The remainder term of the $n+1$point closed NewtonCotes formula

Suppose that $\sum_{i=0}^na_if(x_i)$ denotes the $n+1$-point closed Newton-Cotes formula with $x_0=a,x_n=b$,and $h=\frac{b-a}{n}$.There exists $\xi\in (a,b)$ for which
  \begin{align*}
    \int_a^bf(x)dx=\sum_{i=0}^na_if(x_i)+\frac{h^{n+3}f^{(n+2)}(\xi)}{(n+2)!}\int_0^nt^2(t-1)\cdots (t-n)dt
  \end{align*}
if $n$ is even and $f\in C^{n+2}[a,b]$,and
\begin{align*}
  \int_a^bf(x)dx=\sum_{i=0}^na_if(x_i)+\frac{h^{n+2}f^{(n+1)}(\xi)}{(n+1)!}\int_0^nt(t-1)\cdots (t-n)dt
\end{align*}
if $n$ is odd and $f\in C^{n+1}[a,b]$.



Proof:We first study the relation between $\int_0^nt(t-1)\cdots (t-n)dt$ and $\int_{x_0}^{x_n}(x-x_0)(x-x_1)\cdots (x-x_n)dx$.It is easy to verify that
\begin{align*}
  \int_0^{nh}x(x-h)\cdots (x-nh)dx=h^{n+2}\int_0^nx(x-1)\cdots (x-n)dx
\end{align*}

And it is also easy to verify that
\begin{align*}
  \int_0^{nh}x(x-h)\cdots
  (x-nh)dx=\int_{x_0}^{x_n}(x-x_0)(x-x_1)\cdots (x-x_n)dx
\end{align*}
So when $n$ is odd and $f\in C^{n+1}[a,b]$,we just need to prove that

\begin{align*}
  \int_{x_0}^{x_n}f(x)dx=\sum_{i=0}^na_if(x_i)+\frac{f^{(n+1)}(\xi)}{(n+1)!}\int_{x_0}^{x_n}(x-x_0)(x-x_1)\cdots (x-x_n)dx
\end{align*}
So we just need to prove that

\begin{align*}
  \int_{x_0}^{x_n}[f(x)-\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-x_0)(x-x_1)\cdots
  (x-x_n)]dx=\sum_{i=0}^na_if(x_i)
\end{align*}
This is true according to the remainder of the Newton interpolation.




Now let's see the relation between $\int_0^nt^2(t-1)\cdots (t-n)dt$ and $\int_{x_0}^{x_n}(x-x_0)^2(x-x_1)\cdots (x-x_n)dx$.It is easy to verify that
\begin{align*}
  \int_{x_0}^{x_n}(x-x_0)^2(x-x_1)\cdots
  (x-x_n)dx=\int_0^{nh}x^2(x-h)\cdots (x-nh)dx
\end{align*}
So we just need to investigate the relation between
\begin{align*}
  \int_0^nx^2(x-1)\cdots (x-n)dx
\end{align*}
and
\begin{align*}
  \int_0^{nh}x^2(x-h)\cdots (x-nh)dx
\end{align*}
Let
\begin{align*}
  p=\frac{x}{h}
\end{align*}
Then
\begin{align*}
  \int_0^{nh}x^2(x-h)\cdots (x-nh)dx=\int_0^n (ph)^2(ph-nh)\cdots
  (ph-nh)dx=h^{n+3}\int_0^np^2(p-1)\cdots (p-n)dp
\end{align*}

So we just need to prove that

\begin{align*}
 \int_a^b[f(x)-\frac{f^{(n+2)}(\xi)}{(n+2)!}x^2(x-1)\cdots (x-n)]dx=\sum_{i=0}^na_if(x_i)
\end{align*}

This is true according to the remainder term of the Hermite interpolation.