Find $P(XAfter getting help here, with a similar problem, I tried solving this new problem, but ended up getting stuck yet again.
Previous question with similar problem:
Find $P(X+Y>1)$ probability of density function
I have the following density function:
$$f_{X,Y}(x,y)=\frac{1}{4}xy\text{ when }0<x<2\text{ and }0<y<2$$
Need to calculate the following (which is supposedly $\frac{1}{32}$),
$$P(X<Y<1)=\frac{1}{32}$$

My intuition tells me to do the following double integral,
$$\int_{x}^{1}\int_{0}^{1}\frac{1}{4}xydxdy=\frac{1-x^2}{16}$$
or maby,
$$\int_{x}^{1}\int_{0}^{y}\frac{1}{4}xydxdy=\frac{1-x^4}{32}$$
or maby even this, which gets me the size of the area,
$$\int_{0}^{1}1-xdx=\frac{1}{2}$$
But I am not sure if my approach is correct, and if it is, what I should do from here.
 A: It should be $$\int_{0}^{1}\int_{0}^{y}\frac{1}{4}xydxdy$$ or $$\int_{0}^{1}\int_{x}^{1}\frac{1}{4}xydydx.$$
Let's look at the second integral. Think of it this way: you let $x$ run from $0$ to $1$, and for fixed $x$, the value of $y$ should run from $x$ to $1$. This way you get all points $(x,y)$ with $0 < x < y < 1$.
A: As I'm sure you know there are two ways you can integrate over that range. Your second try is nearly right, but just stop to think what you are integrating over. If you integrate over x first, then the limits will depend on what value of y you are at. In this case, the correct limits are 0 and y.
Imagine drawing the line from x=0 to x=y at a y value of your choice. Then to get the entire shaded region, you need to "add up" all these lines in the y-direction, so the limits on the outer integral will be 0 and 1. If you understand this, try doing it the opposite direction, with vertical lines from y=x to y=1 and then integrating all of these over the range x=0 to x=1. Either way, as you say, the answer should be 1/32. You certainly should NOT get a function of x (or y) after having integrated both of these variables out.
