If $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{pe}}^{2\mathrm{x}}+{\mathrm{qe}}^{\mathrm{x}}+\mathrm{rx}$ satisfies the condition $\mathrm{f}\left(0\right)=-1,\mathrm{f}\text{'}\left(\ln 2\right)=31$ and $\underset{0}{\overset{\ln 4}{\int }}(\mathrm{f}\left(\mathrm{x}\right)-\mathrm{rx})\mathrm{dx}=\frac{39}{2}$, then the value of $(\mathrm{p}+\mathrm{q}+\mathrm{r})$ is equal to 

$$\begin{gathered} \Rightarrow \mathrm{p}+\mathrm{q}=-1 \\ \mathrm{f}\text{'}\left(\ln 2\right)=2{\mathrm{pe}}^{2\ln 2}+2{\mathrm{e}}^{\ln 2}+\mathrm{r}............\left(1\right) \\ =8\mathrm{p}+2\mathrm{q}+\mathrm{r}=31 ................\left(2\right) \\ \underset{0}{\overset{\ln 4}{\int }}(\mathrm{f}\left(\mathrm{x}\right)-\mathrm{rx})\mathrm{dx}=\underset{0}{\overset{\ln 4}{\int }}({\mathrm{pe}}^{2\mathrm{x}}+{\mathrm{qe}}^{\mathrm{x}})\mathrm{dx}=={(\mathrm{p}\left(\frac{{\mathrm{e}}^{2\mathrm{x}}}{2}\right)+{\mathrm{qe}}^{\mathrm{x}})}_0^{\ln 4} \\ =\frac{15}{2}\mathrm{p}+3\mathrm{q}=\frac{39}{2}...............\left(3\right) \\ \mathrm{from} \left(1\right) \mathrm{and} \left(3\right) \\ \mathrm{p}=5,\mathrm{q}=-6\text{\hspace{0.17em} \hspace{0.17em}also \hspace{0.17em}\hspace{0.17em}}\mathrm{r}=3 \\ \therefore \mathrm{p}+\mathrm{q}+\mathrm{r}=2 \end{gathered}$$