Anno 2003 PNI – 7 – Rettangoli di sinistra

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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 rettangoli con altezze di sinistra

nhx_iy_iArea= \displaystyle h\cdot (y_0)
1101= \displaystyle 1\cdot(1)
= 1
Pi greco= 4\cdot Area
= 4\cdot (1)
= 4,0
nhx_iy_iArea= \displaystyle h\cdot (y_0+y_1)
2\displaystyle \frac{1}{2}01= \displaystyle\frac{1}{2}\cdot\left(1+\frac{4}{5}\right)
\displaystyle \frac{1}{2}\displaystyle \frac{4}{5}= \displaystyle\frac{9}{10}
Pi greco= 4\cdot Area
= \displaystyle 4\cdot\left(\frac{9}{10}\right)
= \displaystyle\frac{18}{5}
= 3,6
nhx_iy_iArea= \displaystyle h\cdot (y_0+y_1+y_2)
3\displaystyle \frac{1}{3}01= \displaystyle\frac{1}{3}\cdot\left(1+\frac{9}{10}+\frac{9}{13}\right)
\displaystyle \frac{1}{3}\displaystyle \frac{9}{10}= \displaystyle \frac{337}{390}
\displaystyle \frac{2}{3}\displaystyle \frac{9}{13}Pi greco= 4\cdot Area
= \displaystyle 4\cdot\left(\frac{337}{390}\right)
= \displaystyle \frac{674}{195}
= 3,456410
nhx_iy_iArea= \displaystyle h\cdot (y_0+y_1+y_2+y_3)
4\displaystyle \frac{1}{4}01= \displaystyle\frac{1}{4}\cdot\left(1+\frac{16}{17}+\frac{4}{5}+\frac{16}{25}\right)
\displaystyle\frac{1}{4}\displaystyle\frac{16}{17}= \displaystyle\frac{1437}{1700}
\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{1437}{1700}\right)
= \displaystyle\frac{1437}{425}
= 3,381177
nhx_iy_iArea= \displaystyle h\cdot (y_0+y_1+y_2+y_3+y_4)
5\displaystyle \frac{1}{5}01= \displaystyle\frac{1}{5}\cdot\left(1+\frac{25}{26}+\frac{25}{29}+\frac{25}{34}+\frac{25}{41}\right)
\displaystyle \frac{1}{5}\displaystyle \frac{25}{26}= \displaystyle\frac{1095394}{1313845}
\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{1095394}{1313845}\right)
\displaystyle \frac{4}{5}\displaystyle \frac{25}{41}= \displaystyle\frac{4381576}{1313845}
= 3,334926
nhx_iy_iArea= \displaystyle h\cdot (y_0+y_1+y_2+y_3+y_4+y_5)
6\displaystyle \frac{1}{6}01= \displaystyle\frac{1}{6}\cdot\left(1+\frac{36}{37}+\frac{9}{10}+\frac{4}{5}+\frac{9}{13}+\frac{36}{61}\right)
\displaystyle\frac{1}{6}\displaystyle\frac{36}{37}= \displaystyle\frac{484659}{586820}
\displaystyle\frac{1}{3}\displaystyle\frac{9}{10}Pi greco= 4\cdot Area
\displaystyle\frac{1}{2}\displaystyle\frac{4}{5}= \displaystyle4\cdot\left(\frac{484659}{586820}\right)
\displaystyle\frac{2}{3}\displaystyle\frac{9}{13}= \displaystyle\frac{484659}{146705}
\displaystyle\frac{5}{6}\displaystyle\frac{36}{61}= 3,303630