Definite Integration Question 202
Question: The value of $ \underset{n\to \infty }{\mathop{\lim }}[ \frac{n}{1+n^{2}}+\frac{n}{4+n^{2}}+\frac{n}{9+n^{2}}+….+\frac{1}{2n} ] $ is equal to
[Bihar CEE 1994]
Options:
A) $ \frac{\pi }{2} $
B) $ \frac{\pi }{4} $
C) 1
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
We have, $ \underset{n\to \infty }{\mathop{\lim }}[ \frac{n}{1+n^{2}}+\frac{n}{4+n^{2}}+……+\frac{1}{2n} ] $
$ =\underset{n\to \infty }{\mathop{\lim }}\sum\limits _{r=1}^{n}{\frac{n}{r^{2}+n^{2}}}=\underset{n\to \infty }{\mathop{\lim }}\sum\limits _{r=1}^{n}{\frac{n}{n^{2}( 1+\frac{r^{2}}{n^{2}} )}} $
$ =\underset{n\to \infty }{\mathop{\lim }}\sum\limits _{r=1}^{n}{\frac{1}{n( 1+\frac{r^{2}}{n^{2}} )}=\int_0^{1}{\frac{dx}{1+x^{2}}}} $ , $ { \text{Applying formula, }\underset{n\to \infty }{\mathop{\lim }}\sum\limits _{r=0}^{n-1}{{ f( \frac{r}{n} ) }.\frac{1}{n}=\int_0^{1}{f(x)dx}} } $
$ =[{{\tan }^{-1}}x]_0^{1}={{\tan }^{-1}}1-{{\tan }^{-1}}0=\frac{\pi }{4}. $
BETA
