The area between these graphs would exist from x=0 to x=1 since the graphs intersect at these two points. Being that the radical of a number between 0 and 1 would be greater than the square of such number, then f(x) = x^.5 + 3 would be above the other function for this interval.
Int(0,1) (x^.5 +3 -(x^2 +3))=
Int(0,1) (x^.5 - x^2) = (2/3)x^(3/2) - (1/3)x^3 =
(2/3 - 1/3) - 0 = 1/3
