An exercise asks me to prove this: Numerically integrating $$I=\int_a^b f(x)\ \text{d}x \ \ \text{or}\ \ \int_a^b \rho(x)f(x)\ \text{d}x$$ using any formula of interpolation type $$L_n = \sum_{k=0}^n A_k f(x_k)$$ which has the algebraic accuracy of $m$, its error would be in the form of $$E(f) = I - L_n = K f^{(m+1)}(\xi)$$ where $\xi \in (a,b)$, and $K$ would be a constant independent of $f$. Also, $f$ could be assumed to be smooth, and $(a,b)$ to be a finite interval.
Saying $K$ independent of $f$, I mean, since the formula has the algebraic accuracy of $m$, letting $f(x)=x^{m+1}$, we get $$f^{(m+1)}(x)=(m+1)!\ \ \Rightarrow\ \ K=\dfrac{1}{(m+1)!}\left[\int_a^b \rho(x)x^{m+1}\ \text{d}x - \sum_{k=0}^n A_k x_k^{m+1}\right] $$ which is largely independent of $f$ when integrating $I$ numerically. Also, when proving, $K$ shall also be independent of $f$. For example, for Gaussian quadrature, the error would be $$ R_n[f] = \dfrac{f^{(2n+2)}(\xi)}{(2n+2)!}\int_a^b \omega_{n+1}^2\rho(x)\ \text{d}x\ \ \Rightarrow\ \ K = \dfrac{1}{(2n+2)!}\int_a^b \omega_{n+1}^2\rho(x)\ \text{d}x$$ where $$\omega(x)=\prod_{k=0}^n (x-x_k)$$ and $\rho(x)$ is the weight function.
I have already proved for the case of Gaussian ($m=2n+1$) and Newton-Cotes with even-number equal-length interpolation intervals ($m=n+1$), but I failed to prove for other cases, i.e. the general case for all quadrature of interpolation type.
How can I get rid of the strict requirements for mean value theorem? Many thanks.