Affine change of variables

Change of variables. Set
$$u = y - x, \quad v = 3x - y.$$
The lines become $u = 0$, $u = 4$, $v = 0$, $v = 1$. The image domain is the rectangle $T = [0,4] \times [0,1]$.

Jacobian. The determinant of the Jacobian matrix of $(x,y) \mapsto (u,v)$ is
$$\det \begin{pmatrix} -1 & 1 \\ 3 & -1 \end{pmatrix} = 1 - 3 = -2.$$
So $|J| = 2$, and $dx\,dy = \frac{1}{2} du\,dv$.

Area.
$$\text{Area}(P) = \iint_P dx\,dy = \iint_T \frac12 \, du\,dv = \frac12 \cdot 4 \cdot 1 = 2.$$

Remark. One can check geometrically that the parallelogram has base of length $\sqrt{2}$ and height $\sqrt{2}$, hence area $2$.