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

The integral from 0 to 4 of (sqrt(x)-x/2)^2 is 8/15

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

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

\frac{8}{15}

Decimal

0.53333 \dots

Solution steps

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

Expand

(x - \frac{x}{2})^{2} : x - x^{\frac{3}{2}} + \frac{x ^{2}}{4}

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

\int_{0}^{4} x d x = 8

\int_{0}^{4} x^{\frac{3}{2}} d x = \frac{64}{5}

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

= 8 - \frac{64}{5} + \frac{16}{3}

Simplify

8 - \frac{64}{5} + \frac{16}{3} : \frac{8}{15}

= \frac{8}{15}

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What is the integral from 0 to 4 of (sqrt(x)-x/2)^2 ?
The integral from 0 to 4 of (sqrt(x)-x/2)^2 is 8/15