The area (in sq. units) of the region
`A=[(x,y):0leylexabsx+1" and "-1lexlex]` is

Options

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

Answer

To find the area of the region defined by the inequalities \( A = \{(x,y) : 0 \leq y \leq |x| + 1 \text{ and } -1 \leq x \leq 1\} \), we will follow these steps:

### Step 1: Understand the inequalities
The inequalities define a region in the Cartesian plane. The first inequality \( 0 \leq y \) indicates that we are considering the area above the x-axis. The second inequality \( y \leq |x| + 1 \) describes a V-shaped region that opens upwards, with the vertex at the point (0, 1).

### Step 2: Determine the boundaries
The boundaries of the region can be established as follows:
- The line \( y = |x| + 1 \) can be split into two equations:
- For \( x \geq 0 \): \( y = x + 1 \)
- For \( x < 0 \): \( y = -x + 1 \)
- The vertical lines \( x = -1 \) and \( x = 1 \) will also bound the region.

### Step 3: Sketch the region
To visualize the area, we can sketch the lines:
- The line \( y = x + 1 \) intersects the y-axis at (0, 1) and the line \( x = 1 \) at (1, 2).
- The line \( y = -x + 1 \) intersects the y-axis at (0, 1) and the line \( x = -1 \) at (-1, 2).
- The area of interest is bounded between \( x = -1 \) and \( x = 1 \), above the x-axis, and below the lines defined by \( y = |x| + 1 \).

### Step 4: Set up the integrals
To find the area, we can split the integral into two parts:
1. From \( x = -1 \) to \( x = 0 \): The upper boundary is \( y = -x + 1 \).
2. From \( x = 0 \) to \( x = 1 \): The upper boundary is \( y = x + 1 \).

The area \( A \) can be calculated as:
\[ A = \int_{-1}^{0} (-x + 1) \, dx + \int_{0}^{1} (x + 1) \, dx \]

### Step 5: Calculate the integrals
1. For the first integral:
\[ \int_{-1}^{0} (-x + 1) \, dx = \left[-\frac{x^2}{2} + x\right]_{-1}^{0} = \left[0 - 0\right] - \left[-\frac{1}{2} - 1\right] = \frac{3}{2} \]

2. For the second integral:
\[ \int_{0}^{1} (x + 1) \, dx = \left[\frac{x^2}{2} + x\right]_{0}^{1} = \left[\frac{1}{2} + 1\right] - [0] = \frac{3}{2} \]

### Step 6: Combine the areas
Adding the two areas together:
\[ A = \frac{3}{2} + \frac{3}{2} = 3 \]

### Conclusion
The area of the region \( A \) is \( 3 \) square units.