Reversing the order of integration. (Picture included) I wish to reverse the order of integration.
So, I think the original integration is over the region above.
So, reversing the order of integration, I split it into two cases. 
First region (from 0 to 1) comes first, then second region (from -1/2 to 0) comes next.
But, I get the wrong answer. 
Where did I make a mistake?

Note that when $x < 0$, the integral $$\int_{y=0}^x e^{-y^2} \, dy = - \int_{y=x}^0 e^{-y^2} \, dy,$$ thus this area is negative.  If you simply take the difference of the two double integrals you wrote on the right-hand side, rather than the sum, you would obtain the correct result.

$$\int_{x=-1/2}^1 \int_{y=0}^x f(x,y) \, dy \, dx = \int_{y=0}^1 \int_{x=y}^1 f(x,y) \, dx \, dy - \int_{y=-1/2}^0 \int_{x=-1/2}^y f(x,y) \, dx \, dy.$$  Again, the negative sign is necessary because for $x < 0$, the inner integral on the LHS is REVERSED.
