Using integration, find the area bounded by the curves `y = |x-1|` and `y = 3-|x|`.

Options

A. 2
B. 3
C. 4
D. 1

Answer

To find the area bounded by the curves \( y = |x - 1| \) and \( y = 3 - |x| \), we will follow these steps:

### Step 1: Analyze the curves

The first curve is \( y = |x - 1| \). This can be expressed as:
- \( y = x - 1 \) for \( x \geq 1 \)
- \( y = 1 - x \) for \( x < 1 \)

The second curve is \( y = 3 - |x| \). This can be expressed as:
- \( y = 3 - x \) for \( x \geq 0 \)
- \( y = 3 + x \) for \( x < 0 \)

### Step 2: Find the intersection points

To find the area bounded by the curves, we need to determine the points where they intersect.

1. **Intersection of \( y = 1 - x \) and \( y = 3 + x \)**:
\[ 1 - x = 3 + x \implies 1 - 3 = 2x \implies -2 = 2x \implies x = -1 \]
Substituting \( x = -1 \) into \( y = 1 - x \):
\[ y = 1 - (-1) = 2 \implies \text{Intersection point: } (-1, 2) \]

2. **Intersection of \( y = 1 - x \) and \( y = 3 - x \)**:
\[ 1 - x = 3 - x \implies 1 = 3 \text{ (no solution)} \]

3. **Intersection of \( y = x - 1 \) and \( y = 3 - x \)**:
\[ x - 1 = 3 - x \implies 2x = 4 \implies x = 2 \]
Substituting \( x = 2 \) into \( y = 3 - x \):
\[ y = 3 - 2 = 1 \implies \text{Intersection point: } (2, 1) \]

### Step 3: Set up the integrals

The area \( A \) can be found by integrating the difference between the upper curve and the lower curve over the intervals determined by the intersection points.

1. From \( x = -1 \) to \( x = 0 \):
- Upper curve: \( y = 3 + x \)
- Lower curve: \( y = 1 - x \)

\[ A_1 = \int_{-1}^{0} ((3 + x) - (1 - x)) \, dx = \int_{-1}^{0} (2 + 2x) \, dx \]

2. From \( x = 0 \) to \( x = 2 \):
- Upper curve: \( y = 3 - x \)
- Lower curve: \( y = 1 - x \)

\[ A_2 = \int_{0}^{2} ((3 - x) - (1 - x)) \, dx = \int_{0}^{2} (2) \, dx \]

### Step 4: Calculate the integrals

1. Calculate \( A_1 \):
\[ A_1 = \int_{-1}^{0} (2 + 2x) \, dx = \left[ 2x + x^2 \right]_{-1}^{0} = \left( 0 + 0 \right) - \left( -2 + 1 \right) = 1 \]

2. Calculate \( A_2 \):
\[ A_2 = \int_{0}^{2} 2 \, dx = \left[ 2x \right]_{0}^{2} = 4 - 0 = 4 \]

### Step 5: Total area

The total area \( A \) is:
\[ A = A_1 + A_2 = 1 + 4 = 5 \]

### Final Answer
The area bounded by the curves \( y = |x - 1| \) and \( y = 3 - |x| \) is \( 5 \) square units.