5.3 The Fundamental Theorem of Calculus/10

$g(r)=\int _{0}^{r}{\sqrt {x^{2}+4}}\,dx$

${\frac {d}{dr}}(g(r))={\frac {d}{dr}}\left[\int _{0}^{r}{\sqrt {x^{2}+4}}\,dx\right]=(1)\cdot {\sqrt {(r)^{2}+4}}-(0)\cdot {\sqrt {(0)^{2}+4}}={\sqrt {r^{2}+4}}$

${\text{Therefore, }}g'(r)=={\sqrt {r^{2}+4}}$