The given integration rule is $$I_n(f)=c_0f(x_0)+...+c_nf(x_n)\approx I(f)$$ where $x_0,...,x_n$ are n+1 distinct nodes. $P_n(x)$ is a polynomial that interpolates $f(x)$ at the nodes $x_0,...,x_n$. Show that $I_n(P(x))=I(P(x))$ for the $P(x)$ at most of n degree. It makes sense that it should be true as one approximate the polynomial with another polynomial. As long as the order of polynomial $P(x)$ does not exceed the order of the integration rule. There is a short formal proof but I cannot recall it.
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It is not at all clear what you mean and what you are asking. – copper.hat Dec 14 '19 at 06:52
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Yes, if the integration method is constructed so that $I_n(f)=I(P_n[f])$ where $P_n[f]$ is the interpolation polynomial of $f$ at the sampling points $x_0,...,x_n$, then if $f=P$ is itself a polynomial of degree at most $n$, then $P_n[P]=P$ and $$I_n(P)=I(P_n[P])=I(P).$$
Lutz Lehmann
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