gaussian quadrature weight function x How do I find an orthogonal polynomial for the Gaussian quadrature when the weight function is $x$?
In particular, I need to find $w_0,w_1,x_0, x_1$ such that 
$$\int_0^1xf(x)dx \approx w_0f(x_0)+w_1f(x_1) $$
is exact when $f \in \mathcal{P}_3$.
Which polynomials can I use to find the abscissae $x_0$ and $x_1$?

A shorter way to construct monic orthogonal polynomial $p_2$ is orthogonal to 1 and x, so
$p_2(x)=x^2+ax+b$;
The orthogonality implies
$$
\int_0^1 x p_2(x)dx=\frac{1}{4}-\frac{1}{3}a+\frac{1}{2}b=0;
$$
$$
\int_0 x^2 p_2(x)dx=\frac{1}{5}+\frac{1}{4}a+\frac{1}{3}b=0.
$$
Solving the system with respect to $a$ and $b$ we obtain $a=-\frac{6}{5}$ and $b=\frac{3}{10}$. The nodes are the roots of $p_2$, i.e. $x_0=\frac{3}{5}-\frac{\sqrt{6}}{10}$ and $x_1=\frac{3}{5}+\frac{\sqrt{6}}{10}$.
The weights are the solution of the system
$$w_0+w_1=\int_0^1x dx=\frac{1}{2}$$
$$w_0x_0+w_1x_1=\int_0^1x^2 dx=\frac{1}{3},$$
that is,  $w_0=\frac{1}{4}-\frac{\sqrt{6}}{36}$ and $w_1=\frac{1}{4}+\frac{\sqrt{6}}{36}$.
The error (remainder term)
$$R(f)=\frac{f^{(4)}(\xi)}{4!}\int_0^1 x\left[p_2(x) \right]^2 dx=
\frac{f^{\left(4\right)}\left(\xi \right)}{14400},$$
for some $\xi\in(0,1)$.
