Let $x_0=a,x_n=b,x_i=x_{i-1}+h$ for $i=1,\cdots,n-1$ with $h=\frac{b-a}{n}$ and consider the (n+1)-points Newton-Cotes formula:
\begin{equation}\sum_{i=0}^na_if(x_i), \ \ a_i=\int_{x_0}^{x_n} \prod_{j=0,j\neq i}^n \frac {x-x_j}{x_i-x_j} dx. \end{equation}
Suppose $n$ is an odd integer and $f\in C^{n+1}[a,b]$. Show that there is a $\xi \in (a,b)$ such that \begin{equation}\int_a^b f(x)dx=\sum_{i=0}^na_if(x_i)+\frac{h^{n+2}f^{(n+1)}(\xi)}{(n+1)!} \int_0^n t(t-1)\cdots(t-n)dt. \end{equation}
I know that $$f(x)-\sum_{i=0}^nL_i(x)f(x_i)=\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-x_0)\cdots(x-x_n).$$
Thus, Error term for Numerical Integration is \begin{equation} \int_{a}^b\frac{f^{(n+1)}(\xi(x))}{(n+1)!}(x-x_0)\cdots(x-x_n)dx. \end{equation}
I firstly want to prove that there exist $\eta \in (a,b) $ such that \begin{multline} \frac{f^{(n+1)}(\eta)}{(n+1)!}\int_a^b(x-x_0)\cdots(x-x_n)dx \\=\int_{a}^b\frac{f^{(n+1)}(\xi(x))}{(n+1)!}(x-x_0)\cdots(x-x_n)dx. \end{multline}
I usually used the mean value theorem for integration when I solved these similar problems.
However, I couldn't check that $(x-x_0)\cdots(x-x_n)$ has same sign on $[a,b]$.
Any help is appreciated..
Thank you!!!