Calculus

The area of the region:y = x^1/2 and y=-x+6
enclosed by the two graphs and the x-axis 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- (6-y) dy .....d)
-3

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.