The area bounded by the curves y=lnx, y=ln|x|, y=|lnx| and y=|ln||x| is

Options

A. 5
B. 2
C. 4
D. none of these

Explanation

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

### Step 1: Understand the curves
1. **Curve \( y = \ln x \)**: This curve is defined for \( x > 0 \) and approaches negative infinity as \( x \) approaches 0. It passes through the point \( (1, 0) \).
2. **Curve \( y = \ln |x| \)**: This curve is defined for \( x \neq 0 \) and is symmetric about the y-axis. For \( x < 0 \), it behaves like \( y = \ln (-x) \).
3. **Curve \( y = |\ln x| \)**: This curve is defined for \( x > 0 \) and reflects the negative part of \( y = \ln x \) above the x-axis.
4. **Curve \( y = |\ln |x|| \)**: This curve is defined for \( x \neq 0 \) and reflects both the left and right parts of \( y = \ln |x| \) above the x-axis.

### Step 2: Sketch the curves
- Draw the curves on a coordinate system.
- The curve \( y = \ln x \) will be in the first quadrant, while \( y = \ln |x| \) will extend into the second quadrant as a reflection of \( y = \ln x \).
- The curves \( y = |\ln x| \) and \( y = |\ln |x|| \) will be reflections of the respective logarithmic curves in the first and second quadrants.

### Step 3: Identify the area of interest
- The area we need to calculate is between \( y = |\ln x| \) and \( y = 0 \) from \( x = 0 \) to \( x = 1 \), and the same area will be mirrored in the second quadrant.

### Step 4: Set up the integral for the area
- The area in the first quadrant can be calculated using the integral:
\[ \text{Area} = \int_{0}^{1} |\ln x| \, dx \]
Since \( \ln x < 0 \) for \( 0 < x < 1 \), we have \( |\ln x| = -\ln x \). Thus, the integral becomes:
\[ \text{Area} = \int_{0}^{1} -\ln x \, dx \]

### Step 5: Solve the integral
- To solve the integral \( \int -\ln x \, dx \), we can use integration by parts:
Let \( u = \ln x \) and \( dv = dx \).
Then, \( du = \frac{1}{x} dx \) and \( v = x \).

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

### Step 6: Evaluate the definite integral
- Now we evaluate from 0 to 1:
\[ \left[-x \ln x + x\right]_{0}^{1} \]
- At \( x = 1 \):
\[ -1 \cdot \ln(1) + 1 = 0 + 1 = 1 \]
- At \( x = 0 \):
\[ \lim_{x \to 0} (-x \ln x + x) = 0 \quad (\text{since } -x \ln x \to 0 \text{ as } x \to 0) \]
Thus, the area in the first quadrant is 1.

### Step 7: Calculate total area
- Since the area in the second quadrant is symmetric to that in the first quadrant, the total area is:
\[ \text{Total Area} = 2 \cdot 1 = 2 \]

### Final Answer
The area bounded by the curves is \( \boxed{2} \).