Determine the area bounded by the curves y=(x-1)^2+1 and y=x^2+6?

To determine the area bounded by the given curves, we'll
have to calculate 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. The intercepting points represent the limits of
integration.

For this reason, we'll
equate:

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

We'll shift all terms to the
left 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.

We may 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 indefinite
integral:

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

We'll apply the property of integral to be
additive:

Int (-2x^2 + 2x + 4)dx = Int -2x^2dx + Int 2xdx +
Int4dx

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

Now, we'll apply Leibniz Newton
formula:

Int (-2x^2 + 2x + 4)dx = F(2) -
F(-1)

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