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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.

For two-node Gaussian quadrature with linear weight, the monic node polynomial $q(x)=x^2-sx+p$ must be orthogonal to $1$ and $x$ for weight $x$. Indeed, exactness through degree three makes the quadrature of $q$ and $xq$ vanish at its nodes. The moment equations areso $s=6/5$ and $p=3/10$. Thus the Gaussian quadrature nodes, labelled in increasing order, areIntegrating the linear Lagrange interpolation cardinal functions gives the weightsThey are positive, sum to $1/2$, and reproduce the first moment $1/3$. Any cubic polynomial has remainder of degree at most one after division by $q$; orthogonality annihilates the other part both in the integral and at the nodes. This proves degree-three exactness, not merely the two moment conditions for the weights.