Area bounded by the curve `f(x) ={log_e|x|,|x| >= 1/e |x|-1-1/e, |x|< 1/2` and x axis is

Options

A. `(1)/(e^(2)) + 2 - (2)/(e )`
B. `(1)/(e^(2)) + 2 + (2)/(E )`
C. `(1)/(e^(2)) + (2)/(e )`
D. `(1 - (1)/(e ))^(2) + 1`

Answer

To find the area bounded by the curve \( f(x) \) defined as:

\[ f(x) = \begin{cases} \log_e |x| & \text{if } |x| \geq \frac{1}{e} \\ |x| - 1 - \frac{1}{e} & \text{if } |x| < \frac{1}{2} \end{cases} \]

and the x-axis, we will break down the problem into steps.

### Step 1: Identify the intervals and functions
We need to analyze the function \( f(x) \) in different intervals:
1. For \( |x| \geq \frac{1}{e} \), \( f(x) = \log_e |x| \).
2. For \( |x| < \frac{1}{2} \), \( f(x) = |x| - 1 - \frac{1}{e} \).

### Step 2: Determine the points of intersection with the x-axis
To find the area, we need to determine where \( f(x) = 0 \).

1. **For \( f(x) = \log_e |x| \):**
- Set \( \log_e |x| = 0 \) which gives \( |x| = 1 \).
- Thus, points of intersection are \( x = 1 \) and \( x = -1 \).

2. **For \( f(x) = |x| - 1 - \frac{1}{e} \):**
- Set \( |x| - 1 - \frac{1}{e} = 0 \) which gives \( |x| = 1 + \frac{1}{e} \).
- Thus, points of intersection are \( x = 1 + \frac{1}{e} \) and \( x = -1 - \frac{1}{e} \).

### Step 3: Set up the area calculation
The area \( A \) can be computed as the sum of the areas under the curve from \( -1 - \frac{1}{e} \) to \( -1 \) and from \( 1 \) to \( 1 + \frac{1}{e} \).

\[ A = \int_{-1 - \frac{1}{e}}^{-1} f(x) \, dx + \int_{1}^{1 + \frac{1}{e}} f(x) \, dx \]

### Step 4: Calculate the area for each segment
1. **For \( x \) in \( [-1 - \frac{1}{e}, -1] \)**:
- Here, \( f(x) = |x| - 1 - \frac{1}{e} = -x - 1 - \frac{1}{e} \).
- The area is:
\[ A_1 = \int_{-1 - \frac{1}{e}}^{-1} (-x - 1 - \frac{1}{e}) \, dx \]

2. **For \( x \) in \( [1, 1 + \frac{1}{e}] \)**:
- Here, \( f(x) = \log_e x \).
- The area is:
\[ A_2 = \int_{1}^{1 + \frac{1}{e}} \log_e x \, dx \]

### Step 5: Evaluate the integrals
1. **For \( A_1 \)**:
\[ A_1 = \int_{-1 - \frac{1}{e}}^{-1} (-x - 1 - \frac{1}{e}) \, dx \]
This evaluates to:
\[ A_1 = \left[-\frac{x^2}{2} - x - \frac{x}{e}\right]_{-1 - \frac{1}{e}}^{-1} \]

2. **For \( A_2 \)**:
\[ A_2 = \int_{1}^{1 + \frac{1}{e}} \log_e x \, dx \]
This evaluates to:
\[ A_2 = \left[x \log_e x - x\right]_{1}^{1 + \frac{1}{e}} \]

### Step 6: Combine the areas
Finally, combine \( A_1 \) and \( A_2 \) to find the total area \( A \).