The problem requires the following to be calculated:
$$\int_R f(x)dx$$
where:
$$f(x,y)=e^{-xy}\text{ with region }R=\{ -2 \le x \le -1, xy\ge1 \}$$
The boundaries for the first integral for $x$ are pretty obvious, they are $-2$ and $-1$. However, for the $y$ integral: $xy \ge1$, $y \le \frac1x$ (the $\ge$ changed to $\le$ due to dividing by $x$, while $x$ is negative because it is between $-2$ and $-1$)
As the maximum for $x$ is $-1$ I have concluded that it must be that $y \le -1$ so that $xy \ge1$ is satisfied.
If my conclusions have been correct so far, I guess I get the following integral: $$\int_{-2}^{-1} \int_{-\infty}^{-1} e^{-xy} dydx$$
After trying to solve this, I have come to a part where the negative infinity screws everything up due to a non-existing limit. Have I done something wrong in the solution above? Thanks in advance!
EDIT: As the fellow member from the comment section had noticed, inserting $-1$ as the upper boundary was an error. Instead, I must simply leave $\frac1x$ as the upper boundary. However, I am still stuck:
$$\int_{-2}^{-1} \int_{-\infty}^{-1} e^{-xy} dydx$$
After applying the substitution $-xy=u$ I get the following result:
$$\int_{-2}^{-1} \frac{-1}x [e^{-xy}] \Bigg|_{-\infty}^\frac1x dx$$
and when I put in the $-\infty$ I get a non-existing limit (everything tends to $\infty$).
For each $x\in[-2,-1]$, $xy\geqslant1\iff y\leqslant\frac1x$. Therefore, you have\begin{align}\iint_Re^{-xy}\,\mathrm dx\,\mathrm dy&=\int_{-2}^{-1}\int_{-\infty}^{1/x}e^{-xy}\,\mathrm dy\,\mathrm dx\\&=\int_{-2}^{-1}\left[-\frac{e^{-xy}}x\right]_{-\infty}^{1/x}\,\mathrm dx\\&=-\int_{-2}^{-1}\frac1{ex}\,\mathrm dx\\&=-\frac1e\bigl(\log(1)-\log(2)\bigr)\\&=\frac{\log(2)}e.\end{align}