Question

I'm having trouble on this question:
Find the area of the region in the first quadrant that is bounded above by the curve y= sq rt x and below by the x-axis and the line y=x -2.

Answer 1

Sketch the region. It has vertices at (0,0), (2,0), (4,2)

You can get the area by using vertical or horizontal strips. Using vertical strips, they all lie below the curve √x, but the lower boundary changes from y=0 to y=x-2 at x=2. So, the area is

∫[0,2] √x dx + ∫[2,4] √x - (x-2) dx
= 4√2/3 + 10/3 - 4/3 √2 = 10/3

Or, you could just subtract the area of the triangle from the whole area under the curve:

∫[0,4] √x dx - (1/2)(2*2)
= 16/3 - 2 = 10/3

Using horizontal strips, the left boundary is the parabola x=y^2, and the right boundary is x = y+2, so the area is

∫[0,2] (y+2)-y^2 dy = 10/3