What is the area of the region enclosed between the curves y=x^2-2x+2 and -x^2+6 ?

To determine the area of the region between the given
curves, we'll have to determine the definite integral of the difference between the
expressions of the given curves.

First, we need to find out
the intercepting points of the curves. For this reason, we'll
equate:

x^2-2x+2 =
-x^2+6

We'll move all terms to one
side:

2x^2 - 2x - 4 = 0

We'll
divide by 2:

x^2 - x - 2 =
0

We'll apply quadratic
formula:

x1 = [1 + sqrt(1 +
8)]/2

x1 = (1+3)/2

x1 =
2

x2 = (1-3)/2

x2 =
-1

We'll choose a value for x, between -1 and 2, to verify
what curve is above and what curve is below.

We'll choose x
= 0.

f(x) = x^2-2x+2

f(0) = 
2

g(x) =  -x^2+6

g(0) =
6

We notice that g(x) > f(x), between -1 and
2.

Now, we'll calculate the definite integral of g(x) -
f(x), having as limits of integration x = -1 and x =
2.

g(x) - f(x) =
-x^2+6-x^2+2x-2

g(x) - f(x) = -2x^2 + 2x +
4

We'll calculate the definite
integral:

Int (g(x) - f(x))dx = Int (-2x^2 + 2x +
4)dx

Int (-2x^2 + 2x + 4)dx = -2x^3/3 + 2x^2/2 +
4x

F(2) = -16/3 + 4 + 8

F(2) =
(36-16)/3

F(2) = 20/3

F(-1) =
2/3 + 1 - 4

F(-1) =
(-9+2)/3

F(-1) = -7/3

F(2) -
F(1) = 20/3 + 7/3

F(2) - F(1) =
27/3

F(2) - F(1) =
9

The area enclosed by the given curves is A
= 9 square units.