Anno 2008 Suppletiva – 10 – Parabole

Tenuto conto che $\displaystyle \frac{\pi}{6}=\int_0^{\frac{1}{2}}\frac{1}{\sqrt{1-x^2}}\, dx$ si calcoli un’approssimazione di π, utilizzando uno dei metodi d’integrazione numerica studiati.

Metodo delle parabole

n	h	$x_0$ $y_0$	$x_1$ $y_1$	$x_2$ $y_2$	$x_2$ $y_2$	$x_3$ $y_3$	Area
2	$\displaystyle \frac{1}{4}$	$0$	$\displaystyle \frac{1}{4}$	$\displaystyle \frac{1}{2}$			$\displaystyle \frac{h}{3}\cdot (y_0+4\cdot y_1+y_2)$
		$1$	$\displaystyle \frac{4\sqrt{15}}{15}$	$\displaystyle \frac{2\sqrt{3}}{3}$			= $\displaystyle\frac{1}{12}\cdot\left(1+4\cdot\frac{4\sqrt{15}}{15}+\frac{2\sqrt{3}}{3}}\right)$
							= $\displaystyle \frac{1}{12} +\frac{4\sqrt{15}}{45}+\frac{\sqrt{3}}{18}$
4	$\displaystyle \frac{1}{8}$	$0$	$\displaystyle \frac{1}{8}$	$\displaystyle \frac{1}{4}$	$\displaystyle \frac{3}{8}$	$\displaystyle \frac{1}{2}$	$\displaystyle \frac{h}{3}\cdot (y_0+4\cdot y_1+2\cdot y_2+4\cdot y_3+y_4)$
		$1$	$\displaystyle \frac{8\sqrt{7}}{21}$	$\displaystyle \frac{4\sqrt{15}}{15}$	$\displaystyle \frac{8\sqrt{55}}{55}$	$\displaystyle \frac{2\sqrt{3}}{3}$	= $\displaystyle\frac{1}{24}\cdot\left(1+4\cdot\frac{8\sqrt{7}}{21}+2\cdot\frac{4\sqrt{15}}{15}+4\cdot\frac{8\sqrt{55}}{55}+\frac{2\sqrt{3}}{3}}\right)$
							= $\displaystyle\frac{1}{24} + \frac{4\sqrt{7}}{63}+\frac{\sqrt{15}}{45}+\frac{4\sqrt{55}}{165}+\frac{\sqrt{3}}{36}}$
…

Riepilogo

\displaystyle \frac{1}{12} +\frac{4\sqrt{15}}{45}+\frac{\sqrt{3}}{18}

\displaystyle \frac{1}{2} +\frac{8\sqrt{15}}{15}+\frac{\sqrt{3}}{3}