Solving integral by substitution The question is to solve the integral $$\iint_D e^\frac{x+y}{x-y}dxdy$$
with the substitutions u=x+y and v=x-y,
where D is the trapezoidal region with vertices in (1, 0), (2, 0), (0, −2) and (0, −1).
By solving it in wolfram directly the answer is 4,36895, but I can't get it by solving the integral "by hand". Another problem is that I came up with the integration limits [1,2] to v and [-v,-v+4] to u, but a friend of mine came up with [1,2] to v and [-v,v] to u.
 A: We can write the trapezoid as
$$D = \{(x,y) \in \mathbb{R}^2 : 0 < x < 2 \, , x-2 < y < \min \{x - 1,0\}\}  \, . $$
Now we want to use the linear transformation $\phi : D \rightarrow \phi (D)$ given by
$$ \begin{pmatrix} u \\ v \end{pmatrix} \equiv\phi (x, y) = \begin{pmatrix} x + y \\ x - y \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \equiv M \begin{pmatrix} x \\ y \end{pmatrix} \, . $$
Note that $ \det (M) = -2$. 

We have to find the region $\phi (D)$ to obtain the integration limits for $u$ and $v$ (this is actually the hardest part). For $(x,y) \in D$ we immediately see that 
\begin{align}
v &= x - y > x - \min \{x-1,0\} = \max \{1,x\} \geq 1 \, , \\
v &= x - y < x - (x-2) = 2 \, ,
\end{align}
so $v \in (1,2)$ . The limits for $u$ depend on $v$ and we can write
\begin{align}
u &= x + y = 2x -v > - v \, , \\
u &= x + y = 2y + v < v \, .
\end{align}
This shows that 
$$ \phi(D) \subseteq \{(u,v) \in \mathbb{R}^2 : -v < u < v \, , 1 < v < 2 \} \, .$$
In order to prove that these two regions are indeed equal, take any $v \in (1,2)$ and $u \in (-v,v)$. We have to show that $(x,y) \equiv \phi^{-1} (u,v) \in D$ holds. We have $ M^{-1} = \frac{1}{2} M$ , so
\begin{align}
x &= \frac{u+v}{2} > \frac{-v + v}{2} = 0 \, , \\
x &= \frac{u+v}{2} < \frac{v + v}{2} < 2 \, , \\
y &= \frac{u-v}{2} = x - v >  x - 2 \, , \\
y &= \frac{u-v}{2} < \min \left\{\frac{v - v}{2} , x - 1\right\} = \min \{0,x-1\} \, . \\
\end{align}
But these are exactly the conditions which ensure $(x,y) \in D$. Therefore your friend is right and we have
$$ \phi(D) = \{(u,v) \in \mathbb{R}^2 : -v < u < v \, , 1 < v < 2 \} \, .$$

Your inequality $u < 4 - v$ is also true for $(u,v) \in \phi(D)$, but it is not strong enough. Note that $\phi(D)$ must be a trapezoid as well, since the transformation matrix can be written as
$$ M = \sqrt{2} \begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \, ,  $$
which is a reflection followed by a rotation and a scaling. If you draw the region we just computed, you will find that this is indeed the case, while the region you have found is a slightly larger parallelogram.

Now we can finally compute the integral. Since our definition of the transformation is contrary to the usual convention, we have to divide by the Jacobian determinant when we use the change of variables theorem. We find
\begin{align}
\iint \limits_D  \exp \left(\frac{x+y}{x-y}\right) \, \mathrm{d} x \, \mathrm{d} y &= \iint \limits_{\phi(D)} \exp \left(\frac{u}{v}\right) |\det \phi'(\phi^{-1} (u,v))|^{-1} \, \mathrm{d} u \, \mathrm{d} v \\
&= \frac{1}{|\det (M)|} \int \limits_1^2 \int \limits_{-v}^v \exp \left(\frac{u}{v}\right) \, \mathrm{d} u \, \mathrm{d} v \\
&= \frac{1}{2} \int \limits_1^2 (\mathrm{e} - \mathrm{e}^{-1}) v \, \mathrm{d} v \\
&= \frac{3}{4} (\mathrm{e} - \mathrm{e}^{-1}) \\
&\approx 1.7628 \, .
\end{align} 
