Question:

Consider the functions f(x) = 2x - x^2 and g(x) = x^2 - 2x. The curves C1 and C2 are defined by the equations y = f(x) and y = g(x), respectively.

a) Find the x-coordinate(s) at which the curves C1 and C2 intersect.

b) Find the area of the region bounded by the curves C1 and C2 between their points of intersection.

Answer:

a) To find the x-coordinate(s) at which the curves C1 and C2 intersect, we need to set their equations equal to each other:

f(x) = g(x)

2x - x^2 = x^2 - 2x

Rearranging the equation, we get:

2x - x^2 - x^2 + 2x = 0

Combining like terms, we have:

4x - 2x^2 = 0

Factoring out 2x, we get:

2x(2 - x) = 0

Setting each factor equal to zero, we have:

2x = 0 -> x = 0

2 - x = 0 -> x = 2

Therefore, the curves C1 and C2 intersect at x = 0 and x = 2.

b) To find the area of the region bounded by the curves C1 and C2 between their points of intersection, we need to integrate the difference between the equations of the curves with respect to x, from x = 0 to x = 2:

Area = ∫[0, 2] (f(x) - g(x)) dx

First, let's find the difference function (f(x) - g(x)):

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

Simplifying, we get:

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

         = 4x - 2x^2

Now, we can integrate this difference function with respect to x:

Area = ∫[0, 2] (4x - 2x^2) dx

Splitting the integral into two separate integrals:

Area = ∫[0, 2] 4x dx - ∫[0, 2] 2x^2 dx

Integrating each term separately:

Area = [2x^2] from 0 to 2 - [2/3 x^3] from 0 to 2

Substituting the upper and lower limits:

Area = (2(2)^2 - 2(0)^2) - (2/3(2)^3 - 2/3(0)^3)

Area = (2(4) - 2(0)) - (2/3(8) - 2/3(0))

Area = 8 - (16/3 - 0)

Area = 8 - 16/3

Area = 24/3 - 16/3

Area = 8/3

Therefore, the area of the region bounded by the curves C1 and C2 between their points of intersection is 8/3 square units.