The area (in sq. units) bounded by `y=lnx, y =(x)/(e )` and y - axis is equal to

Options

A. `(e)/(2)-1`
B. `(e)/(2)`
C. `(5e)/(2)`
D. `(3e)/(2)-1`

Answer

To find the area bounded by the curves \( y = \ln x \), \( y = \frac{x}{e} \), and the y-axis, we can follow these steps:

### Step 1: Identify the Points of Intersection
We need to find the points where the two curves intersect. This is done by setting \( \ln x = \frac{x}{e} \).

1. **Set the equations equal**:
\[ \ln x = \frac{x}{e} \]

2. **Rearrange the equation**:
\[ e \ln x = x \]

3. **Let \( x = e^t \)** (where \( t = \ln x \)):
\[ e t = e^t \implies t = e^{t-1} \]

4. **This equation can be solved graphically or numerically**. The intersection occurs at \( x = e \).

### Step 2: Set Up the Integral for Area
The area \( A \) between the curves from \( x = 0 \) to \( x = e \) can be found using the integral:
\[ A = \int_0^e \left( \frac{x}{e} - \ln x \right) dx \]

### Step 3: Calculate the Integral
We can split the integral into two parts:
\[ A = \int_0^e \frac{x}{e} \, dx - \int_0^e \ln x \, dx \]

1. **Calculate the first integral**:
\[ \int_0^e \frac{x}{e} \, dx = \frac{1}{e} \int_0^e x \, dx = \frac{1}{e} \left[ \frac{x^2}{2} \right]_0^e = \frac{1}{e} \cdot \frac{e^2}{2} = \frac{e}{2} \]

2. **Calculate the second integral**:
The integral of \( \ln x \) is given by:
\[ \int \ln x \, dx = x \ln x - x \]
Thus,
\[ \int_0^e \ln x \, dx = \left[ x \ln x - x \right]_0^e \]
Evaluating at the bounds:
- At \( x = e \):
\[ e \ln e - e = e - e = 0 \]
- At \( x = 0 \):
\[ \lim_{x \to 0} (x \ln x - x) = 0 \quad \text{(using L'Hôpital's Rule)} \]
Therefore, the integral evaluates to:
\[ \int_0^e \ln x \, dx = 0 - 0 = 0 \]

### Step 4: Combine the Results
Now we can substitute back into the area formula:
\[ A = \frac{e}{2} - 0 = \frac{e}{2} \]

### Final Answer
The area bounded by the curves \( y = \ln x \), \( y = \frac{x}{e} \), and the y-axis is:
\[ \boxed{\frac{e}{2}} \text{ square units} \]