By the definition of the definite integral,th | vedclass

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

Question

(B) The given expression is $S = \lim _{n \rightarrow \infty} \sum_{r=1}^{n-1} \frac{1}{\sqrt{n^2-r^2}}$.We can rewrite this as $S = \lim _{n \rightar

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(B) The given expression is $S = \lim _{n \rightarrow \infty} \sum_{r=1}^{n-1} \frac{1}{\sqrt{n^2-r^2}}$.We can rewrite this as $S = \lim _{n \rightarrow \infty} \sum_{r=1}^{n-1} \frac{1}{n \sqrt{1-(\frac{r}{n})^2}}$.By the definition of definite integral as the limit of a sum,$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n-1} f(\frac{r}{n}) = \int_0^1 f(x) dx$.Here,$f(x) = \frac{1}{\sqrt{1-x^2}}$.So,$S = \int_0^1 \frac{1}{\sqrt{1-x^2}} dx$.Evaluating the integral,we get $S = [\sin ^{-1} x]_0^1$.$S = \sin ^{-1}(1) - \sin ^{-1}(0) = \frac{\pi}{2} - 0 = \frac{\pi}{2}$.

By the definition of the definite integral,the value of $\lim _{n \rightarrow \infty}\left(\frac{1}{\sqrt{n^2-1^2}}+\frac{1}{\sqrt{n^2-2^2}}+\ldots+\frac{1}{\sqrt{n^2-(n-1)^2}}\right)$ is equal to

Similar Questions

$\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{1}^{99}}+{{2}^{99}}+{{3}^{99}}+......{{n}^{99}}}{{{n}^{100}}}=$

Evaluate $\int_2^3 x^2 dx$ as the limit of a sum.

$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {\frac{1}{n}{e^{\frac{r}{n}}}} $ is

Let $S = \frac{2}{1} {}^{n}C_{0} + \frac{2^{2}}{2} {}^{n}C_{1} + \frac{2^{3}}{3} {}^{n}C_{2} + \ldots + \frac{2^{n+1}}{n+1} {}^{n}C_{n}$. Then, $S$ equals

$\lim _{n \rightarrow \infty} \frac{3}{n} \left\{ 4 + \left( 2 + \frac{1}{n} \right)^2 + \left( 2 + \frac{2}{n} \right)^2 + \dots + \left( 3 - \frac{1}{n} \right)^2 \right\}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module