Solve $\sqrt x = x/2$. If $f(x) = \sqrt x$ and $g(x) = x/2$, find the area of this limited area between $f(x)$ and $g(x)$.
I am having trouble solving this equation $\sqrt x = x/2$ that should give me the x values.
I know that the next step after this one is to solve $\int_a^b f(x)-g(x)~dx$ where $a= x_1$ and $b=x_2$
How do I go by to solve this $\sqrt x = x/2$ equation?
Thank you!

$$\sqrt{x}=\frac{x}{2}\\x\geq 0\\(\sqrt{x}=\frac{x}{2})^2\\x=\frac{x^2}{4}\\4x=x^2\\x(4-x)=0\\x=0\\x=4\\ \int_{0}^{4}(\sqrt{x}-\frac{x}{2})dx=\frac{x^{\frac{3}{2}}}{\frac{3}{2}}-\frac{x^2}{4}=\\(\frac{4^{\frac{3}{2}}}{\frac{3}{2}}-\frac{4^2}{4})-(\frac{0^{\frac{3}{2}}}{\frac{3}{2}}-\frac{0^2}{4})=\frac{4}{3}$$
