Area bounded by the line y=x, curve `y=f(x),(f(x)gtx,AA xgt1)` and the lines x=1,x=t is `(t-sqrt(1+t^2)-(1+sqrt2))` for all `tgt1`. Find `f(x)`.

Answer

To find the function \( f(x) \) given the area bounded by the line \( y = x \), the curve \( y = f(x) \) (where \( f(x) > x \) for \( x > 1 \)), and the vertical lines \( x = 1 \) and \( x = t \), we can follow these steps:

### Step 1: Set up the area integral
The area \( A \) bounded by the curves can be expressed as:
\[ A = \int_{1}^{t} (f(x) - x) \, dx \]
Given that this area is equal to:
\[ A = t - \sqrt{1 + t^2} - (1 + \sqrt{2}) \]

### Step 2: Equate the two expressions for area
We equate the two expressions for the area:
\[ \int_{1}^{t} (f(x) - x) \, dx = t - \sqrt{1 + t^2} - (1 + \sqrt{2}) \]

### Step 3: Compute the integral of \( x \)
The integral of \( x \) from 1 to \( t \) is:
\[ \int_{1}^{t} x \, dx = \left[ \frac{x^2}{2} \right]_{1}^{t} = \frac{t^2}{2} - \frac{1}{2} \]

### Step 4: Substitute the integral into the area equation
Substituting this into the area equation gives:
\[ \int_{1}^{t} f(x) \, dx - \left( \frac{t^2}{2} - \frac{1}{2} \right) = t - \sqrt{1 + t^2} - (1 + \sqrt{2}) \]

### Step 5: Rearrange the equation
Rearranging the equation, we find:
\[ \int_{1}^{t} f(x) \, dx = t - \sqrt{1 + t^2} - (1 + \sqrt{2}) + \frac{t^2}{2} - \frac{1}{2} \]

### Step 6: Differentiate both sides
Differentiating both sides with respect to \( t \) using the Fundamental Theorem of Calculus gives:
\[ f(t) = 1 - \frac{t}{\sqrt{1 + t^2}} + t \]

### Step 7: Simplify the expression
Thus, we can simplify \( f(t) \) as follows:
\[ f(t) = 1 - \frac{t}{\sqrt{1 + t^2}} + t = 1 + t - \frac{t}{\sqrt{1 + t^2}} \]

### Step 8: Change variable from \( t \) to \( x \)
Finally, we replace \( t \) with \( x \) to express \( f(x) \):
\[ f(x) = 1 + x - \frac{x}{\sqrt{1 + x^2}} \]

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