The area of the region bounded by `y = |x| and y =1 - |x|` is

Options

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

Answer

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

### Step 1: Identify the curves
The equations given are:
1. \( y = |x| \)
2. \( y = 1 - |x| \)

### Step 2: Sketch the graphs
- The graph of \( y = |x| \) is a V-shaped graph that opens upwards with its vertex at the origin (0,0).
- The graph of \( y = 1 - |x| \) is an inverted V-shaped graph that opens downwards with its vertex at (0,1).

### Step 3: Find the points of intersection
To find the points where the two curves intersect, we set them equal to each other:
\[ |x| = 1 - |x| \]

This gives us two cases to consider:
1. **Case 1**: \( x \geq 0 \) (where \( |x| = x \)):
\[ x = 1 - x \implies 2x = 1 \implies x = \frac{1}{2} \]
Substituting \( x = \frac{1}{2} \) into \( y = |x| \) gives \( y = \frac{1}{2} \). So, one intersection point is \( \left(\frac{1}{2}, \frac{1}{2}\right) \).

2. **Case 2**: \( x < 0 \) (where \( |x| = -x \)):
\[ -x = 1 + x \implies -2x = 1 \implies x = -\frac{1}{2} \]
Substituting \( x = -\frac{1}{2} \) into \( y = |x| \) gives \( y = \frac{1}{2} \). So, the other intersection point is \( \left(-\frac{1}{2}, \frac{1}{2}\right) \).

### Step 4: Set up the integrals for area calculation
The area between the curves from \( x = -\frac{1}{2} \) to \( x = \frac{1}{2} \) can be calculated using the integral:
\[ \text{Area} = \int_{-\frac{1}{2}}^{\frac{1}{2}} \left((1 - |x|) - |x|\right) \, dx \]

### Step 5: Evaluate the integral
We can split the integral into two parts:
1. From \( -\frac{1}{2} \) to \( 0 \):
\[ \int_{-\frac{1}{2}}^{0} \left(1 + x - (-x)\right) \, dx = \int_{-\frac{1}{2}}^{0} (1 + 2x) \, dx \]
2. From \( 0 \) to \( \frac{1}{2} \):
\[ \int_{0}^{\frac{1}{2}} \left(1 - x - x\right) \, dx = \int_{0}^{\frac{1}{2}} (1 - 2x) \, dx \]

Calculating the first integral:
\[ \int (1 + 2x) \, dx = x + x^2 \Big|_{-\frac{1}{2}}^{0} = \left(0 + 0\right) - \left(-\frac{1}{2} + \frac{1}{4}\right) = \frac{1}{4} \]

Calculating the second integral:
\[ \int (1 - 2x) \, dx = x - x^2 \Big|_{0}^{\frac{1}{2}} = \left(\frac{1}{2} - \frac{1}{4}\right) - (0 - 0) = \frac{1}{4} \]

### Step 6: Add the areas from both integrals
\[ \text{Total Area} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \]

### Final Answer
The area of the region bounded by \( y = |x| \) and \( y = 1 - |x| \) is \( \frac{1}{2} \) square units.

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