The area (in sq. units) bounded by the curve `|y|=|ln|x||` and the coordinate axes is

Options

A. 2
B. 4
C. 6
D. 8

Answer

To find the area bounded by the curve \( |y| = |\ln|x|| \) and the coordinate axes, we will follow these steps:

### Step 1: Understand the curve
The equation \( |y| = |\ln|x|| \) represents two cases:
1. \( y = \ln|x| \) for \( x > 1 \) and \( x < -1 \)
2. \( y = -\ln|x| \) for \( 0 < x < 1 \) and \( -1 < x < 0 \)

### Step 2: Identify the bounded area
The area bounded by the curve and the coordinate axes occurs in the first and second quadrants. We will focus on the area in the first quadrant (where \( x > 0 \) and \( y > 0 \)) and then multiply it by 4 to account for symmetry.

### Step 3: Set up the integral
In the first quadrant, we consider the area under the curve \( y = \ln x \) from \( x = 1 \) to \( x = e \) (where \( \ln x = 1 \)):
\[ \text{Area} = \int_{1}^{e} \ln x \, dx \]

### Step 4: Calculate 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:
\[ \int u \, dv = uv - \int v \, du \]
Substituting:
\[ \int \ln x \, dx = x \ln x - \int x \cdot \frac{1}{x} \, dx = x \ln x - x + C \]

Now, we evaluate the definite integral:
\[ \int_{1}^{e} \ln x \, dx = \left[ x \ln x - x \right]_{1}^{e} \]

### Step 5: Evaluate the limits
Calculating at the upper limit \( x = e \):
\[ e \ln e - e = e \cdot 1 - e = e - e = 0 \]
Calculating at the lower limit \( x = 1 \):
\[ 1 \ln 1 - 1 = 1 \cdot 0 - 1 = -1 \]
Thus,
\[ \int_{1}^{e} \ln x \, dx = 0 - (-1) = 1 \]

### Step 6: Calculate the total area
Since we only calculated the area in the first quadrant, we multiply by 4 (for all quadrants):
\[ \text{Total Area} = 4 \times 1 = 4 \text{ square units} \]

### Final Answer
The area bounded by the curve \( |y| = |\ln|x|| \) and the coordinate axes is \( 4 \) square units.