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

The integral from 0 to 4 of sqrt(4x)-(x^2)/4 is 16/3

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

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

\frac{16}{3}

Decimal

5.33333 \dots

Solution steps

\int_{0}^{4} 4 x - \frac{x ^{2}}{4} 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_{0}^{4} 4 x d x - \int_{0}^{4} \frac{x ^{2}}{4} d x

\int_{0}^{4} 4 x d x = \frac{32}{3}

\int_{0}^{4} \frac{x ^{2}}{4} d x = \frac{16}{3}

= \frac{32}{3} - \frac{16}{3}

Simplify

\frac{32}{3} - \frac{16}{3} : \frac{16}{3}

= \frac{16}{3}

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What is the integral from 0 to 4 of sqrt(4x)-(x^2)/4 ?
The integral from 0 to 4 of sqrt(4x)-(x^2)/4 is 16/3