Definite Integration Question 283
Question: $ \underset{n\to \infty }{\mathop{lim}}
[ \frac{1}{n^{2}}{{\sec }^{2}}\frac{1}{n^{2}}+\frac{2}{n^{2}}{{\sec }^{2}}\frac{4}{n^{2}}+…..+\frac{1}{n}{{\sec }^{2}}1 ] $ equals [AIEEE 2005]
Options:
A) $ \tan 1 $
B) $ \frac{1}{2}\tan 1 $
C) $ \frac{1}{2}\sec 1 $
D) $ \frac{1}{2}cosec1 $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \underset{n\to \infty }{\mathop{\lim }}[ \frac{1}{n^{2}}{{\sec }^{2}}\frac{1}{n^{2}}+\frac{2}{n^{2}}{{\sec }^{2}}\frac{4}{n^{2}}+\frac{3}{n^{2}}{{\sec }^{2}}\frac{9}{n^{2}}+…..+\frac{1}{n}{{\sec }^{2}}1 ] $ is equal to $ \underset{n\to \infty }{\mathop{\lim }}\sum\limits _{r=1}^{n}{\frac{r}{n^{2}}{{\sec }^{2}}\frac{r^{2}}{n^{2}}}=\underset{n\to \infty }{\mathop{\lim }}\frac{1}{n}\sum\limits _{r=1}^{n}{\frac{r}{n}{{\sec }^{2}}\frac{r^{2}}{n^{2}}} $
Given limit is equal to the value of integral $ \int_0^{1}{x{{\sec }^{2}}x^{2}dx} $
= $ \frac{1}{2}\int_0^{1}{2x{{\sec }^{2}}x^{2}dx}=\frac{1}{2}\int_0^{1}{{{\sec }^{2}}t\ dt} $ , [Put $ x^{2}=t $ ] $ =\frac{1}{2}[\tant]_0^{1}=\frac{1}{2}\tan 1 $ .
BETA
