Let f:(0,2)

Question:

Let $f:(0,2) \rightarrow \mathbb{R}$ be defined as $f(\mathrm{x})=\log _{2}\left(1+\tan \left(\frac{\pi \mathrm{x}}{4}\right)\right)$

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:

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

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

replacing $\mathrm{x} \rightarrow 1-\mathrm{x}$

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

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

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

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

$E=\frac{2}{\ell \ln 2} \int_{0}^{1}\left(\ln 2-\ln \left(1+\tan \frac{\pi_{\mathrm{X}}}{4}\right)\right) \mathrm{dx}$

equation

(i) $+$ (ii)

$\mathrm{E}=1$

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