= Solution
Use the weighted <inner product> $\langle u,v\rangle=\int_0^1u(x)v(x)x\,dx$. A monic quadratic <orthogonal polynomial> $\pi_2(x)=x^2+ax+b$, orthogonal to 1 and $x$, satisfies
$$
\frac14+\frac a3+\frac b2=0,\qquad \frac15+\frac a4+\frac b3=0.
$$
Thus $a=-6/5$, $b=3/10$, and its two zeros are $x_-=(6-\sqrt6)/10$ and $x_+=(6+\sqrt6)/10$. Matching the first two weighted moments gives
$$
w_-+w_+=\frac12,\qquad w_-x_-+w_+x_+=\frac13,
$$
so $w_-=(9-\sqrt6)/36$ and $w_+=(9+\sqrt6)/36$. The <two-node Gaussian quadrature with linear weight> is therefore
$$
\boxed{\int_0^1f(x)x\,dx\ \approx\ \frac{9-\sqrt6}{36}f\!\left(\frac{6-\sqrt6}{10}\right)+\frac{9+\sqrt6}{36}f\!\left(\frac{6+\sqrt6}{10}\right).}
$$
For 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>.
Back to article page
