The value of lim _{n rightarrow infty} left( | vedclass

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

Question

(D) Let $I = \lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{1}{\sqrt{4n^2-r^2}}$.We can rewrite the expression as:$I = \lim _{n \rightarrow \infty} \

Vedclass · Accepted Answer

$\frac{\pi}{6}$

Vedclass · Answer

(D) Let $I = \lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{1}{\sqrt{4n^2-r^2}}$.We can rewrite the expression as:$I = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n \frac{1}{\sqrt{4 - (\frac{r}{n})^2}}$.This is a Riemann sum for the definite integral:$I = \int_{0}^{1} \frac{1}{\sqrt{4 - x^2}} dx$.Using the standard integral formula $\int \frac{1}{\sqrt{a^2 - x^2}} dx = \sin^{-1}(\frac{x}{a}) + C$:$I = \left[ \sin^{-1}(\frac{x}{2}) \right]_{0}^{1}$.$I = \sin^{-1}(\frac{1}{2}) - \sin^{-1}(0) = \frac{\pi}{6} - 0 = \frac{\pi}{6}$.

The value of $\lim _{n \rightarrow \infty} \left( \frac{1}{\sqrt{4n^2-1}} + \frac{1}{\sqrt{4n^2-4}} + \dots + \frac{1}{\sqrt{4n^2-n^2}} \right)$ is

Similar Questions

$\lim _{n \rightarrow \infty}\left[\frac{n+1}{n^2+1^2}+\frac{n+2}{n^2+2^2}+\frac{n+3}{n^2+3^2}+\ldots+\frac{n+2 n}{n^2+(2n)^2}\right]=$

By the definition of the definite integral,the value of $\lim _{n \rightarrow \infty}\left[\frac{1^2}{1^3+n^3}+\frac{2^2}{2^3+n^3}+\ldots+\frac{n^2}{n^3+n^3}\right]=$

$\int_0^3 (2+x^2) dx = $

Evaluate the limit: $\lim _{n \rightarrow \infty} \frac{3}{n}\left\{1+\sqrt{\frac{n}{n+3}}+\sqrt{\frac{n}{n+6}}+\sqrt{\frac{n}{n+9}}+\ldots+\sqrt{\frac{n}{n+3(n-1)}}\right\}$

$\lim _{n \rightarrow \infty}\left[\frac{1}{n^2} \sec ^2 \frac{1}{n^2}+\frac{2}{n^2} \sec ^2 \frac{4}{n^2}+\ldots+\frac{1}{n} \sec ^2 1\right]=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module