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

Parabolenhx_iy_iArea= \displaystyle \frac{h}{3}\cdot (y_0+4\cdot y_1+y_2)
12\displaystyle \frac{1}{4}01= \displaystyle\frac{1}{12}\cdot\left(1+4\cdot\frac{4\sqrt{15}}{15}+\frac{2\sqrt{3}}{3}}\right)
\displaystyle \frac{1}{4}\displaystyle \frac{4\sqrt{15}}{15}= \displaystyle \frac{1}{12} +\frac{4\sqrt{15}}{45}+\frac{\sqrt{3}}{18}
\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{8\sqrt{15}}{15}+\frac{\sqrt{3}}{3}
= 3,142941
Parabolenhx_iy_iArea= \displaystyle \frac{h}{3}\cdot (y_0+4\cdot y_1+2\cdot y_2+4\cdot y_3+y_4)
24\displaystyle \frac{1}{8}01= \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}{8}\displaystyle \frac{8\sqrt{7}}{21}= \displaystyle\frac{1}{24} + \frac{4\sqrt{7}}{63}+\frac{\sqrt{15}}{45}+\frac{4\sqrt{55}}{165}+\frac{\sqrt{3}}{36}}
\displaystyle \frac{1}{4}\displaystyle \frac{4\sqrt{15}}{15}Pi greco= 6\cdot Area
\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)
\displaystyle \frac{1}{2}\displaystyle \frac{2\sqrt{3}}{3}= \displaystyle\frac{1}{4} + \frac{8\sqrt{7}}{21}+\frac{2\sqrt{15}}{15}+\frac{8\sqrt{55}}{55}+\frac{\sqrt{3}}{6}}
= 3,141698