CALCULUS:)
posted by Brit .
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
and
y=3
and
2y+2x=5

its not 8

I would take vertical slices, that is, integrate with respect to x
But after looking at the sketch, I realize that we have to find the intersection of the line 2x+2y=5 and 2y = 3√x
equating 3√x = 52x and squaring both sides I got
4x^2  29x + 25=0
(x1)(4x25) =0
x = 1 or x = 25/4
also if y=3, then 6=3√x > x = 4
so we need ∫(3  5/2 + x dx from 0 to 1 + ∫(3  3x^(1/2) ) dx from 1 to 4
Can you finish it? 
4?? that is incorrect..hmm, what am i doing wrong?