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Say we have $a = 0$, $b = 1$ and we want to compute the weights of the interpolatory quadrature formula $I_2$ with the nodes $x_0 = 0$, $x_1 = \frac{2}{5}$ and $x_2 = 1$. I want to show that $I_2(p)$ is equal to \begin{equation}\tag{1} \int_{a}^{b} p(x) dx \end{equation} if p is a quadratic polynomial. And that $I_2(p)$ is in general not equal to (1) if p is a cubic polynomial.

I know that this may be a common proof in Polynomial Interpolation, but I haven't been able to come up with a proof for this result or stumble across. I would appreciate if someone could show me how this can be done or set me on a solid track.

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If you set $p(x)=ax^2+bx+c$ and do the integral considering $a,b,c$ as constants you will get a value. Now assign weights $w_0,w_1,w_2$ to the nodes and compute the formula $I_2$. These must agree for all values of $a,b,c$ which gives three equations in three unknowns, the three weights. Solve them and you have the weights and have proved that the formula is exact for all quadratics. Now let the cubic be $dx^3+ax^2+bx+c$, compute the integral and compute $I_2$. Show they differ any time $d \neq 0$

Ross Millikan
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  • Hi Ross, I'm still confused as to how one computes the formula $I_2$. Could you show a bit more of the working? –  Apr 27 '20 at 09:14
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    You have $I_2=w_0p(0)+w_1p(\frac 25)+w_2p(1)$. Evaluate the function values and you have your formula. – Ross Millikan Apr 27 '20 at 13:27