y= x^2

y= x+2

To find the area will need to find the integral for both equations:

First we need to determine the intersecting points:

==> x^2 = x+ 2

==> x^2 -x -2 = 0

==> (x-2)(x+1) = 0

==> x= 2 x = -1

Then we need to find the integral from x= -1 to x= 2

First let us find the area under y= x^2 from x= -1 to 2

==> A1 = intg ( x^2 ) dx

= x^3/3

= ( 2^3/3) - (-1)^3 / 3

= ( 8 + 1)/3 = 9/3 = 3

Then the area under y=x^2 from (-1 to 2 ) is 3 square units>

Now we will calculate the area under y= x+2:

A2 = intg (x+ 2)

= x^2/2 + 2x

= ( 2^2/2 + 2*2) - ( -1^2/2 + 2*-1)

= 2 + 4) - ( 1/2 - 2)

= 6 + 3/2 = 15/2 = 7.5 square units:

Then the area is:

A = A2 - A1 = 7.5 - 3 = 4.5

**Then, the area between y= x^2 and y= x+2 is 4.5 square units**