# 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?

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 $$9\int_0^1 \sqrt{4-x^2}\; dx = \frac{3}{2}(3\sqrt{3} + 2\pi)\approx 17.219$$

• It's very nice.. (+1) Nov 17, 2018 at 5:39
• Actually, you have a mistake in your very first inequality. notice $i-1 \leq i-1/6 \leq i$ Nov 17, 2018 at 7:06
• @JimmySabater But the expression is subtracted. So, the inequality signs get reversed. Nov 17, 2018 at 7:18
• No mistake as $$\begin{eqnarray*} 0 \leq i-1 \leq i-\frac{1}{6}\leq i & \Leftrightarrow & 0 \leq \left( \frac{i-1}{n}\right)^2 \leq \left( \frac{i-\frac{1}{6}}{n} \right)^2\leq \left( \frac{i}{n} \right)^2 \\ & \Leftrightarrow & 0 \geq -\left( \frac{i-1}{n}\right)^2 \geq -\left( \frac{i-\frac{1}{6}}{n} \right)^2\geq -\left( \frac{i}{n} \right)^2 \\ & \Leftrightarrow & 36-\left( \frac{i-1}{n}\right)^2 \geq 36-\left( \frac{i-\frac{1}{6}}{n} \right)^2\geq 36-\left( \frac{i}{n} \right)^2 \end{eqnarray*}$$ Nov 17, 2018 at 11:55

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 $$\lim_{n\to\infty}\sum_{i=1}^{3n}\frac{1}{n}\sqrt{36-(\frac{i-\frac{1}{6}}{n})^2}=\int_{0}^3\sqrt{36-x^2}dx$$