Double Integral over region bounded by parallel lines I have to calculate the double integral 
$$\iint\limits_{D} {y}{e^{-{(x+y)}}}\,\mathrm{d}x\,\mathrm{d}y\,,$$
where $D$ is the area bounded by the lines $x+y=1$ and $x+y=2$. I tried evaluating the integral by substituting the $y$ integration limits as functions of $x$ (namely $1-x$, $2-x$) and vice versa because the bounded area is a general region, but I seem to get different results for each of the two ways.
The excercise then suggests the substitution $x+y=u$, $y=uv$ and the answer is according to the excercise  $\frac{5(e-2)}{2e^{2}}$ which I didn't get for any of the two ways I tried. Can somebody illustrate a full solution to the excercise with and without the suggested substitution?
Thanks in advance.
 A: Let's first start with a diagram of our bounded region. Assuming that $D$ is also bounded by the lines $x = 0$ and $y = 0$ we have:

Let's take your strategy on integrating $y$ first and then $x$. Notice that the vertical bounds change when we get to $x = 1$. Therefore we must split this region like so:

We can now integrate over both regions separately. Let $I$ be the original integral. Now we can define
$$I_1 = \int_{0}^{1} \int_{1 - x}^{2 - x} y e^{-x - y} dy \, dx = \int_{0}^{1} e^{-x} \int_{1 - x}^{2 - x} y e^{- y} dy \, dx$$
$$I_2 = \int_{1}^{2} \int_{0}^{2 - x} y e^{-x - y} dy \, dx = \int_{1}^{2} e^{-x} \int_{0}^{2 - x} y e^{- y} dy \, dx$$
which can both be calculated using techniques like integration by parts! If we compute the integrals we have
$$I_1 = e^{-2} \int_{0}^{1} \left[ (1 - e)x + 2e - 3 \right] dx = \frac{3e - 5}{2 e^2}$$
$$I_2 = \int_{0}^{1} \left[ e^{-2}(x - 3) + e^{-x} \right] dx = \frac{2e - 5}{2 e^2}$$
If we sum $I_1$ and $I_2$ we have our solution:
$$I = I_1 + I_2 = \frac{3e - 5}{2 e^2} + \frac{2e - 5}{2 e^2} = \frac{5(e - 2)}{2 e^2}$$
Now that we have computed the integral with respect to $x$ and $y$, let's take advantage of the hint! If we change variables to $u$ and $v$, the region is greatly simplified:

The vertical lines were trivial to figure out since $u = x + y$. You can also show that if $x = 0$ then $v = 1$ and if $y = 0$ then $v = 0$. Let's now compute the Jacobian matrix of our transformation by first inverting the equations for $u$ and $v$:
$$ u = x + y, uv = y \implies y = uv, x = u - uv$$
and then computing the necessary partial derivatives:
$$J = \begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{bmatrix} = \begin{bmatrix} 1 - v & -u \\ v & u \end{bmatrix} \implies \mathrm{det}(J) = u$$
With the determinant and our diagram we can compute the now separable integral:
$$I = \int_{1}^{2} \int_{0}^{1} u v e^{-u} u \, dv \, du = \left( \int_{1}^{2} u^2 e^{-u} \, du \right) \left( \int_{0}^{1} v \, dv \right) = \left( \frac{5(e - 2)}{e^2} \right) \left( \frac{1}{2} \right) = \frac{5(e - 2)}{2 e^2}$$
Hope that helps!
