The area of the region bounded by `x=1, x=2, y=logx and y=3^(x)` is

Options

A. `(9)/("log"3)+"log"(e )/(4) `
B. `(6)/("log"3)-2log2+1`
C. `(3)/("log"3)-2log2`
D. `(9)/("log"3)-2log2+1`

Answer

To find the area of the region bounded by the curves \( x = 1 \), \( x = 2 \), \( y = \log x \), and \( y = 3^x \), we will follow these steps:

### Step 1: Identify the curves and the region of integration
We have the curves:
- \( y = \log x \)
- \( y = 3^x \)
- Vertical lines \( x = 1 \) and \( x = 2 \)

### Step 2: Find the points of intersection
To find the area between the curves, we first need to determine where \( y = \log x \) and \( y = 3^x \) intersect. We set:
\[ \log x = 3^x \]
This equation is not straightforward to solve algebraically, so we will evaluate the functions at specific points:
- At \( x = 1 \): \( \log 1 = 0 \) and \( 3^1 = 3 \) (no intersection)
- At \( x = 2 \): \( \log 2 \approx 0.693 \) and \( 3^2 = 9 \) (no intersection)

Since we are only interested in the area between \( x = 1 \) and \( x = 2 \), we will evaluate the area directly without needing the intersection points.

### Step 3: Set up the integral for the area
The area \( A \) between the curves from \( x = 1 \) to \( x = 2 \) can be calculated using the integral:
\[ A = \int_{1}^{2} (3^x - \log x) \, dx \]

### Step 4: Calculate the integral
We will calculate the integral in two parts:
1. \( \int 3^x \, dx \)
2. \( \int \log x \, dx \)

**Integral of \( 3^x \)**:
\[ \int 3^x \, dx = \frac{3^x}{\log 3} + C \]

**Integral of \( \log x \)**:
Using integration by parts, let:
- \( u = \log x \) and \( dv = dx \)
- Then \( du = \frac{1}{x} dx \) and \( v = x \)

Applying integration by parts:
\[ \int \log x \, dx = x \log x - \int x \cdot \frac{1}{x} \, dx = x \log x - x + C \]

### Step 5: Evaluate the definite integrals
Now we evaluate the area:
\[ A = \left[ \frac{3^x}{\log 3} - (x \log x - x) \right]_{1}^{2} \]

Calculating at the bounds:
1. At \( x = 2 \):
\[ A(2) = \frac{3^2}{\log 3} - (2 \log 2 - 2) = \frac{9}{\log 3} - (2 \log 2 - 2) \]
2. At \( x = 1 \):
\[ A(1) = \frac{3^1}{\log 3} - (1 \log 1 - 1) = \frac{3}{\log 3} - (0 - 1) = \frac{3}{\log 3} + 1 \]

### Step 6: Combine the results
Now we combine the results:
\[ A = \left( \frac{9}{\log 3} - (2 \log 2 - 2) \right) - \left( \frac{3}{\log 3} + 1 \right) \]
\[ = \frac{9}{\log 3} - 2 \log 2 + 2 - \frac{3}{\log 3} - 1 \]
\[ = \frac{6}{\log 3} - 2 \log 2 + 1 \]

### Final Area
Thus, the area of the region bounded by the curves is:
\[ A = \frac{6}{\log 3} - 2 \log 2 + 1 \]