posted by Dan .
Sketch the region enclosed by the given curves.
Decide whether to integrate with respect to x or y.
Then find the area of the region.
2y=3sqrtx , y=3 , 2y+2x=5
You will have to sketch the three curves yourself. The curves
y = (3/2)sqrtx and y = (5/2) -x
intersect at (1,3/2)
The enclosed area starts at that point and expands upward to the line y = 3, bounded by the two other curves along the way.
Your best bet is to integrate
(x2 - x1) =
(4/9)y^2 - [(5/2)-y]dy from y= 3/2 to 3