posted by Annie .
find the area of the region enclosed by y=squareroot of x, the line tangent to y= squareroot of x at x=4 and the x-axis
y = √x
y' = 1/2√x
So, at (4,2) the tangent line has slope 1/4.
So, the equation for the line is
(y-2)/(x-4) = 1/4
y = x/4 + 1
which insersects tha x-axis at x = -4.
Now, √x is not defined for x<0, so we need to break the region into two parts:
Int(x/4 + 1)dx [-4,0] + Int(x/4 + 1 - √x)dx [0,4]
= [x^2/8 + x)[-4,0] + (x^2/8 + x - 2/3x^3/2)[0,4]
8 - 16/3 = 8/3