posted by bob .
Compute the area of the region in the fi
rst quadrant bounded
on the left by the curve y = sqrt(x), on the right by the curve y = 6 - x,
and below by the curve y = 1.
well, looking at the curves, The area will be (x2-x1) dy or (6-y - y^2 )dy ...check that.. with the limits from y=1 to ymax, or
ymax occurs when
x are the same, or
so the first quadrant ymax is 2