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 will follow these steps:

### Step 1: Identify the curves and their intersections
The curves we are dealing with are:
1. \( y = \ln x \)
2. \( y = \frac{x}{e} \)

We need to find the points where these two curves intersect. To do this, we set \( \ln x = \frac{x}{e} \).

### Step 2: Solve for the intersection points
To find the intersection, we can rewrite the equation:
\[ \ln x = \frac{x}{e} \]
This is a transcendental equation, and we can solve it by substituting \( x = e \) (since \( \ln e = 1 \) and \( \frac{e}{e} = 1 \)):
\[ \ln e = 1 \quad \text{and} \quad \frac{e}{e} = 1 \]
Thus, the curves intersect at the point \( (e, 1) \).

### Step 3: Set up the area integral
The area we want to find is bounded by the y-axis (which is \( x = 0 \)), the curve \( y = \ln x \), and the line \( y = \frac{x}{e} \). We will integrate with respect to \( y \) from \( y = 0 \) to \( y = 1 \).

The area \( A \) can be expressed as:
\[ A = \int_{0}^{1} (e^y - ey) \, dy \]
where \( e^y \) is the right curve (from \( y = \ln x \)) and \( ey \) is the left curve (from \( y = \frac{x}{e} \)).

### Step 4: Calculate the integral
Now we compute the integral:
\[ A = \int_{0}^{1} (e^y - ey) \, dy \]

Calculating the integral:
\[ = \left[ e^y - \frac{ey^2}{2} \right]_{0}^{1} \]

Evaluating at the limits:
1. At \( y = 1 \):
\[ e^1 - \frac{e(1^2)}{2} = e - \frac{e}{2} = \frac{e}{2} \]
2. At \( y = 0 \):
\[ e^0 - \frac{e(0^2)}{2} = 1 - 0 = 1 \]

Thus, the area \( A \) becomes:
\[ A = \left( \frac{e}{2} - 1 \right) \]

### Step 5: Calculate the area below the x-axis
Next, we need to find the area \( A_2 \) below the x-axis from \( y = -\infty \) to \( y = 0 \):
\[ A_2 = \int_{-\infty}^{0} e^y \, dy = \left[ e^y \right]_{-\infty}^{0} = 1 - 0 = 1 \]

### Step 6: Total area
The total area bounded by the curves is:
\[ \text{Total Area} = A + A_2 = \left( \frac{e}{2} - 1 \right) + 1 = \frac{e}{2} \]

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