The area bounded by the x-axis, the curve y=f(x) and the lines x =1, x =b, is equal to `sqrt(b^2 +1) - sqrt2` for all `b gt 1`, then f(x) is

Options

A. `sqrt(x - 1)`
B. `sqrt(x - 1)`
C. `sqrt(x^2 - 1)`
D. `(x)/(sqrt(1+x^2))`

Answer

To find the function \( f(x) \) given that the area bounded by the x-axis, the curve \( y = f(x) \), and the lines \( x = 1 \) and \( x = b \) is equal to \( \sqrt{b^2 + 1} - \sqrt{2} \) for all \( b > 1 \), we can follow these steps:

### Step 1: Set up the integral for the area
The area \( A \) between the curve \( y = f(x) \), the x-axis, 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 also equal to:
\[ A = \sqrt{b^2 + 1} - \sqrt{2} \]
Thus, we have:
\[ \int_{1}^{b} f(x) \, dx = \sqrt{b^2 + 1} - \sqrt{2} \]

### Step 2: Differentiate both sides with respect to \( b \)
To find \( f(b) \), we differentiate both sides of the equation with respect to \( b \):
\[ \frac{d}{db} \left( \int_{1}^{b} f(x) \, dx \right) = \frac{d}{db} \left( \sqrt{b^2 + 1} - \sqrt{2} \right) \]
Using the Fundamental Theorem of Calculus, the left side simplifies to:
\[ f(b) \]
For the right side, we differentiate \( \sqrt{b^2 + 1} \):
\[ \frac{d}{db} \left( \sqrt{b^2 + 1} \right) = \frac{1}{2\sqrt{b^2 + 1}} \cdot 2b = \frac{b}{\sqrt{b^2 + 1}} \]
Thus, we have:
\[ f(b) = \frac{b}{\sqrt{b^2 + 1}} \]

### Step 3: Conclusion
Therefore, the function \( f(x) \) is:
\[ f(x) = \frac{x}{\sqrt{x^2 + 1}} \]