Let function f(x) is defined in `[-2,2]` as `f(x)={{:({x}",",-2lexle-1),(|sgnx|",",-1lexle1),({-x}",",1ltxle2):}` where {x} and sgn x denotes fractional part and signum functions, respectively. Then the area bounded by the graph of f(x) and x-axis is

Options

A. 2 sq. units
B. 3 sq. units
C. 4 sq. units
D. 5 sq. units

Explanation

To find the area bounded by the graph of the function \( f(x) \) and the x-axis over the interval \([-2, 2]\), we will evaluate the function piecewise and calculate the area under each segment.

### Step 1: Define the function piecewise
The function \( f(x) \) is defined as:
- For \( -2 \leq x < -1 \): \( f(x) = x \)
- For \( -1 \leq x \leq 1 \): \( f(x) = |sgn(x)| = 1 \) (since \( sgn(x) \) is -1 for \( x < 0 \) and +1 for \( x > 0 \))
- For \( 1 < x \leq 2 \): \( f(x) = -x \)

### Step 2: Calculate the area for each segment
1. **From \( -2 \) to \( -1 \)**:
\[ \text{Area}_1 = \int_{-2}^{-1} f(x) \, dx = \int_{-2}^{-1} x \, dx \]
\[ = \left[ \frac{x^2}{2} \right]_{-2}^{-1} = \left( \frac{(-1)^2}{2} - \frac{(-2)^2}{2} \right) = \left( \frac{1}{2} - \frac{4}{2} \right) = \frac{1}{2} - 2 = -\frac{3}{2} \]
Since area cannot be negative, we take the absolute value:
\[ \text{Area}_1 = \frac{3}{2} \]

2. **From \( -1 \) to \( 1 \)**:
\[ \text{Area}_2 = \int_{-1}^{1} f(x) \, dx = \int_{-1}^{1} 1 \, dx = [x]_{-1}^{1} = 1 - (-1) = 2 \]

3. **From \( 1 \) to \( 2 \)**:
\[ \text{Area}_3 = \int_{1}^{2} f(x) \, dx = \int_{1}^{2} -x \, dx \]
\[ = \left[ -\frac{x^2}{2} \right]_{1}^{2} = \left( -\frac{(2)^2}{2} + \frac{(1)^2}{2} \right) = \left( -2 + \frac{1}{2} \right) = -2 + 0.5 = -1.5 \]
Again, taking the absolute value:
\[ \text{Area}_3 = \frac{3}{2} \]

### Step 3: Total area
Now, we sum the absolute areas calculated:
\[ \text{Total Area} = \text{Area}_1 + \text{Area}_2 + \text{Area}_3 = \frac{3}{2} + 2 + \frac{3}{2} = \frac{3}{2} + \frac{4}{2} + \frac{3}{2} = \frac{10}{2} = 5 \]

### Final Answer
The area bounded by the graph of \( f(x) \) and the x-axis is \( 5 \) square units.