$=\int _{4}^{0}-\frac{1}{2\sqrt{u}}du$
$\int_{a}^{b}f(x)dx=-\int_{b}^{a}f(x)dx, a$=-\int _{0}^{4}-\frac{1}{2\sqrt{u}}du$
Take the constant out $\int{a\cdot{f(x)}dx}=a\cdot\int{f(x)dx}$$=-(-\frac{1}{2}\cdot \int _{0}^{4}\frac{1}{\sqrt{u}}du)$
Apply the Power Rule$=-(-\frac{1}{2}[2u^{\frac{1}{2}}]_{0}^{4})$
Simplify $-(-\frac{1}{2}[2u^{\frac{1}{2}}]_{0}^{4}):{\quad}\frac{1}{2}[2\sqrt{u}]_{0}^{4}$$=\frac{1}{2}[2\sqrt{u}]_{0}^{4}$
Compute the boundaries $:{\quad}4$$=\frac{1}{2}\cdot 4$
Simplify$=2$