Anno 2003 PNI – 7 – Parabole

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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 delle parabole

Parabolenhx_iy_iArea= \displaystyle \frac{h}{3}\cdot (y_0+4\cdot y_1+y_2)
12\displaystyle \frac{1}{2}01= \displaystyle\frac{1}{6}\cdot\left(1+4\cdot\frac{4}{5}+\frac{1}{2}\right)
\displaystyle \frac{1}{2}\displaystyle \frac{4}{5}= \displaystyle\frac{47}{60}
1\displaystyle \frac{1}{2}Pi greco= 4\cdot Area
= \displaystyle 4\cdot\left(\frac{47}{60}\right)
= \displaystyle\frac{47}{15}
= 3,133333…
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}{4}01= \displaystyle \frac{1}{12}\cdot\left(1+4\cdot\frac{16}{17}+2\cdot\frac{4}{5}+4\cdot\frac{16}{25}+\frac{1}{2} \right)
\displaystyle\frac{1}{4}\displaystyle\frac{16}{17}= \displaystyle \frac{8011}{10200}
\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{8011}{10200}\right)
1\displaystyle\frac{1}{2}= \displaystyle \frac{8011}{2550}
= 3,141568…
Parabolenhx_iy_iArea= \displaystyle \frac{h}{3}\cdot (y_0+4\cdot y_1+2\cdot y_2+4\cdot y_3+2\cdot y_4+4\cdot y_5 +y_6)
36\displaystyle \frac{1}{6}01= \displaystyle\frac{1}{18}\cdot\left(1+4\cdot\frac{36}{37}+2\cdot\frac{9}{10}+4\cdot\frac{4}{5}+2\cdot\frac{9}{13}+4\cdot\frac{36}{61}+\frac{1}{2} \right)
\displaystyle\frac{1}{6}\displaystyle\frac{36}{37}= \displaystyle\frac{829597}{1056276}
\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{829597}{1056276}\right)
\displaystyle\frac{2}{3}\displaystyle\frac{9}{13}= \displaystyle\frac{829597}{264069}
\displaystyle\frac{5}{6}\displaystyle\frac{36}{61}= 3,141591…
1\displaystyle\frac{1}{2}