What is the integral from 0 to 4 of 4y^2-(y^4)/4 ?

The integral from 0 to 4 of 4y^2-(y^4)/4 is 512/15

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integral from 0 to 4 of 4y^2-(y^4)/4

\int_{0}^{4} 4 y^{2} - \frac{y ^{4}}{4} d y

\frac{512}{15}

Decimal

34.13333 \dots

Solution steps

\int_{0}^{4} 4 y^{2} - \frac{y ^{4}}{4} 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}^{4} 4 y^{2} d y - \int_{0}^{4} \frac{y ^{4}}{4} d y

\int_{0}^{4} 4 y^{2} d y = \frac{256}{3}

\int_{0}^{4} \frac{y ^{4}}{4} d y = \frac{256}{5}

= \frac{256}{3} - \frac{256}{5}

Simplify

\frac{256}{3} - \frac{256}{5} : \frac{512}{15}

= \frac{512}{15}

What is the integral from 0 to 4 of 4y^2-(y^4)/4 ?
The integral from 0 to 4 of 4y^2-(y^4)/4 is 512/15