Question

$\underset{n\to \infty }{lim}\sum _{r=1}^n\frac{1}{\sqrt{4n^2-r^2}}=$

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{6}$
• D. $\frac{\pi }{5}$

Answer 1

Answer: (C)

$\frac{\pi }{6}$

$\underset{n\to \infty }{lim}\sum _{r=1}^n\frac{1}{n\sqrt{4-(\frac{r}{n}{)}^2}}$ is in the form of
$\underset{n\to \infty }{lim}\sum _{r=1}^n\frac{1}{n}\cdot f\left(\frac{r}{n}\right)={\int }_0^{1}f\left(x\right)dx$
$={\int }_0^1\frac{1}{\sqrt{4-x^2}}dx$
$=[{\sin }^{-1}\left(\frac{x}{2}\right)+c]{|}_0^1$
$=[{\sin }^{-1}\left(\frac{1}{2}\right)+c]-\left[{\sin }^{-1}\right(0)+c]$
$=\frac{\pi }{6}$