Evaluating a Limit using Riemann integration We want to evaluate

$$ \lim_{n \to \infty} \sum_{i=1}^{3n} \sqrt{36 - \left(
 \dfrac{i-\frac{1}{6}}{n} \right)^2 } \cdot \frac{1}{n} $$

I have spent almost an hour in this problem. Here is my thought
First, I tried to evaluate by comparing with an integral of the form $\int_0^b f(x) dx $ using right or left-endpoints, but that doesnt work. Now, if we take the midpoint $\bar{x_i} = \dfrac{x_i + x_{i+1}}{2}$ and $\Delta x = \frac{b}{n} $, we can get somewhere, but first, we do $m=3n$ and simplify the expression to 
$$ \lim_{m \to \infty} \sum_{i=1}^m \sqrt{ 9 - \left(\frac{6i-1}{m} \right)^2 } \frac{3}{2m}  $$
With $x_i = 0 + b i\Delta x $, we have $\bar{x_i} =  \frac{2 bi \Delta x +  \Delta x }{2} = b \Delta x \left( \frac{2i + 1}{2} \right) = \frac{2}{b} \left( \frac{2i + 1}{2m} \right) $ 
We are getting closer here but still not what we desired. My feeling is that the integral is integrating over area of a circle of radius $6$, but im having trouble with the transformations that must be done to get to the right form. 
Any suggestion?
 A: You may proceed as follows:
$$\sum_{i=1}^{3n} \sqrt{36 - \left(
 \dfrac{i}{n} \right)^2 } \cdot \frac{1}{n} \leq \sum_{i=1}^{3n} \sqrt{36 - \left(
 \dfrac{i-\frac{1}{6}}{n} \right)^2 } \cdot \frac{1}{n} \leq \sum_{i=1}^{3n} \sqrt{36 - \left(
 \dfrac{i-1}{n} \right)^2 } \cdot \frac{1}{n}$$
$$3\sum_{i=1}^{3n} \sqrt{36 - 9\left(\dfrac{i}{3n} \right)^2 } \cdot \frac{1}{3n} \leq \sum_{i=1}^{3n} \sqrt{36 - \left(
 \dfrac{i-\frac{1}{6}}{n} \right)^2 } \cdot \frac{1}{n} \leq 3\sum_{i=1}^{3n} \sqrt{36 - 9\left(
 \dfrac{i-1}{3n} \right)^2 } \cdot \frac{1}{3n}$$
For the left and right sum you get in the limit
$$9\int_0^1 \sqrt{4-x^2}\; dx = \frac{3}{2}(3\sqrt{3} + 2\pi)\approx 17.219$$
A: Actually by the definition of Riemann integration, if it converges as $n\to\infty$, the value $f(x)$ of the integrand can be assumed arbitrarily with $x$ in every small interval $[a_n,a_{n+1}]$. Therefore it doesn't really matter whether $f(x)$ is taking value on $\frac{i-\frac{1}{6}}{n}$(the point  correspoding to your limit), $\frac{i}{n}$(left endpoint), $\frac{i+1}{n}$(right endpoint) or whatever other points you like.
For this particular problem, let $f(x)=\sqrt{36-x^2}$. We have  $\lim_{n\to\infty}\sum_{i=1}^{3n}\frac{1}{n}\sqrt{36-(\frac{i-\frac{1}{6}}{n})^2}=\int_{0}^3\sqrt{36-x^2}dx$
