I'm trying to look for references on this but I've not found any. I'm probably using the wrong keywords ...
Let's suppose that our domain of integration $\Omega$ is the triangle in $R^2$ with a vertex at the origin, a vertex at $(0,1)$ and one at $(1,0)$. How can I find a quadrature rule for $\int_{\Omega} f \,dx$ if my quadrature points are the 3 vertices and the 3 midpoints of the edges?
Are there any books/lecture notes that you can direct me to for more information on this topic?
I've learned how to do Gauss-Legendre quadrature in an undergrad numerical analysis course but it was in 1 D. Not only is the above in 2 D but the geometry is different too.
I suggest you to approximate $f(x,y)$ with desired order. I leave order of $f$ to you. an example of $f$ is:
$$f(x,y)=a x^2 + b y^2 + c x y + d x +e y +g$$
Once, find the analytical order of $f$ in analytical way.
$$A=\int_\Omega f(x,y) dx dy $$
A is a function of $a,b,c,d,e,g$:
$$A=A(a,b,c,d,e,f,g)$$
Then you need to find out the relationship between A and the points. You have six points: $(0,0), (1,0), (0,1), (\frac12,0), (0,\frac12), (\frac12,\frac12)$.
Let's call them $p_1, \cdots, p_6$ . Function $f$ at these points gives you $f_1, \cdots, f_6$. You have six equations and six variables:
$$f_1=a x_1^2 + b y_1^2 + c x_1 y_1 + d x_1 +e y_1 +g$$ $$\vdots$$ $$f_6=a x_2^6 + b y_6^2 + c x_6 y_6 + d x_6 +e y_6 +g$$
They give you $a, b, c, d, e, g$ as function of $f_1, \cdots, f_6$. Since $A$ is function is $a, b, c, d, e, g$, it is function of $f_1, \cdots, f_6$:
$$A=A(f_1, \cdots, f_6)$$
And done, you can calculate the integrate for any desired function with known values at those six points.