Anno 2003 PNI – 7 – Trapezi

di

Verificare l’uguaglianza \displaystyle \pi = 4\int_{0}^1 \frac{1}{1+x^2}\, dx e utilizzarla per calcolare un’approssimazione di pi greco, applicando un metodo di integrazione numerica.

Metodo dei trapezi

nhx_iy_iArea= \displaystyle \frac{h}{2}\cdot(y_0+y_1)
1101= \displaystyle\frac{1}{2}\cdot\left(1+\frac{1}{2}\right)
1\displaystyle \frac{1}{2}= \displaystyle\frac{3}{4}
Pi greco= 4\cdot Area
= \displaystyle 4\cdot\left(\frac{3}{4}\right)
= 3,0
nhx_iy_iArea= \displaystyle \frac{h}{2}\cdot (y_0+2 y_1+y_2)
2\displaystyle \frac{1}{2}01= \displaystyle\frac{1}{4}\cdot\left(1+2\cdot\frac{4}{5}+\frac{1}{2}\right)
\displaystyle \frac{1}{2}\displaystyle \frac{4}{5}= \displaystyle\frac{31}{40}
1\displaystyle \frac{1}{2}Pi greco= 4\cdot Area
= \displaystyle 4\cdot\left(\frac{31}{40}\right)
= \displaystyle\frac{31}{10}
= 3,1
nhx_iy_iArea= \displaystyle \frac{h}{2}\cdot (y_0+2 y_1+2 y_2+y_3)
3\displaystyle \frac{1}{3}01= \displaystyle\frac{1}{6}\cdot\left(1+2\cdot\frac{9}{10}+2\cdot\frac{9}{13}+\frac{1}{2}\right)
\displaystyle \frac{1}{3}\displaystyle \frac{9}{10}= \displaystyle\frac{203}{260}
\displaystyle \frac{2}{3}\displaystyle \frac{9}{13}Pi greco= 4\cdot Area
1\displaystyle \frac{1}{2}= \displaystyle 4\cdot\left(\frac{203}{260}\right)
= \displaystyle\frac{203}{65}
= 3,123077
nhx_iy_iArea= \displaystyle \frac{h}{2}\cdot (y_0+2\cdot y_1+2\cdot y_2+2\cdot y_3+y_4)
4\displaystyle \frac{1}{4}01= \displaystyle \frac{1}{8}\cdot\left(1+2\cdot\frac{16}{17}+2\cdot\frac{4}{5}+2\cdot\frac{16}{25}+\frac{1}{2} \right)
\displaystyle\frac{1}{4}\displaystyle\frac{16}{17}= \displaystyle \frac{5323}{6800}
\displaystyle\frac{1}{2}\displaystyle\frac{4}{5}Pi greco= 4\cdot Area
\displaystyle\frac{3}{4}\displaystyle\frac{16}{25}= \displaystyle 4\cdot\left(\frac{5323}{6800}\right)
1\displaystyle\frac{1}{2}= \displaystyle \frac{5323}{1700}
= 3,131177
nhx_iy_iArea= \displaystyle \frac{h}{2}\cdot (y_0+2\cdot y_1+2\cdot y_2+2\cdot y_3+2\cdot y_4+y_5)
5\displaystyle \frac{1}{5}01= \displaystyle\frac{1}{10}\cdot\left(1+2\cdot\frac{25}{26}+2\cdot\frac{25}{29}+2\cdot\frac{25}{34}+2\cdot\frac{25}{41}+\frac{1}{2} \right)
\displaystyle \frac{1}{5}\displaystyle \frac{25}{26}= \displaystyle\frac{4118807}{5255380}
\displaystyle \frac{2}{5}\displaystyle \frac{25}{29}Pi greco= 4\cdot Area
\displaystyle \frac{3}{5}\displaystyle \frac{25}{34}= \displaystyle 4\cdot\left(\frac{4118807}{5255380}\right)
\displaystyle \frac{4}{5}\displaystyle \frac{25}{41}= \displaystyle\frac{4118807}{1313845}
1\displaystyle \frac{1}{2}= 3,134926
nhx_iy_iArea= \displaystyle \frac{h}{2}\cdot (y_0+2\cdot y_1+2\cdot y_2+2\cdot y_3+2\cdot y_4+2\cdot y_5 +y_6)
6\displaystyle \frac{1}{6}01= \displaystyle\frac{1}{12}\cdot\left(1+2\cdot\frac{36}{37}+2\cdot\frac{9}{10}+2\cdot\frac{4}{5}+2\cdot\frac{9}{13}+2\cdot\frac{36}{61}+\frac{1}{2} \right)
\displaystyle\frac{1}{6}\displaystyle\frac{36}{37}= \displaystyle\frac{2761249}{3520920}
\displaystyle\frac{1}{3}\displaystyle\frac{9}{10}Pi greco= 4\cdot Area
\displaystyle\frac{1}{2}\displaystyle\frac{4}{5}= \displaystyle 4\cdot\left(\frac{2761249}{3520920}\right)
\displaystyle\frac{2}{3}\displaystyle\frac{9}{13}= \displaystyle \frac{2761249}{880230}
\displaystyle\frac{5}{6}\displaystyle\frac{36}{61}= 3,136963
1\displaystyle\frac{1}{2}