The area of the region bounded by the curves \(y=|x-1|\) and \(y=3-|x|\) is (1) 2 sq. unit (2) 3 sq. unit (3) 4 sq. unit (4) 6 sq. unit

Integral calculus is a branch of mathematics focused on the concept of integration. Integration is essential for calculating areas, especially for regions bounded by curves. When you have functions representing these curves, you use definite integrals to find the area between them.

In the provided exercise, we need to find the bounded area between two absolute value functions. First, we identify the points where these curves intersect. This splits the problem into smaller segments, each handled by a definite integral.

After setting up the integral for each segment, we directly compute the integrals between the points of intersection. Finally, summing these integral results gives us the total bounded area.

Absolute value functions create distinctive V-shaped graphs. The absolute value of x, \(|x|\), measures the distance from zero, making all values non-negative. For instance, \(|x-1|\) shifts the V-shape horizontally to the right by 1 unit. It means the graph's vertex sits at (1, 0).

In the exercise, we work with two such functions: \(|x-1|\) and \(3 - |x|\). The second function \(3 - |x|\) is flipped upside down compared to the standard absolute value function due to the negative sign before \(|x|\). It moves the vertex to (0, 3). This flipping and shifting help us visualize the region enclosed by these curves and set up our integration intervals correctly.

Identifying intersection points of the curves is crucial for setting up integrals properly. To find where the curves \(|x-1|\) and \(|3 - x|\) intersect, set the equations equal:
\(|x-1| = 3 - |x|\).

We consider different cases for x:

• For \(x < 1\): both \(x-1\) and \(3-x\) take negative forms, resulting in points like (-1, 2).
• For \(x > 1\): \(x-1\) and \(x\) are positive, generating points such as (2, 1).

These intersection points divide our problem into two intervals needing separate integration. Solving these integrals on either interval and adding results gives the total area.