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

The integral from-8 to 8 of ((64-x^2))/2 is 1024/3

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integral from-8 to 8 of ((64-x^2))/2

\int_{- 8}^{8} \frac{( 64 - x ^{2} )}{2} d x

\frac{1024}{3}

Decimal Notation

341.33333 \dots

Solution steps

\int_{- 8}^{8} \frac{64 - x ^{2}}{2} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{- 8}^{8} 64 - 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

= \frac{1}{2} (\int_{- 8}^{8} 64 d x - \int_{- 8}^{8} x^{2} d x)

\int_{- 8}^{8} 64 d x = 1024

\int_{- 8}^{8} x^{2} d x = \frac{1024}{3}

= \frac{1}{2} (1024 - \frac{1024}{3})

Simplify

\frac{1}{2} (1024 - \frac{1024}{3}) : \frac{1024}{3}

= \frac{1024}{3}

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What is the integral from-8 to 8 of ((64-x^2))/2 ?
The integral from-8 to 8 of ((64-x^2))/2 is 1024/3