Esame di Stato 2008 Suppletiva -10

SECONDA PROVA

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.

Verifica

$\displaystyle \int_0^{\frac{1}{2}}\frac{1}{\sqrt{1-x^2}}\, dx$ = $\displaystyle \big[\arcsin x\big]_0^{\frac{1}{2}}$ = $\displaystyle \arcsin \frac{1}{2}-\arcsin 0$ = $\displaystyle \frac{\pi}{6}-0$ = $\displaystyle \frac{\pi}{6}$

Calcolo approssimato di pi greco

Utilizza un metodo di integrazione numerica e moltiplica il risultato per 6.

Scegli

il metodo di integrazione numerica (rettangoli, trapezi, parabole)
il numero di intervalli (n=1,2,3,4, …)

Metodo dei rettangoli

$n$	$h$	$x_i$	$y_i$	$\displaystyle h (y_0)$	$\pi$
1	= $\displaystyle \frac{\frac{1}{2}-0}{1}$ = $\displaystyle \frac{1}{2}$	0	1	= $\displaystyle\frac{1}{2}\cdot 1$ = $\displaystyle\frac{1}{2}$	= $\displaystyle 6 \cdot ...$ = 3
$n$	$h$	$x_i$	$y_i$	$\displaystyle h (y_0+y_1)$	$\pi$
2	= $\displaystyle \frac{\frac{1}{2}-0}{2}$ = $\displaystyle \frac{1}{4}$	0 $\displaystyle \frac{1}{4}$	1 $\displaystyle \frac{4\sqrt{15}}{15}$	= $\displaystyle\frac{1}{4}\left(1+\frac{4\sqrt{15}}{15}}\right)$ = $\displaystyle\frac{1}{4}+\frac{\sqrt{15}}{15}}$ = $0,508198889747161$	= $\displaystyle 6 \cdot ...$ = $\displaystyle\frac{3}{2}+\frac{2\sqrt{15}}{5}}$ = $3,04919333848297$
$n$	$h$	$x_i$	$y_i$	$\displaystyle h (y_0+y_1+y_2)$	$\pi$
3	= $\displaystyle \frac{\frac{1}{2}-0}{3}$ = $\displaystyle \frac{1}{6}$	0 $\displaystyle \frac{1}{6}$ $\displaystyle \frac{1}{3}$	1 $\displaystyle \frac{6\sqrt{35}}{35}$ $\displaystyle \frac{3\sqrt{2}}{4}$	= $\displaystyle\frac{1}{6}\left(1+\frac{6\sqrt{35}}{35}+\frac{3\sqrt{2}}{4}\right)$ = $\displaystyle\frac{1}{6}+ \frac{\sqrt{35}}{35}+\frac{\sqrt{2}}{8}\right)$ = $0,512474212909007$	= $6\cdot ...$ = $\displaystyle 1+\frac{6\sqrt{35}}{35}+\frac{3\sqrt{2}}{4}$ = $3,07484527745404$
$n$	$h$	$x_i$	$y_i$	$\displaystyle h (y_0+y_1+y_2+y_3)$	$\pi$
4	= $\displaystyle \frac{\frac{1}{2}-0}{4}$ = $\displaystyle \frac{1}{8}$	0 $\displaystyle \frac{1}{8}$ $\displaystyle \frac{1}{4}$ $\displaystyle \frac{3}{8}$	1 $\displaystyle \frac{8\sqrt{7}}{21}$ $\displaystyle \frac{4\sqrt{15}}{15}$ $\displaystyle \frac{8\sqrt{55}}{55}$	= $\displaystyle \frac{1}{8}\left(1 + \frac{8\sqrt{7}}{21} + \frac{4\sqrt{15}}{15} + \frac{8\sqrt{55}}{55}\right)$ = $\displaystyle \frac{1}{8} + \frac{\sqrt{7}}{21} + \frac{\sqrt{15}}{30} + \frac{\sqrt{55}}{55}\right)$ = $0,514927575035971$	= $6\cdot ...$ = $\displaystyle \frac{3}{4} + \frac{2\sqrt{7}}{7} + \frac{\sqrt{15}}{5} + \frac{6\sqrt{55}}{55}\right)$ = $3,08956545021583$

Metodo dei trapezi

$n$	$h$	$x_i$	$y_i$	$\displaystyle \frac{h}{2} (y_0+ y_1)$	$\pi$
1	= $\displaystyle \frac{\frac{1}{2}-0}{1}$ = $\displaystyle \frac{1}{2}$	0 $\displaystyle \frac{1}{2}$	1 $\displaystyle \frac{2\sqrt{3}}{3}$	= $\displaystyle\frac{1}{4}\left(1+\frac{2\sqrt{3}}{3}}\right)$ = $\displaystyle\frac{1}{4}+\frac{\sqrt{3}}{6}}$ = $0,538675134594813$	= $6\cdot ...$ = $\displaystyle\frac{3}{2}+\sqrt{3}$ = $3,23205080756888$
$n$	$h$	$x_i$	$y_i$	$\displaystyle \frac{h}{2} (y_0+2 y_1+y_2)$	$\pi$
2	= $\displaystyle \frac{\frac{1}{2}-0}{2}$ = $\displaystyle \frac{1}{4}$	0 $\displaystyle \frac{1}{4}$ $\displaystyle \frac{1}{2}$	1 $\displaystyle \frac{4\sqrt{15}}{15}$ $\displaystyle \frac{2\sqrt{3}}{3}$	= $\displaystyle\frac{1}{8}\left(1+2\cdot\frac{4\sqrt{15}}{15}+\frac{2\sqrt{3}}{3}}\right)$ = $\displaystyle\frac{1}{8} + \frac{\sqrt{15}}{15}+\frac{\sqrt{3}}{12}$ = $0,527536457044568$	= $6\cdot ...$ = $\displaystyle\frac{3}{4} + \frac{2\sqrt{15}}{5}+\frac{\sqrt{3}}{2}$ = $3,16521874226741$
$n$	$h$	$x_i$	$y_i$	$\displaystyle \frac{h}{2} (y_0+2y_1+2y_2+y_3)$	$\pi$
3	= $\displaystyle \frac{\frac{1}{2}-0}{3}$ = $\displaystyle \frac{1}{6}$	0 $\displaystyle \frac{1}{6}$ $\displaystyle \frac{1}{3}$ $\displaystyle \frac{1}{2}$	1 $\displaystyle \frac{6\sqrt{35}}{35}$ $\displaystyle \frac{3\sqrt{2}}{4}$ $\displaystyle \frac{2\sqrt{3}}{3}$	= $\displaystyle\frac{1}{12}\left(1+2\cdot\frac{6\sqrt{35}}{35}+2\cdot\frac{3\sqrt{2}}{4}+\frac{2\sqrt{3}}{3}\right)$ = $\displaystyle\frac{1}{12} + \frac{\sqrt{35}}{35} + \frac{\sqrt{2}}{8}+\frac{\sqrt{3}}{18}$ = $0,525365924440611$	= $6\cdot ...$ = $\displaystyle\frac{1}{2} + \frac{6\sqrt{35}}{35} + \frac{3\sqrt{2}}{4}+\frac{\sqrt{3}}{3}$ = $3,15219554664367$
$n$	$h$	$x_i$	$y_i$	$\displaystyle \frac{h}{2} (y_0+2 y_1+2 y_2+2 y_3+ y_4)$	$\pi$
4	= $\displaystyle \frac{\frac{1}{2}-0}{4}$ = $\displaystyle \frac{1}{8}$	0 $\displaystyle \frac{1}{8}$ $\displaystyle \frac{1}{4}$ $\displaystyle \frac{3}{8}$ $\displaystyle \frac{1}{2}$	1 $\displaystyle \frac{8\sqrt{7}}{21}$ $\displaystyle \frac{4\sqrt{15}}{15}$ $\displaystyle \frac{8\sqrt{55}}{55}$ $\displaystyle \frac{2\sqrt{3}}{3}$	= $\displaystyle\frac{1}{16}\left( 1 + 2\cdot\frac{8\sqrt{7}}{21} + 2\cdot\frac{4\sqrt{15}}{15} +2\cdot \frac{8\sqrt{55}}{55} + \frac{2\sqrt{3}}{3}\right)$ = $\displaystyle\frac{1}{16} + \frac{\sqrt{7}}{21} + \frac{\sqrt{15}}{30} + \frac{\sqrt{55}}{55} + \frac{\sqrt{3}}{24}$ $= 0,524596358684675$	= $6\cdot ...$ = $\displaystyle\frac{3}{8} + \frac{2\sqrt{7}}{7} + \frac{\sqrt{15}}{5} + \frac{6\sqrt{55}}{55} + \frac{\sqrt{3}}{4}$ = $3,14757815210805$

Metodo delle parabole

$n$	$h$	$x_i$	$y_i$	$\displaystyle \frac{h}{3} (y_0+4 y_1+y_2)$	$\pi$
2	= $\displaystyle \frac{\frac{1}{2}-0}{2}$ = $\displaystyle \frac{1}{4}$	0 $\displaystyle \frac{1}{4}$ $\displaystyle \frac{1}{2}$	1 $\displaystyle \frac{4\sqrt{15}}{15}$ $\displaystyle \frac{2\sqrt{3}}{3}$	= $\displaystyle\frac{1}{12}\left(1+4\cdot\frac{4\sqrt{15}}{15}+\frac{2\sqrt{3}}{3}}\right)$ = $\displaystyle \frac{1}{12} +\frac{4\sqrt{15}}{45}+\frac{\sqrt{3}}{18}$ = $0,523823564527819$	= $6\cdot ...$ = $\displaystyle \frac{1}{2} +\frac{8\sqrt{15}}{15}+\frac{\sqrt{3}}{3}$ = $3,14294138716691$
$n$	$h$	$x_i$	$y_i$	$\displaystyle \frac{h}{3} (y_0+4 y_1+2 y_2+4 y_3+y_4)$	$\pi$
4	= $\displaystyle \frac{\frac{1}{2}-0}{4}$ = $\displaystyle \frac{1}{8}$	0 $\displaystyle \frac{1}{8}$ $\displaystyle \frac{1}{4}$ $\displaystyle \frac{3}{8}$ $\displaystyle \frac{1}{2}$	1 $\displaystyle \frac{8\sqrt{7}}{21}$ $\displaystyle \frac{4\sqrt{15}}{15}$ $\displaystyle \frac{8\sqrt{55}}{55}$ $\displaystyle \frac{2\sqrt{3}}{3}$	= $\displaystyle\frac{1}{24}\left(1+4\cdot\frac{8\sqrt{7}}{21}+2\cdot\frac{4\sqrt{15}}{15}+4\cdot\frac{8\sqrt{55}}{55}+\frac{2\sqrt{3}}{3}}\right)$ = $\displaystyle\frac{1}{24} + \frac{4\sqrt{7}}{63}+\frac{\sqrt{15}}{45}+\frac{4\sqrt{55}}{165}+\frac{\sqrt{3}}{36}}$ = $0,523616325898044$	= $6\cdot ...$ = $\displaystyle\frac{1}{4} + \frac{8\sqrt{7}}{21}+\frac{2\sqrt{15}}{15}+\frac{8\sqrt{55}}{55}+\frac{\sqrt{3}}{6}}$ $= 3,14169795538826$

Riepilogo

n	Rettangoli	Trapezi	Parabole
1	3,0	3,232050807568877
2	3,049193338482967	3,165218742267405	$3,14294138716691$
3	3,0748452774540405	3,152195546643666
4	3,0895654502158276	3,147578152108048	$3,14169795538826$
…	…	…	…
100	3,1392817679437535	3,1416022760194413	…