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

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### 1 Answer

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**