Let `f(x)` be a derivable function satisfying `f(x)=int_0^x e^t sin(x-t) dt` and `g(x)=f '' (x)-f(x)` Then the possible integers in the range of `g(x)` is_______

Explanation

To solve the problem, we need to analyze the function \( f(x) \) defined by the integral:

\[ f(x) = \int_0^x e^t \sin(x - t) \, dt \]

We also have the function \( g(x) \) defined as:

\[ g(x) = f''(x) - f(x) \]

### Step 1: Differentiate \( f(x) \)

Using Leibniz's rule for differentiation under the integral sign, we can find the first derivative of \( f(x) \):

\[ f'(x) = \frac{d}{dx} \left( \int_0^x e^t \sin(x - t) \, dt \right) = e^x \sin(0) + \int_0^x e^t \cos(x - t) \, dt \]

Since \( \sin(0) = 0 \), we have:

\[ f'(x) = \int_0^x e^t \cos(x - t) \, dt \]

### Step 2: Differentiate \( f'(x) \) to find \( f''(x) \)

Now, we differentiate \( f'(x) \) again:

\[ f''(x) = \frac{d}{dx} \left( \int_0^x e^t \cos(x - t) \, dt \right) = e^x \cos(0) + \int_0^x e^t (-\sin(x - t)) \, dt \]

This simplifies to:

\[ f''(x) = e^x + \int_0^x e^t (-\sin(x - t)) \, dt \]

### Step 3: Substitute \( f(x) \) and \( f''(x) \) into \( g(x) \)

Now we substitute \( f(x) \) and \( f''(x) \) into \( g(x) \):

\[ g(x) = f''(x) - f(x) = \left( e^x + \int_0^x e^t (-\sin(x - t)) \, dt \right) - \int_0^x e^t \sin(x - t) \, dt \]

This can be simplified to:

\[ g(x) = e^x + \int_0^x e^t (-\sin(x - t) - \sin(x - t)) \, dt = e^x - \int_0^x e^t \sin(x - t) \, dt \]

### Step 4: Analyze the range of \( g(x) \)

We know that \( g(x) = e^x - f(x) \). To find the range of \( g(x) \), we need to analyze the behavior of \( f(x) \).

Using the properties of the sine function, we can say:

\[ \sin(x - t) \text{ oscillates between } -1 \text{ and } 1 \]

Thus, \( f(x) \) will be bounded. More specifically, we can find the maximum and minimum values of \( g(x) \).

### Step 5: Determine the possible integers in the range of \( g(x) \)

The range of \( g(x) \) can be determined by evaluating the limits of \( g(x) \) as \( x \) approaches certain values. Given that \( e^x \) grows exponentially while \( f(x) \) is bounded, we can conclude that:

\[ g(x) \text{ will take values between } -\sqrt{2} \text{ and } \sqrt{2} \]

Thus, the possible integers in the range of \( g(x) \) are:

\[ -1, 0, 1 \]

### Final Answer

The possible integers in the range of \( g(x) \) are:

\[ \boxed{3} \]