area under the curve?

Question

find the area of the region R in the 1st quadrant that is bounded above by y = sqrt x and below by the x axis and the line y = x-2

Answer 1

To get interception point between y=sqrt(x) and y=x-2
sqrt(x) = x-2
x = (x-2)^2
x = x^2 + 4 - 4x
x^2 -5x + 4 = 0
(x - 4) (x -1) = 0
x = 4, or x = 1
But, when x = 1, y = 1-2 = -1 <> y= sqrt(1) which must be > 0 in 1st quadrant.

Area between y=sqrt(x) and y=x-2
= ∫ sqrt(x) - (x-2) dx (from x=0 to x=4)
= (2/3) x^(3/2) - 1/2 x^2 + 2x + C (from x=0 to x=4)
= (2/3) (4)^(3/2) - 1/2 (4)^2 + 2(4)
= 16/3

Area between x-axis and y=x-2
= ∫ -(x-2) dx (from x=0 to x=2)
= -(1/2x^2 -2x) + C (from x=0 to x=2)
= -(1/2 (2)^2 - 2(2))
= 2

Area R
= 16/3 - 2 = 10/3

Answer 2

I'd plug it into a graphing calc first so you can visualize whats going on.

Find the interesection point.
x^.5 = x-2
x = x^2-4x+4
0=x^2-5x+4
you get (x-4)(x-1). So x = 1 or 4. The square root on the left side causes problems,
Plug values back into original equation
x=1....you get 1 = -1 <--obviously not right
x=4 you get 2 =2 So x=4 is the right intersection point.

Now you need y>0
equation 1. x = 0, y=0
equation 2 x=2 , y=0

So when you integrate the first equation, it'll go form 0->4
Second equation will go from 2->4 since you are only subtracting this area out from under the first equation

based on the problems statment, and a graph if you made one. You know you want the area between equation 1 and 2.
So you're oging to integrate equation 1 and subract the integral of equation 2.

x^(1.5) /(3/2) from 0->4 = 5 1/3
x^(2) / 2 - 2x from 2->4 = 2
16/3 - 6/3 = 10/3