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

Options

A. `(5)/(2)` square units
B. `(1)/(2)` square units
C. 1 square units
D. `(3)/(2)` square units

Explanation

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 can follow these steps:

### Step 1: Identify the Intersection Points
First, we need to find the points where the curves intersect. The curves \(y = x\) and \(y = \frac{1}{x}\) intersect when:
\[ x = \frac{1}{x} \]
Multiplying both sides by \(x\) (assuming \(x \neq 0\)):
\[ x^2 = 1 \implies x = 1 \quad (\text{since we are only considering positive } x) \]
Thus, the intersection point is \((1, 1)\).

### Step 2: Set Up the Area Calculation
The area we want to find is bounded by the curves from \(x = 1\) to \(x = e\). We can divide the area into two parts:
1. The area of the triangle formed by the points \((1, 1)\), \((1, 0)\), and \((e, 0)\).
2. The area under the curve \(y = \frac{1}{x}\) from \(x = 1\) to \(x = e\).

### Step 3: Calculate the Area of the Triangle
The area \(A_T\) of the triangle can be calculated using the formula:
\[ A_T = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the base is \(e - 1\) and the height is \(1\):
\[ A_T = \frac{1}{2} \times (e - 1) \times 1 = \frac{e - 1}{2} \]

### Step 4: Calculate the Area Under the Curve
The area \(A_C\) under the curve \(y = \frac{1}{x}\) from \(x = 1\) to \(x = e\) is given by the integral:
\[ A_C = \int_{1}^{e} \frac{1}{x} \, dx \]
Calculating the integral:
\[ A_C = [\ln x]_{1}^{e} = \ln e - \ln 1 = 1 - 0 = 1 \]

### Step 5: Total Area Calculation
Now, we sum the areas of the triangle and the area under the curve:
\[ \text{Total Area} = A_T + A_C = \frac{e - 1}{2} + 1 \]
To combine these:
\[ \text{Total Area} = \frac{e - 1 + 2}{2} = \frac{e + 1}{2} \]

### Final Answer
Thus, the area of the region enclosed by the curves is:
\[ \frac{e + 1}{2} \]

---