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?


2 Answers 2


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

  • $\begingroup$ It's very nice.. (+1) $\endgroup$
    – James
    Nov 17, 2018 at 5:39
  • $\begingroup$ Actually, you have a mistake in your very first inequality. notice $i-1 \leq i-1/6 \leq i$ $\endgroup$
    – James
    Nov 17, 2018 at 7:06
  • $\begingroup$ @JimmySabater But the expression is subtracted. So, the inequality signs get reversed. $\endgroup$ Nov 17, 2018 at 7:18
  • $\begingroup$ No mistake as $$\begin{eqnarray*} 0 \leq i-1 \leq i-\frac{1}{6}\leq i & \Leftrightarrow & 0 \leq \left( \frac{i-1}{n}\right)^2 \leq \left( \frac{i-\frac{1}{6}}{n} \right)^2\leq \left( \frac{i}{n} \right)^2 \\ & \Leftrightarrow & 0 \geq -\left( \frac{i-1}{n}\right)^2 \geq -\left( \frac{i-\frac{1}{6}}{n} \right)^2\geq -\left( \frac{i}{n} \right)^2 \\ & \Leftrightarrow & 36-\left( \frac{i-1}{n}\right)^2 \geq 36-\left( \frac{i-\frac{1}{6}}{n} \right)^2\geq 36-\left( \frac{i}{n} \right)^2 \end{eqnarray*}$$ $\endgroup$ 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 $\lim_{n\to\infty}\sum_{i=1}^{3n}\frac{1}{n}\sqrt{36-(\frac{i-\frac{1}{6}}{n})^2}=\int_{0}^3\sqrt{36-x^2}dx$


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