If `f'(x) = f(x)+ int _(0)^(1)f (x)` dx and given `f (0) =1,` then `int f (x) dx` is equal to :

Options

A. `(2)/(3-e) e ^(x) +((3-e)/(1-e))x+C`
B. `(2)/(3-e ) e ^(x) +((1-e)/(3-e))x +C`
C. `(3)/(2-e)e ^(x)+ ((1+e)/(3+e))x +C`
D. `(2)/(2-e) e ^(x) +((1-e)/(3+e))x +C`

Answer

To solve the problem, we start with the given differential equation:

\[ f'(x) = f(x) + \int_0^1 f(x) \, dx \]

and the initial condition:

\[ f(0) = 1. \]

### Step 1: Differentiate both sides with respect to \( x \)

Differentiating both sides gives us:

\[ f''(x) = f'(x) + 0. \]

The integral \( \int_0^1 f(x) \, dx \) is a constant with respect to \( x \), so its derivative is zero.

### Step 2: Substitute \( f'(x) \)

From the original equation, we can substitute \( f'(x) \):

\[ f''(x) = f(x) + \int_0^1 f(x) \, dx. \]

### Step 3: Set \( f'(x) = t \)

Let \( f'(x) = t \). Then \( f''(x) = \frac{dt}{dx} \).

### Step 4: Rewrite the equation

Using our substitution, we rewrite the equation as:

\[ \frac{dt}{dx} = t + \int_0^1 f(x) \, dx. \]

### Step 5: Solve the differential equation

This is a first-order linear differential equation. We can separate variables:

\[ \frac{dt}{t + C} = dx, \]

where \( C = \int_0^1 f(x) \, dx \).

### Step 6: Integrate both sides

Integrating both sides gives:

\[ \ln |t + C| = x + K, \]

where \( K \) is a constant of integration.

### Step 7: Solve for \( t \)

Exponentiating both sides yields:

\[ t + C = e^{x + K} \Rightarrow t = e^{x + K} - C. \]

### Step 8: Substitute back for \( f'(x) \)

Since \( t = f'(x) \), we have:

\[ f'(x) = e^{x + K} - C. \]

### Step 9: Integrate to find \( f(x) \)

Integrating \( f'(x) \):

\[ f(x) = \int (e^{x + K} - C) \, dx = e^{x + K} - Cx + D, \]

where \( D \) is another constant of integration.

### Step 10: Use the initial condition

Using the initial condition \( f(0) = 1 \):

\[ f(0) = e^{K} - 0 + D = 1 \Rightarrow D = 1 - e^{K}. \]

### Step 11: Final expression for \( f(x) \)

Thus, we can express \( f(x) \):

\[ f(x) = e^{x + K} - Cx + (1 - e^{K}). \]

### Step 12: Find \( \int f(x) \, dx \)

Now we need to find \( \int f(x) \, dx \):

\[ \int f(x) \, dx = \int (e^{x + K} - Cx + (1 - e^{K})) \, dx. \]

This can be computed as:

\[ \int e^{x + K} \, dx - C \int x \, dx + \int (1 - e^{K}) \, dx. \]

### Final Result

The final expression for \( \int f(x) \, dx \) will depend on the constants \( C \) and \( K \), which can be determined based on the context or additional conditions provided.