What is the integral from 0 to 4 of sqrt(x)-(x^2)/8 ?

The integral from 0 to 4 of sqrt(x)-(x^2)/8 is 8/3

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integral from 0 to 4 of sqrt(x)-(x^2)/8

\int_{0}^{4} x - \frac{x ^{2}}{8} d x

\frac{8}{3}

Decimal

2.66666 \dots

Solution steps

\int_{0}^{4} x - \frac{x ^{2}}{8} 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} x d x - \int_{0}^{4} \frac{x ^{2}}{8} d x

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

\int_{0}^{4} \frac{x ^{2}}{8} d x = \frac{8}{3}

= \frac{16}{3} - \frac{8}{3}

Simplify

\frac{16}{3} - \frac{8}{3} : \frac{8}{3}

= \frac{8}{3}

What is the integral from 0 to 4 of sqrt(x)-(x^2)/8 ?
The integral from 0 to 4 of sqrt(x)-(x^2)/8 is 8/3