Anno 2003 PNI – 7 – Rettangoli di destra

di
Pi greco

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 destra

nhixiyiArea= \displaystyle h\cdot (y_1)
1111\displaystyle \frac{1}{2}= \displaystyle 1\cdot \left(\frac{1}{2}\right)
= \displaystyle \frac{1}{2}
Pi greco= 4\cdot Area
= \displaystyle 4\cdot \left(\frac{1}{2}\right)
= 2,0
nhixiyiArea= \displaystyle h\cdot (y_1+y_2)
2\displaystyle \frac{1}{2}1\displaystyle \frac{1}{2}\displaystyle \frac{4}{5}—
21
\displaystyle \frac{1}{2}		—
Pi greco= 4\cdot Area
—
—
—
nhixiyiArea= \displaystyle h\cdot (y_1+y_2+y_3)
3\displaystyle \frac{1}{3}1\displaystyle \frac{1}{3}\displaystyle \frac{9}{10}—
2\displaystyle \frac{2}{3}\displaystyle \frac{9}{13}—
31\displaystyle \frac{1}{2}Pi greco= 4\cdot Area
—
—
—
nhixiyiArea= \displaystyle h\cdot (y_1+y_2+y_3+y_4)
4\displaystyle \frac{1}{4}1\displaystyle\frac{1}{4}\displaystyle\frac{16}{17}—
2\displaystyle\frac{1}{2}\displaystyle\frac{4}{5}—
3\displaystyle\frac{3}{4}\displaystyle\frac{16}{25}Pi greco= 4\cdot Area
41\displaystyle\frac{1}{2}—
—
—
nhixiyiArea= \displaystyle h\cdot (y_1+y_2+y_3+y_4+y_5)
5\displaystyle \frac{1}{5}1\displaystyle \frac{1}{5}\displaystyle \frac{25}{26}—
2\displaystyle \frac{2}{5}\displaystyle \frac{25}{29}—
3\displaystyle \frac{3}{5}\displaystyle \frac{25}{34}Pi greco= 4\cdot Area
4\displaystyle \frac{4}{5}\displaystyle \frac{25}{41}—
51\displaystyle \frac{1}{2}—
—
nhixiyiArea= \displaystyle h\cdot (y_1+y_2+y_3+y_4+y_5+y_6)
6\displaystyle \frac{1}{6}1\displaystyle\frac{1}{6}\displaystyle\frac{36}{37}—
2\displaystyle\frac{1}{3}\displaystyle\frac{9}{10}—
3\displaystyle\frac{1}{2}\displaystyle\frac{4}{5}Pi greco= 4\cdot Area
4\displaystyle\frac{2}{3}\displaystyle\frac{9}{13}—
5\displaystyle\frac{5}{6}\displaystyle\frac{36}{61}—
61\displaystyle\frac{1}{2}—