The area bounded by `y=|log_(e)|x||` and `y=0` is

Explanation

To find the area bounded by the curve \( y = |\log_e |x|| \) and the line \( y = 0 \), we will follow these steps:

### Step 1: Understand the function
The function \( y = |\log_e |x|| \) can be split into two cases:
- For \( x > 0 \), \( y = \log_e x \)
- For \( x < 0 \), \( y = \log_e (-x) \)

### Step 2: Determine the points of intersection with \( y = 0 \)
To find the area, we need to determine where the curve intersects the line \( y = 0 \):
- For \( x > 0 \): Set \( \log_e x = 0 \) which gives \( x = 1 \).
- For \( x < 0 \): Set \( \log_e (-x) = 0 \) which gives \( -x = 1 \) or \( x = -1 \).

Thus, the points of intersection are \( x = -1 \) and \( x = 1 \).

### Step 3: Set up the integral for the area
The area \( A \) can be calculated as the integral of the function from \( -1 \) to \( 1 \):
\[ A = \int_{-1}^{1} |\log_e |x|| \, dx \]

Since the function is symmetric about the y-axis, we can calculate the area from \( 0 \) to \( 1 \) and then double it:
\[ A = 2 \int_{0}^{1} \log_e x \, dx \]

### Step 4: Calculate the integral
To calculate the integral \( \int \log_e x \, dx \), we can use integration by parts:
Let \( u = \log_e x \) and \( dv = dx \).
Then, \( du = \frac{1}{x} dx \) and \( v = x \).

Using integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
\[ \int \log_e x \, dx = x \log_e x - \int x \cdot \frac{1}{x} \, dx = x \log_e x - x + C \]

### Step 5: Evaluate the definite integral
Now we evaluate from \( 0 \) to \( 1 \):
\[ \int_{0}^{1} \log_e x \, dx = \left[ x \log_e x - x \right]_{0}^{1} \]

At \( x = 1 \):
\[ 1 \cdot \log_e 1 - 1 = 0 - 1 = -1 \]

At \( x = 0 \):
As \( x \) approaches \( 0 \), \( x \log_e x \) approaches \( 0 \) (using L'Hôpital's Rule or recognizing the limit).

Thus:
\[ \int_{0}^{1} \log_e x \, dx = -1 - 0 = -1 \]

### Step 6: Calculate the total area
Now, substituting back into our area calculation:
\[ A = 2 \cdot (-1) = 2 \]

### Final Answer
The area bounded by the curve \( y = |\log_e |x|| \) and \( y = 0 \) is \( 2 \).

---