Find the double integral of f (x, y) = (x^7)y over the region between the curves y = x^2 and y = x(3 - x).
The curves intersect at (0,0) and (3/2,9/4)
So, we want
Int[0,3/2] Int[x^2,3x-x^2] (x^7 y) dy dx
= Int[0,3/2] (x^7 y^2)/2 [x^2,3x-x^2] dx
= Int[0,3/2] (1/2)(9x^9 - 6x^10) dx
= 1/2(9/10 x^10 - 6/11 x^11)[0,3/2]
= 1/2 (3/2)^10 (9/10 - 6/11 * 3/2)
= 1/2 (3/2)^10 (9/110)
= 531441/225280 = 2.359