Geometricus geometriorum (HP II, secret "semi-tyre")

Suppose the semicircle has radius r and center (0,0), then the points that determine the divisions are: $P_k = (r \cos \left(\frac{k\pi}{n +1} \right), r \sin \left(\frac{k\pi}{n + 1} \right)), \space \text { for k= 1,... ,n}$ And the arithmetic mean of the areas are: $M_n = \frac{1}{n} \displaystyle \sum_{ k = 1}^n r^2 \sin \left( \frac{k\pi}{n + 1} \right)$ . So, $\displaystyle \lim_{n \to \infty} M_n = \frac{r^2}{\pi} \lim_{n \to \infty} \frac{n + 1}{n} (\sum_{k = 1}^{n} \frac{\pi}{n + 1} \sin \left(\frac{k\pi}{n + 1} \right)) =$ $= \frac{r^2}{\pi} \displaystyle \int_0^{\pi} \sin (x) dx = \frac{r^2}{\pi} \left( - \cos(x) \right)_0^{\pi} = \frac{2r^2}{\pi}$ Now, substituing $r = \pi$ the limit of arithmetic mean of the areas is $2\pi \approx 6.2831...$

Detail: $\displaystyle \lim_{n \to \infty} \frac{n + 1}{n} (\sum_{k = 1}^{n} \frac{\pi}{n + 1} \sin \left(\frac{k\pi}{n + 1} \right)) = \lim_{n \to \infty} \frac{n + 1}{n} \cdot \lim_{n \to \infty} (\sum_{k = 1}^{n} \frac{\pi}{n + 1} \sin \left(\frac{k\pi}{n + 1} \right)) =$ $= 1 \cdot \lim_{n \to \infty} (\sum_{k = 1}^{n} \frac{\pi}{n + 1} \sin \left(\frac{k\pi}{n + 1} \right)) = \int_0^{\pi} \sin (x) dx$