Esame di Stato 2008 Suppletiva -10

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}}

=

= \displaystyle 6 \cdot ...

= \displaystyle\frac{3}{2}+\frac{2\sqrt{15}}{5}}

=

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)

=

= 6\cdot ...

= \displaystyle 1+\frac{6\sqrt{35}}{35}+\frac{3\sqrt{2}}{4}

=

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)

=

= 6\cdot ...

= \displaystyle \frac{3}{4} + \frac{2\sqrt{7}}{7} + \frac{\sqrt{15}}{5} + \frac{6\sqrt{55}}{55}\right)

=

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}}

=

= 6\cdot ...

= \displaystyle\frac{3}{2}+\sqrt{3}

=

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}

=

= 6\cdot ...

= \displaystyle\frac{3}{4} + \frac{2\sqrt{15}}{5}+\frac{\sqrt{3}}{2}

=

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}

= 6\cdot ...

= \displaystyle\frac{1}{2} + \frac{6\sqrt{35}}{35} + \frac{3\sqrt{2}}{4}+\frac{\sqrt{3}}{3}

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}

= 6\cdot ...

= \displaystyle\frac{3}{8} + \frac{2\sqrt{7}}{7} + \frac{\sqrt{15}}{5} + \frac{6\sqrt{55}}{55} + \frac{\sqrt{3}}{4}

=

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}

=

= 6\cdot ...

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

=

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}}

=

= 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}}

Riepilogo

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