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?
 A: 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 $x=\sin\theta$:
$$\int_{-1}^{1}\sqrt{1-x^2}\,dx = 2\int_{0}^{1}\sqrt{1-x^2}\,dx=2\int_{0}^{\pi/2}\cos^2\theta\,d\theta=\int_{0}^{\pi/2}\left(\cos(2\theta)+1\right)d\theta = \int_{0}^{\pi/2}1\,d\theta = \frac{\pi}{2}.$$
A: 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.
