Anno 2008 Suppletiva – 10 – Parabole

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 delle parabole

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2\displaystyle \frac{1}{4}0\displaystyle \frac{1}{4}\displaystyle \frac{1}{2}\displaystyle \frac{h}{3}\cdot (y_0+4\cdot y_1+y_2)
1\displaystyle \frac{4\sqrt{15}}{15}\displaystyle \frac{2\sqrt{3}}{3}= \displaystyle\frac{1}{12}\cdot\left(1+4\cdot\frac{4\sqrt{15}}{15}+\frac{2\sqrt{3}}{3}}\right)
= \displaystyle \frac{1}{12} +\frac{4\sqrt{15}}{45}+\frac{\sqrt{3}}{18}
4\displaystyle \frac{1}{8}0\displaystyle \frac{1}{8}\displaystyle \frac{1}{4}\displaystyle \frac{3}{8}\displaystyle \frac{1}{2}\displaystyle \frac{h}{3}\cdot (y_0+4\cdot y_1+2\cdot y_2+4\cdot y_3+y_4)
1\displaystyle \frac{8\sqrt{7}}{21}\displaystyle \frac{4\sqrt{15}}{15}\displaystyle \frac{8\sqrt{55}}{55}\displaystyle \frac{2\sqrt{3}}{3}= \displaystyle\frac{1}{24}\cdot\left(1+4\cdot\frac{8\sqrt{7}}{21}+2\cdot\frac{4\sqrt{15}}{15}+4\cdot\frac{8\sqrt{55}}{55}+\frac{2\sqrt{3}}{3}}\right)
= \displaystyle\frac{1}{24} + \frac{4\sqrt{7}}{63}+\frac{\sqrt{15}}{45}+\frac{4\sqrt{55}}{165}+\frac{\sqrt{3}}{36}}

Riepilogo

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21\displaystyle \frac{1}{12} +\frac{4\sqrt{15}}{45}+\frac{\sqrt{3}}{18}\displaystyle \frac{1}{2} +\frac{8\sqrt{15}}{15}+\frac{\sqrt{3}}{3}3,142941
42\displaystyle\frac{1}{24} + \frac{4\sqrt{7}}{63}+\frac{\sqrt{15}}{45}+\frac{4\sqrt{55}}{165}+\frac{\sqrt{3}}{36}}\displaystyle\frac{1}{4} + \frac{8\sqrt{7}}{21}+\frac{2\sqrt{15}}{15}+\frac{8\sqrt{55}}{55}+\frac{\sqrt{3}}{6}}3,141698
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