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