That would not be correct. A good tip is to draw the bounds you have (in two dimensions) to see what the domain (and region you are integrating over) is. You are constrained by

So, combined with your previous bounds of , your domain for your function f is actually a triangle in the plane with vertices at (0,0),(0,1),(1,0).

Armed with that, you then need to put in the new bounds for the probability they want you to find. The problem is, if you sketch the region, you will see that its not so simple as finding a double integral. You need to break it into two regions,and take double integrals over BOTH regions.