Recall that the derivative of a square root will produce its reciprocal, i.e. $(\sqrt{x})'=-\frac{1}{2}\frac{1}{\sqrt{x}}$. It can be find by trying that 

$$f(x)=\sqrt{x}\sqrt{4-x} \implies f'(x)=\frac{1}{2}\left(\sqrt{\frac{4-x}{x}}-\sqrt{\frac{x}{4-x}}\right)$$

Therefore, the given integral must equal

$$2\sqrt{x}\sqrt{4-x}|_0^2=\boxed{4}$$