# Let f : (0,2) → R be defined as

Question:

Let $f:(0,2) \rightarrow \mathbb{R}$ be defined as

$f(x)=\log _{2}\left(1+\tan \left(\frac{\pi x}{4}\right)\right)$

Then, $\lim _{n \rightarrow \infty} \frac{2}{n}\left(f\left(\frac{1}{n}\right)+f\left(\frac{2}{n}\right)+\ldots .+f(1)\right)$ is equal

to______.

Solution:

$E=2 \lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{1}{n} f\left(\frac{r}{n}\right)$

$\mathrm{E}=\frac{2}{\ell \mathrm{n} 2} \int_{0}^{1} \ln \left(1+\tan \frac{\pi \mathrm{x}}{4}\right) \mathrm{dx}$.......(i)

replacing $x \rightarrow 1-x$

$E=\frac{2}{\ln 2} \int_{0}^{1} \ln \left(1+\tan \frac{\pi}{4}(1-x)\right) d x$

$\mathrm{E}=\frac{2}{\ell \mathrm{n} 2} \int_{0}^{1} \ell \mathrm{n}\left(1+\tan \left(\frac{\pi}{4}-\frac{\pi}{4} \mathrm{x}\right)\right) \mathrm{dx}$

$E=\frac{2}{\ell n 2} \int_{0}^{1} \ln \left(1+\frac{1+\tan \frac{\pi}{4} x}{1+\tan \frac{\pi}{4} x}\right) d x$

$E=\frac{2}{\ell \mathrm{n} 2} \int_{0}^{1} \ln \left(\frac{2}{1+\tan \frac{\pi \mathrm{x}}{4}}\right) \mathrm{dx}$

$\mathrm{E}=\frac{2}{\ell \mathrm{n} 2} \int_{0}^{1}\left(\ell \ln 2-\ell \mathrm{n}\left(1+\tan \frac{\pi \mathrm{x}}{4}\right)\right) \mathrm{dx}$......(ii)

equation (i) $+$ (ii)

$\mathrm{E}=1$