Anno 2008 Suppletiva – 10 – Rettangoli di destra

di
Esame
Integrazione numerica
Pi greco

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 dei rettangoli con altezze di destra

nhixiyiArea= \displaystyle h\cdot (y_1)
1\displaystyle \frac{1}{2}1\displaystyle \frac{1}{2}\displaystyle \frac{2\sqrt{3}}{3}= \displaystyle\frac{1}{2}\cdot \left(\frac{2\sqrt{3}}{3}\right)
= \displaystyle\frac{\sqrt{3}}{3}
Pi greco= 6\cdot Area
= \displaystyle 6\cdot\left(\frac{\sqrt{3}}{3}\right)
= \displaystyle 2\sqrt{3}
= 3,464102
nhixiyiArea= \displaystyle h\cdot (y_1+y_2)
2\displaystyle \frac{1}{4}1\displaystyle \frac{1}{4}\displaystyle \frac{4\sqrt{15}}{15}= \displaystyle\frac{1}{4}\cdot\left(\frac{4\sqrt{15}}{15}}+\frac{2\sqrt{3}}{3}\right)
2\displaystyle \frac{1}{2}\displaystyle \frac{2\sqrt{3}}{3}= \displaystyle \frac{\sqrt{3}}{6}+\frac{\sqrt{15}}{15}}
Pi greco= 6\cdot Area
= \displaystyle 6\cdot\left(\frac{\sqrt{3}}{6}+\frac{\sqrt{15}}{15}}\right)
= \displaystyle \sqrt{3}+\frac{2\sqrt{15}}{5}}
= 3,281244
nhixiyiArea= \displaystyle h\cdot (y_1+y_2+y_3)
3\displaystyle \frac{1}{6}1\displaystyle \frac{1}{6}\displaystyle \frac{6\sqrt{35}}{35}= \displaystyle\frac{1}{6}\cdot\left(\frac{6\sqrt{35}}{35}+\frac{3\sqrt{2}}{4}+\frac{2\sqrt{3}}{3}\right)
2\displaystyle \frac{1}{3}\displaystyle \frac{3\sqrt{2}}{4}= \displaystyle\frac{\sqrt{2}}{8}+\frac{\sqrt{3}}{9}+\frac{\sqrt{35}}{35}
3\displaystyle \frac{1}{2}\displaystyle \frac{2\sqrt{3}}{3}Pi greco= 6\cdot Area
= \displaystyle 6\cdot\left(\frac{\sqrt{2}}{8}+\frac{\sqrt{3}}{9}+\frac{\sqrt{35}}{35}\right)
= \displaystyle \frac{3\sqrt{2}}{4}+\frac{2\sqrt{3}}{3}+\frac{6\sqrt{35}}{35}\right)
= 3,229546
nhixiyiArea= \displaystyle h\cdot (y_1+y_2+y_3+y_4)
4\displaystyle \frac{1}{8}1\displaystyle \frac{1}{8}\displaystyle \frac{8\sqrt{7}}{21}= \displaystyle \frac{1}{8}\cdot\left( \frac{8\sqrt{7}}{21} + \frac{4\sqrt{15}}{15} + \frac{8\sqrt{55}}{55}+\frac{2\sqrt{3}}{3}\right)
2\displaystyle \frac{1}{4}\displaystyle \frac{4\sqrt{15}}{15}= \displaystyle \frac{\sqrt{3}}{12}+\frac{\sqrt{7}}{21} + \frac{\sqrt{15}}{30} + \frac{\sqrt{55}}{55}
3\displaystyle \frac{3}{8}\displaystyle \frac{8\sqrt{55}}{55}Pi greco= 6\cdot Area
4\displaystyle \frac{1}{2}\displaystyle \frac{2\sqrt{3}}{3}= \displaystyle 6\cdot\left(\frac{\sqrt{3}}{12}+\frac{\sqrt{7}}{21} + \frac{\sqrt{15}}{30} + \frac{\sqrt{55}}{55}\right)
= \displaystyle \frac{\sqrt{3}}{2}+\frac{2\sqrt{7}}{7} + \frac{\sqrt{15}}{5} + \frac{6\sqrt{55}}{55}
= 3,205591