Question

The value of $\underset{n\to \infty }{lim}\frac{1}{n}\sum _{r=0}^{2n-1}\frac{n^2}{n^2+4r^2}$ is

• A. $\frac{1}{2}{\tan }^{-1}\left(2\right)$
• B. $\frac{1}{2}{\tan }^{-1}\left(4\right)$
• C. ${\tan }^{-1}\left(4\right)$
• D. $\frac{1}{4}{\tan }^{-1}\left(4\right)$

Answer 1

The correct option is B $\frac{1}{2}{\tan }^{-1}\left(4\right)$

$\underset{n\to \infty }{lim}\frac{1}{n}\sum _{r=0}^{2n-1}\frac{1}{1+4{\left(\frac{r}{n}\right)}^2}$
$=\underset{0}{\overset{2}{\int }}\frac{dx}{1+4x^2}=\frac{1}{2}{\tan }^{-1}\left(2x\right){|}_0^2=\frac{1}{2}{\tan }^{-1}4$