# Solving an integral where the bounds are max and min functions

I'm trying to solve Problem 1.13 in An intermediate Course in Probability, 2nd Ed. (Gut, 2009):

Let $$X$$ and $$Y$$ have a joint density function given by

$$f_{X,Y} (x, y) = \begin{cases} 1, & \text{for } 0 \le x \le 2, \text{max}(0, x − 1) \le y \le \text{min}(1, x),\\ 0, & \text{otherwise.} \end{cases}$$ Determine the marginal density functions and the joint and marginal distribution functions.

Earlier in the chapter, Gut shows us how to find these marginal density functions $$f_X(x)$$ and $$f_Y(y)$$:

$$f_X(x) = \int_{-\infty}^{\infty}f_{X,Y}(x,y)dy\\ f_Y(y) = \int_{-\infty}^{\infty}f_{X,Y}(x,y)dx$$

So in this case we have

$$\begin{split} f_X(x) &= \int_{-\infty}^{\infty}f_{X,Y}(x,y)dy\\ &= \int_{\text{max}(0,x-1)}^{\text{min}(1,x)}1dy\\ \end{split}$$

This is where I'm stuck. How do I solve this integral?

$$\min(1,x)=1$$ if $$x\geqslant 1$$ and $$x$$ if $$x\leqslant 1$$, $$\max(0,x-1)=0$$ if $$x\leqslant 1$$ and $$x-1$$ if $$x\geqslant 1$$. Calculate the integral when $$x\geqslant 1$$ first and then when $$x\leqslant 1$$. You can also say that $$\int_a^b{1dy}=b-a$$ for all $$(a,b)\in\mathbb{R}^2$$