= Solution
Use the <vertex degree>[degree]-ordered <orthogonal> <polynomials> identified in (b); otherwise the printed nodes need not exist. Let $\ell_j$ be the <vertex degree>[degree]-$n$ <Lagrange interpolation polynomial> with $\ell_j(x_i)=\delta_{ij}$ and set $w_j=\int_a^b\ell_j(x)w(x)dx$.
For any $f\in\Pi_{2n+1}$, <polynomial division> gives $f=p_{n+1}s+r$ with $\deg s\le n$ and $\deg r\le n$. <Orthogonality> gives $\int p_{n+1}sw=0$, while this product also vanishes at every node. The interpolation formula for $r$ then proves
$$
\boxed{\int_a^bf(x)w(x)dx=\sum_{j=0}^nw_jf(x_j).}
$$
Apply this exactness to $\ell_j^2$, of <vertex degree>[degree] at most $2n$. It gives $w_j=\int_a^b\ell_j(x)^2w(x)dx>0$, because $\ell_j$ is not the zero <polynomial>. Thus all <quadrature weights> are actually strictly positive, stronger than the requested nonnegativity. Here exactness is for the weighted integral defined at the start of the question.
Back to article page
