• You have to do an integral: Let y1(x) = 1 + sqrt(x), and y2(x) = 1 + x/3 The area of the region is the area under y1(x) minus the area under y2(x), between 0 < x < 9. they intersect each other at points 0 and 9... set y1=y2 and solve for x, you find that x=0 and x=9 (if my algebra is right). So do the integral of [y1(x)-y2(x)]dx over the range 0<x<9 and you have your answer.
  • In this answer I assume that you have some knowledge of integral calculus since this is a problem whose soultion requires calculus. First determine what values of x will give both equations the same value of y. These are the points of intersection and also the upper and lower limits for the integral you will need to set up. Next you have to determine which function is "on top" in the region between the two limits of integration. You can do this graphically or numerically. The answer to this problem is the intergral from a to b of the function on top minus the function underneath, where a and b are the points of intersection of the two graphs and a is the lesser of the two. Hope this helps.

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