Could somebody check my solution if it is correct.
I got the answer -4.
I don't know if it is right or not.
First observe that the integrand (4-y) is positive on D. Therefore the integral
over D cannot be negative.
The region of integration is 0<x<2, x^2/4<y<1, so our integral may be written:
I = integral_{x=0,2} [integral_{y=x^2/4,1} (4-y) dy] dx
.. = integral_{x=0,2} [7/2 - x^2 + x^4/32] dx
.. = 68/15 = 4.533..
Which agrees with my numerical result using MonteCarlo methods of 4.535.
RonL
Trying to sort out time zones can be a problem online, you have yours
set up 3 hours east of me, but your ISP appears to be in the Netherlands!?
But to be brief - When was 10 AM? My post is time stamped 08:11 (AM)
and your reply 14:09 (02:09 PM) in my time zone.
OK repeat search - Estonia!
RonL