Sketch the curves and identify the region bounded by `x=1//2, x=2,y=log_(e)x` and
`y=2^(x)`. The area of this region is …..........

Answer

To find the area of the region bounded by the curves \( y = \log_e x \), \( y = 2^x \), and the vertical lines \( x = \frac{1}{2} \) and \( x = 2 \), we will follow these steps:

### Step 1: Sketch the Curves
1. **Plot the curves**:
- The curve \( y = \log_e x \) is defined for \( x > 0 \) and increases slowly.
- The curve \( y = 2^x \) is an exponential function that increases rapidly.
- Draw the vertical lines \( x = \frac{1}{2} \) and \( x = 2 \).

### Step 2: Determine the Points of Intersection
2. **Find the points of intersection**:
- Set \( \log_e x = 2^x \) to find the points where the curves intersect. This may require numerical methods or graphing to find approximate intersection points, but we will focus on the area between \( x = \frac{1}{2} \) and \( x = 2 \).

### Step 3: Set Up the Integral
3. **Set up the integral for the area**:
- The area \( A \) between the curves from \( x = \frac{1}{2} \) to \( x = 2 \) can be found using the integral:
\[ A = \int_{\frac{1}{2}}^{2} (2^x - \log_e x) \, dx \]
- Here, \( 2^x \) is the upper function and \( \log_e x \) is the lower function in the given interval.

### Step 4: Compute the Integral
4. **Calculate the integral**:
- We will compute:
\[ A = \int_{\frac{1}{2}}^{2} 2^x \, dx - \int_{\frac{1}{2}}^{2} \log_e x \, dx \]

- **First integral**:
\[ \int 2^x \, dx = \frac{2^x}{\log 2} \]
- Evaluating from \( \frac{1}{2} \) to \( 2 \):
\[ \left[ \frac{2^x}{\log 2} \right]_{\frac{1}{2}}^{2} = \frac{2^2}{\log 2} - \frac{2^{\frac{1}{2}}}{\log 2} = \frac{4 - \sqrt{2}}{\log 2} \]

- **Second integral**:
\[ \int \log_e x \, dx = x \log_e x - x \]
- Evaluating from \( \frac{1}{2} \) to \( 2 \):
\[ \left[ x \log_e x - x \right]_{\frac{1}{2}}^{2} = \left( 2 \log_e 2 - 2 \right) - \left( \frac{1}{2} \log_e \frac{1}{2} - \frac{1}{2} \right) \]
- Simplifying:
\[ = 2 \log_e 2 - 2 - \left( -\frac{1}{2} \log_e 2 + \frac{1}{2} \right) = 2 \log_e 2 - 2 + \frac{1}{2} \log_e 2 - \frac{1}{2} \]
\[ = \frac{5}{2} \log_e 2 - \frac{5}{2} \]

### Step 5: Combine Results
5. **Combine the results**:
- The area \( A \) becomes:
\[ A = \left( \frac{4 - \sqrt{2}}{\log 2} \right) - \left( \frac{5}{2} \log_e 2 - \frac{5}{2} \right) \]

### Final Answer
The area of the region bounded by the curves is:
\[ A = \frac{4 - \sqrt{2}}{\log 2} - \left( \frac{5}{2} \log_e 2 - \frac{5}{2} \right) \]