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
y^2=6-y
y^2+y-6=0
(y+3)(y-2)=0
so the first quadrant ymax is 2
check that.
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.
1 answer