Let r be the region bounded by the graphs of y= sqrt(x) and y=x/2, how do you find the area of R?

Nov 7, 2016

Answer:

The area of R $= \frac{10}{3}$

Explanation:

Let's find the intercepts of the two graphs.
$\sqrt{x} = \frac{x}{2}$$\implies$$x = {x}^{2} / 4$ $\implies$$x \left(\frac{x}{4} - 1\right)$
so, $x = 0$ and $x = 4$
a small area is $\mathrm{dA} = \left(\sqrt{x} - \frac{x}{4}\right) \mathrm{dx}$
$\therefore A = {\int}_{0}^{4} \left(\sqrt{x} - \frac{x}{4}\right) \mathrm{dx}$
$= {\left[\frac{2}{3} {x}^{\frac{3}{2}} - {x}^{2} / 8\right]}_{0}^{4} = \frac{16}{3} - 2 = \frac{10}{3}$
graph{(y-sqrtx)(y-x/2)=0 [-4.19, 6.907, -2.38, 3.167]}