Let $f:\left(0, 2\right)\to R$ be defined as $f\left(x\right)={\log }_2\left(1+\tan \left(\frac{\pi x}{4}\right)\right).$ Then, $\underset{n\to \infty }{\lim }\frac{2}{n}\left(f\left(\frac{{\displaystyle 1}}{{\displaystyle n}}\right)+f\left(\frac{{\displaystyle 2}}{{\displaystyle n}}\right)+\mathrm{........}+f\left(1\right)\right)$ is equal to 

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)={\log }_2\left(1+\tan \frac{\mathrm{\pi x}}{4}\right) \\ \underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{Lt}}\frac{2}{\mathrm{n}}\left(\mathrm{f}\left(\frac{{\displaystyle 1}}{{\displaystyle \mathrm{n}}}\right)+\mathrm{f}\left(\frac{{\displaystyle 2}}{{\displaystyle \mathrm{n}}}\right)+........+\mathrm{f}\left(\frac{\mathrm{n}}{\mathrm{n}}\right)\right) \\ =\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{Lt}}\frac{2}{\mathrm{n}}\sum _{\mathrm{r}=1}^{\mathrm{n}} \mathrm{f}\left(\frac{\mathrm{r}}{\mathrm{n}}\right) \\ =2{\int }_0^1\mathrm{f}\left(\mathrm{x}\right) \mathrm{dx} \\ \mathrm{I}=2{\int }_0^1{\log }_2\left(1+\tan \frac{\mathrm{\pi x}}{4}\right)\mathrm{dx} \\ =2{\int }_0^1{\log }_2\left(1+\tan \frac{\mathrm{\pi }}{4} (1-\mathrm{x})\right)\mathrm{dx} \\ =2{\int }_0^1 {\log }_2\left(1+\frac{1-\tan {\displaystyle \frac{\mathrm{\pi }}{4}}\mathrm{x}}{1+\tan {\displaystyle \frac{\mathrm{\pi }}{4}}\mathrm{x}}\right)\mathrm{dx} \\ \mathrm{I}=2{\int }_0^1 {\log }_2\left(\frac{2}{1+\tan {\displaystyle \frac{\mathrm{\pi }}{4}}\mathrm{x}}\right)\mathrm{dx} \\ 2\mathrm{I}=2{\int }_0^1 {\log }_2 2 \mathrm{dx} \\ \mathrm{I}={\log }_2 2 =1 \end{gathered}$$