Use the change of variables to evaluate the double-integral $\iint_R4(x+y)e^{x-y}\;dA$ 
Use the change of variables indicated to evaluate the double-integral $\iint_R4(x+y)e^{x-y}\;dA$ where $R$ is the interior of the triangle whose vertices are $(-1,1)$, $(0,0)$, and $(1,1)$; $x=\frac{1}{2}(u+v)$ and $y=\frac{1}{2}(u-v)$

I know I have to find the following components for the formula:
$$\iint_S f(g(u,v),h(u,v)) \left|\frac{\partial(x,y)}{\partial(u,v,)}\right| \ du \; dv$$
The Jacobian is $\frac{1}{2}$
Next I replaced $x$ and $y$ in my integral with the change of variable to give me the following double integral:
$$\iint 4\left[\frac{u}{2}+\frac{v}{2}+\frac{u}{2}-\frac{v}{2}\right]
e^{\frac{u}{2}+\frac{v}{2}-\frac{u}{2}+\frac{v}{2}}\;\left(\frac{1}{2}\right)\;du\;dv$$
After simplifying the above I got:
$$2\iint ue^v\;du\;dv$$
I'm completely confused with how to map $(x,y)$ to $(u,v)$ so that I can identify my limits of integration.  

If I'm not mistaken:
Notice how 
$$u = x +y \qquad \text{and} \qquad v = x-y.$$
Then notice how $v = \alpha$ implies parallell lines to the first bisector of the plane. This means $v = 0$ results in the first bisector.
Furthermore $u =0$ results in the second bisector.
All that is remaining is the straight line $y = 1$. You can write this using the variables $u$ and $v$ as $1 = \dfrac{1}{2}(u-v)$ or $2 = u-v$.
This all implies the following:
Let $$v: 0\to u-2 \qquad \text{and} \qquad u = 0 \to 2.$$
Added picture:

By using the change of variables you integrate along the red arrows. 
These red arrows start at $v = 0$, but and stop at $v = y-2$. Integration goes as follow:
$$\int_0^2 \operatorname du \int_0^{u-2} \operatorname dv$$
For example: At $u=0$ (the first red arrow) integration goes from $v = 0$ to $v = 0-2 = -2$ (or $x-y = -2$ or $y = x+2$ the blue line).
At $u =1$ (the second red arrow) integrations goes from $v=0$ to $v = 1-2 = -1$ (or $x-y = -1$ or $y = x+1$ the pink line).
When $u \approx 2$ then integration goes from $v=0$ to $v \approx 2-2 =0$ or the red arrow is getting tiny small.
