Find $x_i$ and weights such that the following integration rule is exact for all polynomials of degree $\leq 5$ I'm going over an exam I failed. I was told that I can't use the method I used to solve the following question, and I don't know why.
Can you please explain and suggest a correct solution?
Question
Find $x_i$ and $A_i$ such that the following integration rule is exact for all polynomials of degree $\leq 5$:
$$
\int_{1}^{-1}f(x)|x| dx \approx A_0 f(x_0) + A_1 f'(x_1) + A_2 f'(x_2)
$$
My Attempt
I tried to choose $x_i$ as a Gaussian Quadrature (PDF, section 6.6) relative to the weight function $\omega(x) = |x|$ (which satisfies $\omega(x) \geq 0$ for all $x$ so we can use it to define a proper inner product).
Then I wrote a system of linear equations by applying the rule to $P_n(x) = x^n, n \in \{0..5\}$ and trying to solve it for $A_i$. I didn't have time to finish that part so I'm not sure why it doesn't work.
I assume my approach fails because the integration rule uses derivatives of $f$, is that correct? 
What would be the correct approach for this problem?
Edit: Extended explanation for my approach
To find $x_i$ we look for a Gaussian Quadrature:


*

*We define the inner product $<f,g> = \int_{-1}^{1}f(x)g(x)|x|dx$. We can do this since $\omega(x) \geq 0$ for all $x$.

*Then we find a basis of orthogonal polynomials relative to that inner product using Gram-Schmidt.

*$x_i$ are the roots of the 3rd degree polynomial we get using this process.


Then we want to find $A_i$:


*

*For $n \in {0..5}$ we write the equation:
$$
P_n(x) = x^n \\
\int_{-1}^{1}P_n(x)|x|dx = A_0P_n(x_0) + A_1P'(x_1) + A_2P'(x_2)
$$

*The only thing we don't know how to calculate is the $A_i$'s so we get a system of linear equations that we can solve to find them.


Thanks! :)
 A: A quick work would be as follows.  By symmetry ($x\mapsto -x$ in the integral), we can see that $x_0=0$ and $x_1+x_2=0$.  By integrating with $P_0$, we obtain $A_0=1$.  With $P_1$, we get $A_0+A_1=0$.  By integrating with $P_2$, $\frac{1}{2}=2A_1x_1+2A_2x_2$.  With $P_4$, we get $\frac13=4A_1x_1^3+4A_2x_2^3$.  Thus, $A_1x_1=\frac{1}{8}$ and $A_1x_1^3=\frac{1}{24}$, or $x_1^2=\frac{1}{3}$.  We can assume without loss of generality that $x_1=+\frac1{\sqrt3}$ and $x_2=-\frac1{\sqrt3}$.  Hence, $A_1=+\frac{\sqrt{3}}{8}$ and $A_2=-\frac{\sqrt{3}}{8}$.
A: Due to the fact that both the integration segment $[-1,1]$ and $\omega(x)$ are symmetric to the transformation $x \to -x$, one might look for a quadrature with the same symmetry. That is weights $A_i$ should satisfy (the minus comes from the fact that replacing $x$ with $-x$ also negates the first derivatives):
$$
A_1 = -A_2
$$
and the abscissae should satisfy
$$
x_1 = -x_2 \equiv, \quad x_0 = -x_0 = 0.
$$
Thus letting $A = A_0, B = A_1, \xi = x_1$
$$
\int_{-1}^{1} f(x) |x|dx = A f(0) + B \left(f'(\xi) - f'(-\xi)\right).
$$
Also, since we are already restricting ourselves to symmetric quadratures, we don't need to check every polynomial $x^{k}, k = 0,\dots, 5$, but only those, that are symmetric (there is a theorem for that fact, but it is quite obvious that $\int_{-1}^{1}x^{2m+1}|x|dx = 0$ both for the exact integration and the quadrature rule).
As I said, it is necessary to check only symmetric (even for the case) polynomials $P_1 = x^0 = 1, P_2 = x^2$ and $P_3 = x^4$. That will give three equations for three parameters - quite sane.
$$
\int_{-1}^1 |x| dx = 2\int_0^1 xdx = 1 = A \cdot P_1(0) = A,\quad \Rightarrow A = 1\\
\int_{-1}^1 x^2|x| dx = 2\int_0^1 x^3dx = \frac{1}{2} = B(2\xi + 2\xi) = 4\xi B,\quad \Rightarrow 4\xi B = \frac{1}{2}\\
\int_{-1}^1 x^4|x| dx = 2\int_0^1 x^5dx = \frac{1}{3} = B(4\xi^3 + 4\xi^3) = 8\xi^3 B,\quad \Rightarrow 8\xi^3 B = \frac{1}{3}\\
$$
Thus either $\xi = \frac{1}{\sqrt{3}}, B = \frac{\sqrt{3}}{8}$ or 
$\xi = -\frac{1}{\sqrt{3}}, B = -\frac{\sqrt{3}}{8}$. Both gives essentially the same rule
$$
\int_{-1}^1 f(x) |x| dx \approx f(0) + \frac{\sqrt{3}}{8} \left(
f'\left(\frac{1}{\sqrt{3}}\right) - f'\left(-\frac{1}{\sqrt{3}}\right)
\right)
$$
