Anno 2001 PNI – 6 – Parabole

Con uno dei metodi di quadratura studiati, si calcoli un’approssimazione dell’integrale definito $\displaystyle \int_{0}^\pi\ \sin{x}\ dx$ e …

Metodo delle parabole

Parabole	n	h	i	x_i	y_i	Area	= $\displaystyle \frac{h}{3}\cdot (y_0+4\cdot y_1+y_2)$
1	2	$\displaystyle \frac{\pi}{2}$	0	$0$	$0$		= $\displaystyle\frac{\pi}{6}\cdot(0+4\cdot 1+0)$
			1	$\displaystyle \frac{\pi}{2}$	$1$		= $\displaystyle \frac{2}{3}\ \pi$
			2	$\pi$	$0$		= 2,094395


Parabole	n	h	i	x_i	y_i	Area	= $\displaystyle \frac{h}{3}\cdot (y_0+4\cdot y_1+2\cdot y_2+4\cdot y_3+y_4)$
2	4	$\displaystyle \frac{\pi}{4}$	0	$0$	$0$		= $\displaystyle\frac{\pi}{12}\cdot\left(0+4\cdot\frac{\sqrt{2}}{2}+2\cdot 1+4\cdot\frac{\sqrt{2}}{2}+0\right)$
			1	$\displaystyle\frac{\pi}{4}$	$\displaystyle\frac{\sqrt{2}}{2}$		= $\displaystyle \left(\frac{1}{6}+\frac{\sqrt{2}}{3}\right) \pi$
			2	$\displaystyle\frac{\pi}{2}$	$1$		= 2,004560
			3	$\displaystyle\frac{3}{4}\,\pi$	$\displaystyle\frac{\sqrt{2}}{2}$
			4	$\pi$	$0$


Parabole	n	h	i	x_i	y_i	Area	= $\displaystyle \frac{h}{3}\cdot (y_0+4\cdot y_1+2\cdot y_2+4\cdot y_3+2\cdot y_4+4\cdot y_5 +y_6)$
3	6	$\displaystyle \frac{\pi}{6}$	0	$0$	$0$		= $\displaystyle\frac{\pi}{18}\cdot\left(0+4\cdot\frac{1}{2}+2\cdot\frac{\sqrt{3}}{2}+4\cdot 1+2\cdot\frac{\sqrt{3}}{2}+4\cdot\frac{1}{2}+0\right)$
			1	$\displaystyle\frac{\pi}{6}$	$\displaystyle\frac{1}{2}$		= $\displaystyle \left(\frac{4}{9}+\frac{\sqrt{3}}{9}\right) \pi$
			2	$\displaystyle\frac{\pi}{3}$	$\displaystyle\frac{\sqrt{3}}{2}$		= 2,000863
			3	$\displaystyle\frac{\pi}{2}$	$1$
			4	$\displaystyle\frac{2}{3}\ \pi$	$\displaystyle\frac{\sqrt{3}}{2}$
			5	$\displaystyle\frac{5}{6}\ \pi$	$\displaystyle\frac{1}{2}$
			6	$\pi$	$0$