The area of region bounded by `y=abs(x-1) and y=1` is

Options

A. 1
B. 2
C. `1//2`
D. None

Answer

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

### Step 1: Understand the Functions
The function \( y = |x - 1| \) can be expressed in piecewise form:
- For \( x \geq 1 \): \( y = x - 1 \)
- For \( x < 1 \): \( y = -(x - 1) = -x + 1 \)

The line \( y = 1 \) is a horizontal line that intersects the \( y \)-axis at 1.

### Step 2: Find Points of Intersection
To find the area between the curves, we need to determine where they intersect. We set the two equations equal to each other:

1. For \( x \geq 1 \):
\[ x - 1 = 1 \implies x = 2 \]

2. For \( x < 1 \):
\[ -x + 1 = 1 \implies x = 0 \]

Thus, the points of intersection are \( x = 0 \) and \( x = 2 \).

### Step 3: Set Up the Integral
The area \( A \) between the curves from \( x = 0 \) to \( x = 2 \) can be calculated using the integral:
\[ A = \int_{0}^{1} (1 - (-x + 1)) \, dx + \int_{1}^{2} (1 - (x - 1)) \, dx \]

### Step 4: Calculate the Area for Each Segment
1. **From \( x = 0 \) to \( x = 1 \)**:
\[ A_1 = \int_{0}^{1} (1 - (-x + 1)) \, dx = \int_{0}^{1} (x) \, dx \]
\[ A_1 = \left[ \frac{x^2}{2} \right]_{0}^{1} = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2} \]

2. **From \( x = 1 \) to \( x = 2 \)**:
\[ A_2 = \int_{1}^{2} (1 - (x - 1)) \, dx = \int_{1}^{2} (2 - x) \, dx \]
\[ A_2 = \left[ 2x - \frac{x^2}{2} \right]_{1}^{2} = \left( 2(2) - \frac{(2)^2}{2} \right) - \left( 2(1) - \frac{(1)^2}{2} \right) \]
\[ = (4 - 2) - (2 - \frac{1}{2}) = 2 - 1.5 = 0.5 \]

### Step 5: Total Area
The total area \( A \) is the sum of \( A_1 \) and \( A_2 \):
\[ A = A_1 + A_2 = \frac{1}{2} + \frac{1}{2} = 1 \]

### Final Answer
The area of the region bounded by \( y = |x - 1| \) and \( y = 1 \) is \( 1 \) square unit.