5.3 The Fundamental Theorem of Calculus/10

${\begin{aligned}g(r)=\int _{0}^{r}{\sqrt {x^{2}+4}}dx\\[2ex]{\frac {d}{dx}}\int _{0}^{r}{\sqrt {x^{2}+4}}dx\\[2ex]1\cdot {\sqrt {r^{2}+4}}-0\cdot {\sqrt {0^{2}+4}}\\[2ex]&={\sqrt {r^{2}}}+{4}\end{aligned}}$