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For the closed Newton-Cotes quadrature over $[x_1, x_n]$, the coefficients $H_{n,i}$ for $$ \int_{x_1}^{x_n} f(x)\:\text{d}x = h \sum_{i=1}^n H_{n,i} \; f(x_i) $$ are given explicitly by $$ H_{n,r+1} =\frac{(-1)^{n-r}}{r!(n-r)!}\int_0^n \frac{\prod_{k=0}^n (t-k)}{t-r}\:\text{d}t; $$ see http://mathworld.wolfram.com/Newton-CotesFormulas.html.

Is there a similar formula for the weights of the open Newton-Cotes scheme?

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The general expression is $$ H_{n,k} = \frac{1}{\Psi'(x_k)} \int_a^b \frac{\Psi(x)}{x-x_k}\;\text{d}x $$ with $$ \Psi(x) = \prod_{k\in S_n} (x-x_k) $$ with $S_n$ in the set of quadrature points; see, e.g., http://www.doiserbia.nb.rs/img/doi/0354-5180/2013/0354-51801304649M.pdf (2).

Choosing $a=0, b=n$ and $$ \text{closed:}\quad S_n = \{k: 0\le k \le n\}\\ \text{open:}\quad S_n = \{k: 1\le k \le n-1\} $$ makes clear that this is an entirely rational expression. For open Newton-Cotes, it is $$ H_{n,r}=\frac{(−1)^{n−r+1}}{(r-1)!(n−r-1)!}\int_{0}^n \frac{\prod_1^n(t-k)}{t-r}\;\text{d}t. $$

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