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07242008  #1 
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Calculus
The area of the region:y = x^1/2 and y=x+6
enclosed by the two graphs and the xaxis is given by: 2 ∫− x +6 − x^1/2 dx .....a) 0 2 ∫− x +6 − x^1/2 dx .....b) 3 2 ∫ 6− y − y^2 dy ....c) 0 2 ∫y2 (6y) dy .....d) 3 
07282008  #2 
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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 xaxis at (0,0) and (6,0). The region is bound by: the xaxis 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+6y^2. The users of this forum would approximate that area by adding up the areas of thin rectangles of base y+6y^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+6y^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. 
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