Answer
$\frac{2}{3}$
Work Step by Step
$\int^4_1 \frac{u-2}{\sqrt u}du$
$\int^4_1 u^{\frac{1}{2}} - 2u^{-\frac{1}{2}}du$, Rewrite
$ {\frac{2}{3}u^{\frac{3}{2}} - 4u^{\frac{1}{2}}}]^4_1$, Integrate
$(\frac{16}{3}-8)-(\frac{2}{3}-4) $, Take Definite Integral
$\frac{2}{3}$, Simplify
