Yes, they will be the same (if the integrals in question are not improper) - reversing the order of integration should make no difference.
Show your working and I'll see if I can work out what you're doing wrong.
When I have a problem such as
if f(x,y) = 2e^(-x-2y) , x>0, y>0 and 0 elsewhere find the integral when x< 2y
after drawing the picture, it seems like I could either first integrate over x x from 0 to 2 and then over y from x/2 to x or reverse. But only this one gives the right answer. What am I getting the answer to when I switch it?
The stuff in red was the first clue - this is just the style in which a pdf is defined.
Then of course when you calculate on a hunch and get 1 as the answer, the evidence gets stronger.
And the stuff in blue is just the sort of thing you're asked to do with a bivariate pdf ....
So lots of little things add up to a strong hunch.
if f(x,y) = 2e^(-x-2y) , x>0, y>0 and 0 elsewhere find the integral when x< 2y
after drawing the picture, it seems like I could either first integrate over x x from 0 to 2 and then over y from x/2 to x or reverse. But only this one gives the right answer. What am I getting the answer to when I switch it?
So if I first integrated from 0 to 2 over x that would give me the probability function for y, and then if I integrated that from y= x/2 to x does that make any kind of sense? I know it doesn't, but I don't know exactly why.
I cannot see why you would want to do this ...... Why do you think it would work? Where has "first integrated from 0 to 2 over x" come from? And where has "then if I integrated that from y= x/2 to x" come from?
I don't understand the logic behind your thinking .....
By the way, doing what you say gives a final answer with x in it. Surely you're meant to get a numerical value since you're finding a probability .....