Let f:(0,2) rightarrow R be defined as f(x) = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The given limit is $E = 2 \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} f\left(\frac{r}{n}\right)$.This is a Riemann sum,which can be wri

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) The given limit is $E = 2 \lim_{n ightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} f\left(\frac{r}{n}ight)$.This is a Riemann sum,which can be written as the definite integral $E = 2 \int_{0}^{1} f(x) dx$.Substituting $f(x) = \log_{2}\left(1+	an\left(\frac{\pi x}{4}ight)ight) = \frac{\ln(1+	an(\pi x/4))}{\ln 2}$,we get:$E = \frac{2}{\ln 2} \int_{0}^{1} \ln\left(1+	an\frac{\pi x}{4}ight) dx \quad \dots(i)$Using the property $\int_{0}^{a} g(x) dx = \int_{0}^{a} g(a-x) dx$,we replace $x$ with $1-x$:$E = \frac{2}{\ln 2} \int_{0}^{1} \ln\left(1+	an\left(\frac{\pi}{4}(1-x)ight)ight) dx$$E = \frac{2}{\ln 2} \int_{0}^{1} \ln\left(1+\frac{1-	an(\pi x/4)}{1+	an(\pi x/4)}ight) dx$$E = \frac{2}{\ln 2} \int_{0}^{1} \ln\left(\frac{2}{1+	an(\pi x/4)}ight) dx$$E = \frac{2}{\ln 2} \int_{0}^{1} (\ln 2 - \ln(1+	an(\pi x/4))) dx \quad \dots(ii)$Adding $(i)$ and $(ii)$:$2E = \frac{2}{\ln 2} \int_{0}^{1} \ln 2 dx = \frac{2}{\ln 2} \cdot \ln 2 = 2$Therefore,$E = 1$.

Let $f:(0,2) \rightarrow R$ 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

Similar Questions

$\int_0^\pi x \sin^3 x \, dx = $

If $I_n = \int_0^{\pi / 4} \tan^n x \, dx$,then $I_2+I_4, I_3+I_5, I_4+I_6, \ldots$ are in

The value of $\int_{0}^{\sin^2 x} \sin^{-1} \sqrt{t} \, dt + \int_{0}^{\cos^2 x} \cos^{-1} \sqrt{t} \, dt$ for $x \in (0, \pi/2)$ is:

$\int_{0}^{\pi} [\cot x] dx = $

$ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{dx}{e^{\sin x}+1} $ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module