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September 22, 2014

September 22, 2014

Posted by **Dan** on Wednesday, August 17, 2011 at 1:49am.

Decide whether to integrate with respect to x or y.

Then find the area of the region.

2y=3sqrtx , y=3 , 2y+2x=5

- Calculus -
**drwls**, Wednesday, August 17, 2011 at 7:46amYou 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

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