{2} d x = \frac{256}{3}

= 256 - \frac{256}{3}

Simplify

256 - \frac{256}{3} : \frac{512}{3}

= \frac{512}{3}

What is the integral from-4 to 4 of [(32-x^2)-(x^2)] ?
The integral from-4 to 4 of [(32-x^2)-(x^2)] is 512/3