Yes, and now you can evaluate that single integral using a u substitution.
The problem is to compute the double integral sin(y^2)dydx from 3x<=y<=12, 0<=x<=4.
I wanted to check to see if I was on the right track in solving this. I figured I should switch the order of integration, so that it becomes sin(y^2)dxdy from 0<=y<=12, 0<=x<=y/3. And since sin(y^2) is constant with respect to x, then the double integral would be the same as a single integral of sin(y^2)*(y/3) dy from 0<=y<=12.
Is this correct?