A continuous function `f(x)` satisfies the relation `f(x)=e^x+int_0^1 e^xf(t)dt` then `f(1)=`

Options

A. `f(0)lt0`
B. `f(x)` is a decreasing function
C. `f(x)` is increasing function
D. `int_(0)^(1)f(x)dxgt0`

Explanation

To solve the problem, we need to find the value of \( f(1) \) given the relation:

\[ f(x) = e^x + \int_0^1 e^x f(t) \, dt \]

### Step 1: Simplify the integral expression

We can factor out \( e^x \) from the integral since it does not depend on \( t \):

\[ f(x) = e^x + e^x \int_0^1 f(t) \, dt \]

Let \( k = \int_0^1 f(t) \, dt \). Then we can rewrite the equation as:

\[ f(x) = e^x + e^x k \]

### Step 2: Factor out \( e^x \)

Now, we can factor \( e^x \) out of the right-hand side:

\[ f(x) = e^x (1 + k) \]

### Step 3: Substitute \( f(t) \) into the integral

Now, we substitute \( f(t) \) back into the definition of \( k \):

\[ k = \int_0^1 f(t) \, dt = \int_0^1 e^t (1 + k) \, dt \]

### Step 4: Evaluate the integral

We can factor out \( (1 + k) \) from the integral:

\[ k = (1 + k) \int_0^1 e^t \, dt \]

The integral \( \int_0^1 e^t \, dt \) can be calculated as follows:

\[ \int_0^1 e^t \, dt = [e^t]_0^1 = e - 1 \]

Thus, we have:

\[ k = (1 + k)(e - 1) \]

### Step 5: Rearranging the equation

Expanding this gives:

\[ k = (e - 1) + k(e - 1) \]

Rearranging terms yields:

\[ k - k(e - 1) = e - 1 \]

Factoring out \( k \):

\[ k(1 - (e - 1)) = e - 1 \]

This simplifies to:

\[ k(2 - e) = e - 1 \]

### Step 6: Solve for \( k \)

Now, we can solve for \( k \):

\[ k = \frac{e - 1}{2 - e} \]

### Step 7: Substitute \( k \) back into \( f(x) \)

Now that we have \( k \), we can substitute it back into the expression for \( f(x) \):

\[ f(x) = e^x \left(1 + \frac{e - 1}{2 - e}\right) \]

This simplifies to:

\[ f(x) = e^x \left(\frac{(2 - e) + (e - 1)}{2 - e}\right) = e^x \left(\frac{1}{2 - e}\right) \]

### Step 8: Find \( f(1) \)

Finally, we substitute \( x = 1 \) to find \( f(1) \):

\[ f(1) = e^1 \cdot \frac{1}{2 - e} = \frac{e}{2 - e} \]

Thus, the value of \( f(1) \) is:

\[ \boxed{\frac{e}{2 - e}} \]