Area bounded by the curve `y= log_(e)x, x=0, y le 0` and x-axis is:

Options

A. 1 sq. unit
B. 2 sq. unit
C. `(1)/(2)` sq. unit
D. None of these

Answer

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

### Step 1: Identify the points of intersection
The curve \( y = \ln x \) intersects the x-axis where \( y = 0 \). This occurs when:
\[ \ln x = 0 \implies x = 1 \]
Thus, the area we are interested in is bounded between \( x = 0 \) and \( x = 1 \).

### Step 2: Set up the integral for the area
The area \( A \) under the curve from \( x = 0 \) to \( x = 1 \) can be expressed as:
\[ A = \int_{0}^{1} \ln x \, dx \]

### Step 3: Evaluate the integral using integration by parts
To evaluate the integral \( \int \ln x \, dx \), we will use integration by parts. We choose:
- \( u = \ln x \) (thus \( du = \frac{1}{x} \, dx \))
- \( dv = dx \) (thus \( v = x \))

Using the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
we have:
\[ \int \ln x \, dx = x \ln x - \int x \cdot \frac{1}{x} \, dx = x \ln x - \int 1 \, dx \]
This simplifies to:
\[ x \ln x - x + C \]

### Step 4: Evaluate the definite integral
Now we evaluate the definite integral from \( 0 \) to \( 1 \):
\[ \int_{0}^{1} \ln x \, dx = \left[ x \ln x - x \right]_{0}^{1} \]

Calculating at the upper limit \( x = 1 \):
\[ 1 \ln 1 - 1 = 0 - 1 = -1 \]

Calculating at the lower limit \( x = 0 \):
As \( x \to 0 \), \( x \ln x \) approaches \( 0 \) (this can be shown using L'Hôpital's Rule). Thus:
\[ \lim_{x \to 0} (x \ln x) - 0 = 0 \]

Putting it all together:
\[ \int_{0}^{1} \ln x \, dx = (-1) - (0) = -1 \]

### Step 5: Find the area
Since the area is defined as a positive quantity, we take the absolute value:
\[ A = |-1| = 1 \]

### Final Answer
The area bounded by the curve \( y = \ln x \), the x-axis, and the line \( x = 0 \) is:
\[ \boxed{1} \text{ square unit} \]