To find the area of the region bounded by the curves y = |x – 1| and y = 3 - |x|, we need to first determine the points of intersection between the two curves.

Step 1: Finding the points of intersection
To find the points of intersection, we equate the two equations and solve for x:
|x – 1| = 3 - |x|

Since the absolute value function can be written as two separate equations, we have two cases:

Case 1: When x – 1 ≥ 0, we have |x – 1| = x – 1
So the equation becomes:
x – 1 = 3 - |x|

Case 2: When x – 1 < 0,="" we="" have="" |x="" –="" 1|="-(x" –="" 1)="-x" +="" />
So the equation becomes:
-x + 1 = 3 - |x|

Solving each case separately:

Case 1:
x – 1 = 3 - |x|
x – 1 = 3 - x (since x ≥ 1, |x| = x)
2x = 4
x = 2

Case 2:
-x + 1 = 3 - |x|
-x + 1 = 3 - (-x + 1) (since x < 1,="" |x|="-x" +="" />
-x + 1 = 3 + x - 1
-2x = 2
x = -1

So the curves intersect at x = 2 and x = -1.

Step 2: Determining the limits of integration
To find the area between the curves, we need to determine the limits of integration. The curves y = |x – 1| and y = 3 - |x| intersect at x = 2 and x = -1. So the limits of integration for x will be -1 to 2.

Step 3: Evaluating the integral
The area between the curves can be found by taking the integral of the difference between the two curves with respect to x:

Area = ∫[from -1 to 2] [(3 - |x|) - |x – 1|] dx

To simplify the integral, we can break it up into two parts, one for each case:

Area = ∫[from -1 to 1] [(3 - (-x + 1)) - (-x + 1)] dx + ∫[from 1 to 2] [(3 - x) - (x – 1)] dx

Simplifying each integral:

Area = ∫[from -1 to 1] (4 + x) dx + ∫[from 1 to 2] (2 - 2x) dx

Area = [4x + (x^2)/2] [from -1 to 1] + [2x - x^2] [from 1 to 2]

Area = [(4 + 1/2) - (-4 + 1/2)] + [(4 - 2) - (2 - 1)]

Area = [9/2 - 7/2] + [2 - 1]

Area = 2 + 1

Area = 3 sq. units

Therefore, the