Do the inequalities matter in this case of the marginal probability distribution?

I have this probability density of $f(x,y)$ if $0\le x \lt y\le 1$ and I can't seem to get the right answer. It says in the back of my book that the bound's on the integral for say $f_X(x)$ is from $0$ to $x$. Shouldn't the integral to find the marginal density of $X$ be $$f_X(x)=\int_x^1 f(x,y)dy$$ Then it says the marginal for $Y$ is $$f_Y(y)=\int_y^1f(x,y)dx$$ Shouldnt this one be $$f_Y(y)=\int_0^yf(x,y)dx$$?? Im slightly confused now. Is it because $x$ is strictly less than $y$?

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Books can have typographical errors in them, and the sections titled "answers to odd-numbered problems" or "solutions to odd-numbered problems" often have more such errors than the main text, perhaps because these sections are prepared last, in a rush by the author(s) or their teaching assistants, after the main text has been carefully written and re-written. Your integral for $f_Y(y)$ has the correct limits; the book's integral does not. – Dilip Sarwate Dec 21 '12 at 16:02
@DilipSarwate Thanks that's what I thought. I knew that didn't look right. So glad I have a mathematical community to correct me! – TheHopefulActuary Dec 21 '12 at 16:05

If $f(x,y)$ is the joint density of the random variables $X,Y$, then the marginal density of $X$ is $$f_X(x)=\int_{\mathbb{R}}{f(x,y)dy}$$ Now, you know that $f(x,y)$ equals zero if $y<x$, or $x$ or $y$ falls outside the interval $[0,1]$ (or so I interpret your question) and thus the formula simplifies to $$f_X(x)=\int_x^1{f(x,y)dy}.$$
Similarly one finds $$f_Y(y)=\int_0^y{f(x,y)dx}.$$