Find the area bounded by `y=-(log)_e x , y=(log)_e x ,y=(log)_e(-x),a n dy=-(log)_e(-x)dot`

Explanation

To find the area bounded by the curves \( y = -\log_e x \), \( y = \log_e x \), \( y = \log_e(-x) \), and \( y = -\log_e(-x) \), we can follow these steps:

### Step 1: Understand the curves
The curves \( y = \log_e x \) and \( y = -\log_e x \) are symmetric about the x-axis. Similarly, \( y = \log_e(-x) \) and \( y = -\log_e(-x) \) are also symmetric about the x-axis.

### Step 2: Identify the points of intersection
The curves intersect at \( x = 1 \) and \( x = -1 \). This is because:
- For \( y = \log_e x \) and \( y = -\log_e x \), they intersect at \( x = 1 \) where both equal zero.
- For \( y = \log_e(-x) \) and \( y = -\log_e(-x) \), they intersect at \( x = -1 \) where both also equal zero.

### Step 3: Set up the area calculation
The area bounded by these curves can be calculated by finding the area between \( x = 0 \) and \( x = 1 \) for the first pair of curves and multiplying it by 4 (due to symmetry).

The area \( A \) can be expressed as:
\[ A = 4 \int_0^1 \left( \log_e x - (-\log_e x) \right) \, dx = 4 \int_0^1 2\log_e x \, dx = 8 \int_0^1 \log_e x \, dx \]

### Step 4: Integrate using integration by parts
To integrate \( \log_e x \), we use integration by parts:
Let \( u = \log_e x \) and \( dv = dx \).
Then, \( du = \frac{1}{x}dx \) and \( v = x \).

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

### Step 5: Evaluate the definite integral
Now, we evaluate the integral 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 \( x = 0 \), we need to evaluate the limit:
\[ \lim_{x \to 0^+} (x \log_e x - x) = \lim_{x \to 0^+} (x \log_e x) - \lim_{x \to 0^+} x \]
The term \( x \log_e x \) approaches 0 as \( x \) approaches 0. Thus:
\[ \int_0^1 \log_e x \, dx = -1 - 0 = -1 \]

### Step 6: Calculate the total area
Now substituting back into the area formula:
\[ A = 8 \int_0^1 \log_e x \, dx = 8 \cdot (-1) = -8 \]
Since area cannot be negative, we take the absolute value:
\[ A = 8 \text{ square units} \]

### Final Answer
The area bounded by the curves is \( 8 \) square units.

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