# Determining boundaries of Probability Density Function integral for a requested probability

This isn't one specific homework question, but a concept I'm having trouble with in class. We were asked on a couple of questions recently on homework dealing with the probability density function of two random variables, $f_{X,Y}(x,y)$. We were given a joint probability density function and asked to find a given probability, like $P[X + Y > 5]$. One solid example being:

Given a Joint PDF $f_{X,Y}(x,y)$ let A be the event that $X + Y \leq 1$. Now, I know that $$P[A] = \int_A\int f_{X,Y}(x,y)dxdy$$ (at least, that's how my book writes it). My question is, how do I determine the bounds of those two integrals? In the solution manual for this particular problem, it is stated that the solution is $$P[X+Y\leq 1] = \int_0^1\int_0^{1-x}f_{X,Y}(x,y)dxdy$$

I can see pretty clearly that $x + (1-x)$ will always be less than or equal to one for $0 \leq x \leq 1$, so I understand why those bounds work. I'm just not sure I could have come up with them. Is there a process for figuring this out that I am missing? A similar question posed to us was given another joint PDF, let A be the event that $X+Y>5$. I have no idea what the bounds would be for this one. Can someone show me how to find the integral bounds for questions like this?

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You should read about multiple integrals –  Artem Dec 9 '12 at 23:21

In your example, it seems the density $f_{X,Y}(x,y)$ is zero when $x\leqslant0$ or when $y\leqslant0$ hence, to compute $\mathbb P(X+Y\leqslant1)$, the domain of integration is defined by the inequalities $$x\geqslant0,\qquad y\geqslant0,\qquad x+y\leqslant1.$$ This is indeed equivalent to $$0\leqslant x\leqslant1,\qquad 0\leqslant y\leqslant 1-x,$$ but I know no systematic way of deducing the latter from the former, except drawing a rough sketch of the domain of interest.