If `A_(1)` is the area enclosed by the curve `xy=1,` x-axis and the ordinates `x=1,x=2,and A_(2)` is the area enclosed by the curve `xy=1,` x-axis and the ordinates `x=2, x=4,` then

Options

A. `A_(1)=2A_(2)`
B. `A_(2)=2A_(1)`
C. `A_(2)=3A_(1)`
D. `A_(1)=A_(2)`

Answer

To solve the problem, we need to find the areas \( A_1 \) and \( A_2 \) as defined by the given curves and limits.

### Step-by-Step Solution:

1. **Understanding the Curve**:
The curve given is \( xy = 1 \). We can express this in terms of \( y \):
\[ y = \frac{1}{x} \]

2. **Finding Area \( A_1 \)**:
The area \( A_1 \) is enclosed by the curve \( y = \frac{1}{x} \), the x-axis, and the ordinates \( x = 1 \) and \( x = 2 \). We can calculate this area using integration:
\[ A_1 = \int_{1}^{2} \frac{1}{x} \, dx \]

3. **Calculating the Integral for \( A_1 \)**:
The integral of \( \frac{1}{x} \) is \( \log x \):
\[ A_1 = \left[ \log x \right]_{1}^{2} = \log 2 - \log 1 \]
Since \( \log 1 = 0 \):
\[ A_1 = \log 2 \]

4. **Finding Area \( A_2 \)**:
The area \( A_2 \) is enclosed by the same curve \( y = \frac{1}{x} \), the x-axis, and the ordinates \( x = 2 \) and \( x = 4 \). We calculate this area similarly:
\[ A_2 = \int_{2}^{4} \frac{1}{x} \, dx \]

5. **Calculating the Integral for \( A_2 \)**:
Again, using the integral of \( \frac{1}{x} \):
\[ A_2 = \left[ \log x \right]_{2}^{4} = \log 4 - \log 2 \]
We can simplify \( \log 4 \) as \( \log (2^2) = 2 \log 2 \):
\[ A_2 = 2 \log 2 - \log 2 = \log 2 \]

6. **Conclusion**:
We find that:
\[ A_1 = \log 2 \quad \text{and} \quad A_2 = \log 2 \]
Therefore, we conclude that:
\[ A_1 = A_2 \]

### Final Answer:
Thus, \( A_1 \) is equal to \( A_2 \).