The area of figure bounded by the curves `y=a^x(a gt 1) and y=b^-x (b gt 1)` and the straight line `x = 1` is:

Options

A. (a) `(1)/(log a)""(a-1) + (1)/(log b) ((1)/(b) - 1)`
B. (b) `log a (a - 1) + log ((1)/(b) - 1)`
C. (c) `(1)/(log a) (a-1)+ (1)/(log b) (b-1)`
D. (d) `log (a-1) + log (b-1)`

Answer

To find the area bounded by the curves \( y = a^x \) (where \( a > 1 \)), \( y = b^{-x} \) (where \( b > 1 \)), and the vertical line \( x = 1 \), we can follow these steps:

### Step 1: Identify Points of Intersection
We need to find the points where the curves intersect.

1. **Intersection of \( y = a^x \) and \( x = 1 \)**:
- At \( x = 1 \), \( y = a^1 = a \).
- So, the point is \( (1, a) \).

2. **Intersection of \( y = b^{-x} \) and \( x = 1 \)**:
- At \( x = 1 \), \( y = b^{-1} = \frac{1}{b} \).
- So, the point is \( (1, \frac{1}{b}) \).

3. **Intersection of \( y = a^x \) and \( y = b^{-x} \)**:
- Set \( a^x = b^{-x} \).
- Taking logarithm on both sides gives us \( x \log a + x \log b = 0 \).
- This simplifies to \( x(\log a + \log b) = 0 \), which gives \( x = 0 \) as the only solution.
- At \( x = 0 \), \( y = a^0 = 1 \) (and also \( y = b^0 = 1 \)).
- So, the point is \( (0, 1) \).

### Step 2: Set Up the Area Calculation
The area \( A \) we want to find is between the curves from \( x = 0 \) to \( x = 1 \).

The area under the curve \( y = a^x \) from \( x = 0 \) to \( x = 1 \):
\[ \text{Area under } y = a^x = \int_0^1 a^x \, dx \]

The area under the curve \( y = b^{-x} \) from \( x = 0 \) to \( x = 1 \):
\[ \text{Area under } y = b^{-x} = \int_0^1 b^{-x} \, dx \]

### Step 3: Calculate the Integrals
1. **Integral of \( a^x \)**:
\[ \int a^x \, dx = \frac{a^x}{\log a} + C \]
Evaluating from 0 to 1:
\[ \left[ \frac{a^x}{\log a} \right]_0^1 = \frac{a^1}{\log a} - \frac{a^0}{\log a} = \frac{a}{\log a} - \frac{1}{\log a} = \frac{a - 1}{\log a} \]

2. **Integral of \( b^{-x} \)**:
\[ \int b^{-x} \, dx = -\frac{b^{-x}}{\log b} + C \]
Evaluating from 0 to 1:
\[ \left[ -\frac{b^{-x}}{\log b} \right]_0^1 = -\frac{b^{-1}}{\log b} + \frac{b^0}{\log b} = \frac{1}{\log b} - \frac{1}{b \log b} = \frac{1 - \frac{1}{b}}{\log b} = \frac{b - 1}{b \log b} \]

### Step 4: Calculate the Required Area
The required area \( A \) is:
\[ A = \text{Area under } a^x - \text{Area under } b^{-x} = \left( \frac{a - 1}{\log a} \right) - \left( \frac{b - 1}{b \log b} \right) \]

### Final Expression
Thus, the area bounded by the curves is:
\[ A = \frac{a - 1}{\log a} + \frac{1 - \frac{1}{b}}{\log b} = \frac{a - 1}{\log a} + \frac{1 - b^{-1}}{\log b} \]

### Conclusion
The area of the figure bounded by the curves is:
\[ \frac{1}{\log a}(a - 1) + \frac{1}{\log b}(1 - \frac{1}{b}) \]