what is the area of the region enclosed by the graphs of y=2- lx-3l and y= x^2-2x?

To find the area of the region enclosed by the graphs of these two functions, we need to determine the points of intersection first. Let's solve the equation y = 2 - |x - 3| and y = x^2 - 2x simultaneously.

Step 1: Solve for the intersection points:
Setting the two equations equal to each other, we have:
2 - |x - 3| = x^2 - 2x

Step 2: Analyze the absolute value:
Since |x - 3| can be either x - 3 or -(x - 3), we will need to solve for both cases.

Case 1: x - 3 = 2 - x + 3
Simplifying this, we get: 2x = 4 -> x = 2

Case 2: -(x - 3) = 2 - x + 3
Simplifying this, we get: -x + 3 = 5 -> x = -2

So, the intersection points are (2, 2) and (-2, -2).

Step 3: Determine the limits of integration:
To find the area, we need to know the range over which we will integrate. In this case, it will be between the x-values of the intersection points, -2 and 2.

Step 4: Set up the integral:
To calculate the area between the two curves, we integrate the top function minus the bottom function with respect to x over the given range.
A = ∫[from -2 to 2] (x^2 - 2x - (2 - |x - 3|)) dx

Step 5: Evaluate the integral:
We can break this integral into two parts based on the absolute value function:
A = ∫[-2 to 2] (x^2 - 2x - (2 - (3 - x))) dx
= ∫[-2 to 2] (x^2 - 2x - 1) dx

Integrating each term independently:
A = ∫[-2 to 2] (x^2) dx - ∫[-2 to 2] (2x) dx - ∫[-2 to 2] (1) dx

Evaluating each integral:
A = [(x^3)/3] [-2 to 2] - [x^2] [-2 to 2] - [x] [-2 to 2]
= (8/3 - (-8/3)) - (4 - 4) - (2 - (-2))
= (16/3) - 0 - 4
= (16/3) - 12/3
= 4/3

So, the area of the region enclosed by the graphs of y = 2 - |x - 3| and y = x^2 - 2x is 4/3.