 XP Math - Forums - View Single Post - Calculus 07-28-2008 #2 MyriamK Guest   Posts: n/a Looks like a Calculus question, but ... I did study some calculus once, and know what that integral symbol means. The square root function does not exist for x<0, so b) is not an option. The two functions exist for x=0 and x>0. The functions intersect at (4,2). They intersect the x-axis at (0,0) and (6,0). The region is bound by: the x-axis from (0,0) to (6,0), the line y=-x+6 (which we could call x=-y+6) between (4,2) and (6,0), and the parabola y=x^1/2 (or x=y^2) between (0,0) and (4,2) The "horizontal" distance between the two curves (difference in x values for the same y) is -y+6-y^2. The users of this forum would approximate that area by adding up the areas of thin rectangles of base -y+6-y^2 and small height d, inscribed in that region between y=0 and y=2. The limit of those approximations as d tends to zero is the integral, and that's all the Calculus you need to know for this problem. Integrating the function -y+6-y^2 between y=0 and y=2 is the easiest way to find the area (choice c). The other choices don't make much sense.