If the area bounded by the x-axis, the curve `y=f(x), (f(x)gt0)" and the lines "x=1, x=b " is equal to "sqrt(b^(2)+1)-sqrt(2)" for all "bgt1,` then find f(x).

Explanation

To solve the problem, we need to find the function \( f(x) \) given that the area bounded by the x-axis, the curve \( y = f(x) \) (where \( f(x) > 0 \)), and the lines \( x = 1 \) and \( x = b \) is equal to \( \sqrt{b^2 + 1} - \sqrt{2} \) for all \( b > 1 \).

### Step-by-step Solution:

1. **Understand the Area Expression**:
The area \( A \) bounded by the x-axis, the curve \( y = f(x) \), and the vertical lines \( x = 1 \) and \( x = b \) can be expressed as:
\[ A = \int_{1}^{b} f(x) \, dx \]
According to the problem, this area is given by:
\[ A = \sqrt{b^2 + 1} - \sqrt{2} \]

2. **Set Up the Equation**:
We can set the two expressions for the area equal to each other:
\[ \int_{1}^{b} f(x) \, dx = \sqrt{b^2 + 1} - \sqrt{2} \]

3. **Differentiate Both Sides**:
To find \( f(x) \), we differentiate both sides with respect to \( b \):
\[ \frac{d}{db} \left( \int_{1}^{b} f(x) \, dx \right) = \frac{d}{db} \left( \sqrt{b^2 + 1} - \sqrt{2} \right) \]
By the Fundamental Theorem of Calculus, the left-hand side simplifies to \( f(b) \):
\[ f(b) = \frac{d}{db} \left( \sqrt{b^2 + 1} - \sqrt{2} \right) \]

4. **Differentiate the Right Side**:
Now we differentiate the right side:
\[ \frac{d}{db} \left( \sqrt{b^2 + 1} \right) = \frac{1}{2\sqrt{b^2 + 1}} \cdot 2b = \frac{b}{\sqrt{b^2 + 1}} \]
Therefore, we have:
\[ f(b) = \frac{b}{\sqrt{b^2 + 1}} \]

5. **Replace \( b \) with \( x \)**:
Since \( b \) is just a variable, we can replace it with \( x \):
\[ f(x) = \frac{x}{\sqrt{x^2 + 1}} \]

### Final Answer:
Thus, the function \( f(x) \) is:
\[ f(x) = \frac{x}{\sqrt{x^2 + 1}} \]