Anno 2008 Suppletiva – 10 – Parabole

Pi greco

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

\displaystyle \frac{h}{3}\cdot (y_0+4\cdot y_1+y_2)

Parabole	n	h	i	x_i	y_i	Area	= $\displaystyle \frac{h}{3}\cdot (y_0+4\cdot y_1+y_2)$
1	2	$\displaystyle \frac{1}{4}$	0	$0$	$1$		= $\displaystyle\frac{1}{12}\cdot\left(1+4\cdot\frac{4\sqrt{15}}{15}+\frac{2\sqrt{3}}{3}}\right)$
			1	$\displaystyle \frac{1}{4}$	$\displaystyle \frac{4\sqrt{15}}{15}$		= $\displaystyle \frac{1}{12}+\frac{\sqrt{3}}{18} +\frac{4\sqrt{15}}{45}$
			2	$\displaystyle \frac{1}{2}$	$\displaystyle \frac{2\sqrt{3}}{3}$	Pi greco	= $6\cdot Area$
							= $\displaystyle 6\cdot\left(\frac{1}{12} +\frac{4\sqrt{15}}{45}+\frac{\sqrt{3}}{18}\right)$
							= $\displaystyle \frac{1}{2}+\frac{\sqrt{3}}{3} +\frac{8\sqrt{15}}{15}$
							= 3,142941


Parabole	n	h	i	x_i	y_i	Area	= $\displaystyle \frac{h}{3}\cdot (y_0+4\cdot y_1+2\cdot y_2+4\cdot y_3+y_4)$
2	4	$\displaystyle \frac{1}{8}$	0	$0$	$1$		= $\displaystyle\frac{1}{24}\cdot\left(1+4\cdot\frac{8\sqrt{7}}{21}+2\cdot\frac{4\sqrt{15}}{15}+4\cdot\frac{8\sqrt{55}}{55}+\frac{2\sqrt{3}}{3}}\right)$
			1	$\displaystyle \frac{1}{8}$	$\displaystyle \frac{8\sqrt{7}}{21}$		= $\displaystyle\frac{1}{24}+\frac{\sqrt{3}}{36}} + \frac{4\sqrt{7}}{63}+\frac{\sqrt{15}}{45}+\frac{4\sqrt{55}}{165}$
			2	$\displaystyle \frac{1}{4}$	$\displaystyle \frac{4\sqrt{15}}{15}$	Pi greco	= $6\cdot Area$
			3	$\displaystyle \frac{3}{8}$	$\displaystyle \frac{8\sqrt{55}}{55}$		= $\displaystyle 6\cdot\left(\frac{1}{24} + \frac{4\sqrt{7}}{63}+\frac{\sqrt{15}}{45}+\frac{4\sqrt{55}}{165}+\frac{\sqrt{3}}{36}}\right)$
			4	$\displaystyle \frac{1}{2}$	$\displaystyle \frac{2\sqrt{3}}{3}$		= $\displaystyle\frac{1}{4}+\frac{\sqrt{3}}{6}} + \frac{8\sqrt{7}}{21}+\frac{2\sqrt{15}}{15}+\frac{8\sqrt{55}}{55}$
							= 3,141698