What is the integral from 0 to 2 of (-8y^2+256/3+8y) ?

The integral from 0 to 2 of (-8y^2+256/3+8y) is 496/3

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integral from 0 to 2 of (-8y^2+256/3+8y)

\int_{0}^{2} (- 8 y^{2} + \frac{256}{3} + 8 y) d y

\frac{496}{3}

Decimal Notation

165.33333 \dots

Solution steps

\int_{0}^{2} - 8 y^{2} + \frac{256}{3} + 8 y 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}^{2} 8 y^{2} d y + \int_{0}^{2} \frac{256}{3} d y + \int_{0}^{2} 8 y d y

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

\int_{0}^{2} \frac{256}{3} d y = \frac{512}{3}

\int_{0}^{2} 8 y d y = 16

= - \frac{64}{3} + \frac{512}{3} + 16

Simplify

- \frac{64}{3} + \frac{512}{3} + 16 : \frac{496}{3}

= \frac{496}{3}

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What is the integral from 0 to 2 of (-8y^2+256/3+8y) ?
The integral from 0 to 2 of (-8y^2+256/3+8y) is 496/3