The area of the region enclosed by the curves `y=x, x=e,y=(1)/(x)` and the positive x-axis is

Options

A. 1 sq unit
B. `3/2` sq units
C. `5/2`sq units
D. `1/2` sq unit

Answer

To find the area of the region enclosed by the curves \( y = x \), \( x = e \), \( y = \frac{1}{x} \), and the positive x-axis, we will follow these steps:

### Step 1: Identify the curves and their intersections
The curves involved are:
1. \( y = x \) (a straight line)
2. \( x = e \) (a vertical line)
3. \( y = \frac{1}{x} \) (a hyperbola)

We need to find the points of intersection between these curves.

### Step 2: Find the intersection points
- **Intersection of \( y = x \) and \( y = \frac{1}{x} \)**:
\[ x = \frac{1}{x} \implies x^2 = 1 \implies x = 1 \quad (\text{since } x \text{ is positive}) \]
Thus, the point of intersection is \( (1, 1) \).

- **Intersection of \( y = \frac{1}{x} \) and \( x = e \)**:
\[ y = \frac{1}{e} \]
Thus, the point of intersection is \( \left(e, \frac{1}{e}\right) \).

### Step 3: Set up the area integral
The area can be computed using vertical strips. The area \( A \) can be expressed as:
\[ A = \int_0^1 y \, dx + \int_1^e y \, dx \]
Where:
- From \( x = 0 \) to \( x = 1 \), \( y = x \).
- From \( x = 1 \) to \( x = e \), \( y = \frac{1}{x} \).

### Step 4: Calculate the integrals
1. **First integral** from \( 0 \) to \( 1 \):
\[ \int_0^1 x \, dx = \left[ \frac{x^2}{2} \right]_0^1 = \frac{1^2}{2} - 0 = \frac{1}{2} \]

2. **Second integral** from \( 1 \) to \( e \):
\[ \int_1^e \frac{1}{x} \, dx = \left[ \ln x \right]_1^e = \ln e - \ln 1 = 1 - 0 = 1 \]

### Step 5: Combine the results
Now, we can combine the results of the two integrals:
\[ A = \frac{1}{2} + 1 = \frac{3}{2} \]

### Step 6: Conclusion
Thus, the area of the region enclosed by the curves is:
\[ \text{Area} = \frac{3}{2} \text{ square units} \]