 Solution

Use the weighted inner product $⟨u,v⟩={\int }_0^{1}u(x)v(x)xdx$. A monic quadratic orthogonal polynomial ${\pi }_2(x)=x^2+ax+b$, orthogonal to 1 and $x$, satisfiesThus $a=-6/5$, $b=3/10$, and its two zeros are $x_{-}=(6-\sqrt{6})/10$ and $x_{+}=(6+\sqrt{6})/10$. Matching the first two weighted moments givesso $w_{-}=(9-\sqrt{6})/36$ and $w_{+}=(9+\sqrt{6})/36$. The two-node Gaussian quadrature with linear weight is thereforeFor completeness, a polynomial $P$ of degree at most three can be divided as $P=q{\pi }_2+r$, where both $q$ and $r$ have degree at most one. Orthogonality makes the weighted integral of $q{\pi }_2$ zero, and this term vanishes at both nodes. The two moment equations integrate $r$ exactly. Thus the Gaussian quadrature is exact for every cubic polynomial. No two-node rule can integrate every quartic: the square of its nodal polynomial has quadrature value zero and strictly positive weighted integral.