Joint distribution functions and finding the bounds 
I am trying to find the marginals of this function, however, I am not sure how to determine the limits of integration. The answer key says that for $f_1(y_1)$, the limit is from $y_1$ to 1, and for $f_2(y_2)$, the limits of integration are from 0 to $y_2$, can anyone explain why?
Here is the answer key:

 A: I will call the random variables $X$ and $Y$ instead of $Y_1$ and $Y_2$, and use the associated letters $x$ and $y$ instead of $y_1$ and $y_2$.  
We are told that $f(x,y)=k(1-y)$ when $0\le x\le y\le 1$ and $0$ elsewhere.
Let us identify the region where the density function lives. Draw the usual square with corners $(0,0)$, $(1,0)$, $(1,1)$ and $(0,1)$.  Draw the line $y=x$. We are told that $y\ge x$, so our region is above the line $y=x$.
So the joint density function lives in the triangle with corners $(0,0)$, $(1,1)$, and $(0,1)$.  Carefully identity that triangle, and perhaps shade it in.
Now we are ready to compute. For the marginal density of $X$, we must, for any fixed value of $x$, "integrate out" $y$. 
So take a point $(x,0)$ on the $x$-axis, and draw a vertical line upwards, The line first meets our triangle at $y=x$, and leaves it at $y=1$. Thus
$$f_X(x)=\int_{y=x}^1 k(1-y)\,dy.$$
Similarly, to find $f_Y(y)$, we integrate out $x$. For any fixed value of $y$, draw a horizontal line at height $y$. This line enters the triangle at $x=0$ and leaves it at $x=y$. So
$$f_Y(y)=\int_0^y k(1-y)\,dy.$$
A: The key is really to just draw out the support of the jdf. Think of $y_1$ and $y_2$ as the usual $x$ and $y$ in the plane. What does the region $0\leq x \leq y \leq 1$ look like? Think of the inequality $x\leq y$, which is just the region above the line $x=y$. Then cut off the regions where $x,y\leq 0$ and $x,y\geq 1$, because you know that the jdf is 0 when evaluated there. This gives you the support of the jdf, which looks like a triangle bounded by the lines $x=0$, $y=1$ and $x=y$. This should give you the bounds of integration.
