What is the integral from 0 to 8 of (8-(y^2)/8) ?

The integral from 0 to 8 of (8-(y^2)/8) is 128/3

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integral from 0 to 8 of (8-(y^2)/8)

\int_{0}^{8} (8 - \frac{y ^{2}}{8}) d y

\frac{128}{3}

Decimal

42.66666 \dots

Solution steps

\int_{0}^{8} 8 - \frac{y ^{2}}{8} d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{8} 8 d y - \int_{0}^{8} \frac{y ^{2}}{8} d y

\int_{0}^{8} 8 d y = 64

\int_{0}^{8} \frac{y ^{2}}{8} d y = \frac{64}{3}

= 64 - \frac{64}{3}

Simplify

64 - \frac{64}{3} : \frac{128}{3}

= \frac{128}{3}

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What is the integral from 0 to 8 of (8-(y^2)/8) ?
The integral from 0 to 8 of (8-(y^2)/8) is 128/3