Yea I made a mistake on my last post. I actually did write my second bound as {0}\leq{y}\leq{2-x}
We've never seen a problem like that before so I got so nervous I just crossed the entire answer out. The answer I decided to write down in the last couple seconds was actually...
So x=2-y and x=y^{2} so y=2-x y=\sqrt{x}
intersect at: 2-x=\sqrt{x} x=1 y=1
y at 0 is (0,2)
I got confused because at y=1,x=1 each side can be explained differently so my first attempt at a solutions was:
\int_{0}^{1}\int_{0}^{\sqrt{x}} f(x,y) \ dy\ dx +...
Homework Statement
Reverse the order of integration:
\int_{0}^{1}\int_{y^2}^{2-y} f(x,y) \ dx\ dy
This was on a test I took today. I was having trouble with it and was wondering how I should have gone about it?