Estimate the value $ \int_{1}^{2}\int_{3}^{4}{f(x,y)dydx} $ Using the Gauss-Legendre form. Estimate the value of$$\displaystyle\int_{1}^{2}\displaystyle\int_{3}^{4}{f(x,y)dydx}$$
where, $f(x,y)=x^3y$.
My approach: We can approximate the integral $\int_{-1}^{1}{f(x)dx}=\sum_{i=1}^{n}{A_{i}f(x_{i})}$ with the form. Gauss-Legendre, by orthogonals polynomials $$p_{n}(x)=\dfrac{(-1)^{n}}{2^{n}n!}\dfrac{d^{n}}{dx^{n}}[(1-x^2)^n]$$ $$A_{i}=\dfrac{2}{(1-x_{i}^2)[p'_{n}(x_{i})]^{2}}$$ If n=2, then $x_{1}=0.577350=-x_{2}$, and $A_{i}=1$. And ,for instance $[1,2]\to [-1,1]$, this implies that $y=2x-3$ as I continue?

Some pretty good progress. Since the integrand is at most cubic in $x$ and $y$ the result should be exact for the Gauss $2$-point formula. In the $x$-dimension, we want $u=-1$ when $x=1$ and $u=1$ when $x=2$. So
$$\frac{x-1}{u+1}=\frac{2-1}{1+1}=\frac12$$
So $x=\frac12u+\frac32$. In the $y$-dimension we want $v=-1$ when $y=3$ and $v=1$ when $y=4$:
$$\frac{y-3}{v+1}=\frac{4-3}{1+1}=\frac12$$
And we have $y=\frac12v+\frac72$. Then
$$\begin{align}\int_1^2\int_3^4x^3y\,dy\,dx&=\int_{-1}^1\int_{-1}^1\left(\frac{u+3}2\right)^3\left(\frac{v+2}{2}\right)\frac{dv}2\frac{du}2\\
&=\sum_{i=1}^2\sum_{j=1}^2A_iA_j\frac1{64}(x_i+3)^3(x_j+7)\\
&=\frac1{64}(0.577350+3)^3(0.577350+7)+\frac1{64}(0.577350+3)^3(-0.577350+7)\\
&\quad+\frac1{64}(-0.577350+3)^3(0.577350+7)+\frac1{64}(-0.577350+3)^3(-0.577350+7)\\
&=5.420279376+4.594291848+1.683485587+1.426941966\\
&=13.12499878\end{align}$$
This compares well with the exact value of $\frac{105}8$.
