# Interesing, hard limit with sum, involving $\pi$

Yesterday I was boring so I decided to derive formula for area of circle with integrals. Very good exercise, I think, because I forgot many, many things about integrals. So I started with: $$\int_{-r}^{r} \sqrt{r^2-x^2}dx$$ but I didn't have any clue how to count indefinite integral $\int\sqrt{r^2-x^2}dx$ (is it even possible? today I only found method for counting definite integral above with trigonometric substitution, but this does not apply in general), so I decided to use Riemann's theorem, since I only need to count definite integral. And everything was going well, till something extremely interesting happend. The last step I need to do is to find this limit: $$\lim_{n \to +\infty}\frac{1}{n}\sum_{k=1}^{n}\sqrt{\frac{k}{n}\cdot\frac{n-k}{n}}$$ Surprisingly it is equal to $\frac{\pi}{8}$, and it is mindblowing ;-) but I only know that because I know formula for area of circle which I'm trying to derive. But without knowing it, is it possible to calculate this limit with relatively simple methods? I really, really want to to this in order to award my attempts. Can anybody help?

• the riemann sum of that integration should be $\int_0^1 \sqrt{x(1-x}dx$ for integral of area of circle the Riemann sum should be $\lim_{n\to\infty} \frac 1 n \sum_{k=1}^n \sqrt{1 - k^2/n^2}$
– S L
Sep 30, 2013 at 9:51
• Yes, it is possible to find the indefinite integral $\int\sqrt{r^2-x^2}\,dx$, and it is done by starting with the substitution $x=r\sin u$, $dx=r\cos u$. Sep 30, 2013 at 9:58
• Thank you Gerry, I can see now. @experimentX, I'm pretty sure I didn't make a mistake. Moreover approximation with wolframalpha seems to confirm 'my' limit is equal to $\frac{\pi}{8}$, so I am hoping it can be calculated in some way, so the question is really about this limit ;-)
– xan
Sep 30, 2013 at 10:12
• Interesting, +1. Can you put the steps you took to get to that limit though? Maybe there's another way even using that method. Dec 19, 2013 at 14:00
• @xan: I can only arrive at $\;\lim_{n \to \infty}\frac{1}{n}\sum_{k=1}^{n}\sqrt{\frac{n+k}{n}\cdot\frac{n-k}{n}}\;$ and that limit must be $\pi/4$ . Please tell us what other steps you have taken. Jan 14, 2014 at 10:49

As pointed in the comments, your limit, regared as a Riemann sum, is just (by only using simple changes of variable): $$L=\int_{0}^{1}\sqrt{x(1-x)}\,dx=\int_{0}^{1}\sqrt{\frac{1}{4}-\left(\frac{1}{2}-x\right)^2}\,dx=\int_{-1/2}^{1/2}\sqrt{\frac{1}{4}-x^2}\,dx=\frac{1}{4}\int_{-1}^{1}\sqrt{1-x^2}dx = \frac{1}{4}\cdot\frac{\pi}{2}=\frac{\pi}{8}.$$ $\int_{-1}^{1}\sqrt{1-x^2}dx$ is clearly half the area of the unit circle. In another fashion, by putting $x=\sin\theta$: $$\int_{-1}^{1}\sqrt{1-x^2}\,dx = 2\int_{0}^{1}\sqrt{1-x^2}\,dx=2\int_{0}^{\pi/2}\cos^2\theta\,d\theta=\int_{0}^{\pi/2}\left(\cos(2\theta)+1\right)d\theta = \int_{0}^{\pi/2}1\,d\theta = \frac{\pi}{2}.$$
Your last step is wrong. Let $$A=2\int_{-r}^r\sqrt{r^2-x^2}dx$$ be the area of circle with radius $r$. Then \begin{align}\\ A&=2\int_{\style{color:red}{0}}^\style{color:red}{2r}\sqrt{r^2-(\style{color:red}{x-r})^2}\style{color:red}{dx}\\ &=2\int_{0}^{2r}\sqrt{x(2r-x)}dx\\ &=2\int_\style{color:red}{0}^\style{color:red}{1}\sqrt{\style{color:red}{2rx}(2r-\style{color:red}{2rx})}\style{color:red}{2rdx}\\ &=8r^2\int_0^1\sqrt{x(1-x)}dx\\ \end{align} So, the correct limit will be $$\lim_{n\to\infty}8r^2\sum_{k=1}^n\frac1n\sqrt{\frac{k}{n}\cdot\frac{n-k}{n}}$$ This limit converges to $r^2\pi$ which is the real area of circle. If you want to compute this limit algebraically, I think there is no other way than converting it back to definite integral which means that you must use trigonometric substitution.