Anno 2001 PNI – 6 – Rettangoli di destra

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

Metodo dei rettangoli con altezze di destra

nhixiyiArea= \displaystyle h\cdot (y_1)
1\pi1\pi0= \pi\cdot (0)
= 0,0
nhixiyiArea= \displaystyle h\cdot (y_1+y_2)
2\displaystyle \frac{\pi}{2}1\displaystyle \frac{\pi}{2}1= \displaystyle\frac{\pi}{2}\cdot(1+0)
2\pi0= \displaystyle \frac{\pi}{2}
= 1,570796
nhixiyiArea= \displaystyle h\cdot (y_1+y_2+y_3)
3\displaystyle \frac{\pi}{3}1\displaystyle\frac{\pi}{3}\displaystyle\frac{\sqrt{3}}{2}= \displaystyle\frac{\pi}{3}\cdot\left(\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}+0\right)
2\displaystyle\frac{2}{3}\ \pi\displaystyle\frac{\sqrt{3}}{2}= \displaystyle\frac{\sqrt{3}}{3}\ \pi
3\pi0= 1,813799
nhixiyiArea= \displaystyle h\cdot (y_1+y_2+y_3+y_4)
4\displaystyle \frac{\pi}{4}1\displaystyle\frac{\pi}{4}\displaystyle\frac{\sqrt{2}}{2}= \displaystyle\frac{\pi}{4}\cdot\left(\frac{\sqrt{2}}{2}+1+\frac{\sqrt{2}}{2}+0\right)
2\displaystyle\frac{\pi}{2}1= \displaystyle \left(\frac{1}{4}+\frac{\sqrt{2}}{4}\right) \pi
3\displaystyle\frac{3}{4}\,\pi\displaystyle\frac{\sqrt{2}}{2}= 1,896119
4\pi0
nhixiyiArea= \displaystyle h\cdot (y_1+y_2+y_3+y_4+y_5+y_6)
6\displaystyle \frac{\pi}{6}1\displaystyle\frac{\pi}{6}\displaystyle\frac{1}{2}= \displaystyle\frac{\pi}{6}\cdot\left(\frac{1}{2}+\frac{\sqrt{3}}{2}+1+\frac{\sqrt{3}}{2}+\frac{1}{2}+0\right)
2\displaystyle\frac{\pi}{3}
\displaystyle\frac{\sqrt{3}}{2}= \displaystyle \left(\frac{1}{3}+\frac{\sqrt{3}}{6}\right) \pi
3\displaystyle\frac{\pi}{2}1= 1,954097
4\displaystyle\frac{2}{3}\,\pi\displaystyle\frac{\sqrt{3}}{2}
5\displaystyle\frac{5}{6}\,\pi\displaystyle\frac{1}{2}
6\pi0