If `int_(0)^(x)f(t)dt=e^(x)-ae^(2x)int_(0)^(1)f(t)e^(-t)dt`, then

Options

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

Explanation

To solve the given equation

\[ \int_{0}^{x} f(t) dt = e^{x} - a e^{2x} \int_{0}^{1} f(t) e^{-t} dt, \]

we can follow these steps:

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

We start by differentiating both sides of the equation with respect to \(x\).

Using the Fundamental Theorem of Calculus on the left side, we have:

\[ \frac{d}{dx} \left( \int_{0}^{x} f(t) dt \right) = f(x). \]

For the right side, we differentiate:

\[ \frac{d}{dx} \left( e^{x} - a e^{2x} \int_{0}^{1} f(t) e^{-t} dt \right) = e^{x} - 2a e^{2x} \int_{0}^{1} f(t) e^{-t} dt. \]

Thus, we have:

\[ f(x) = e^{x} - 2a e^{2x} \int_{0}^{1} f(t) e^{-t} dt. \]

### Step 2: Rearranging the equation

Now, we can rearrange the equation to isolate \(f(x)\):

\[ f(x) = e^{x} - 2a e^{2x} C, \]

where \(C = \int_{0}^{1} f(t) e^{-t} dt\).

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

Next, we substitute \(f(x)\) back into the original integral to find \(C\):

\[ C = \int_{0}^{1} \left( e^{t} - 2a e^{2t} C \right) e^{-t} dt. \]

This simplifies to:

\[ C = \int_{0}^{1} e^{t} e^{-t} dt - 2a C \int_{0}^{1} e^{2t} e^{-t} dt. \]

### Step 4: Calculate the integrals

Now we calculate the integrals:

1. \(\int_{0}^{1} e^{t} e^{-t} dt = \int_{0}^{1} 1 dt = 1\).
2. \(\int_{0}^{1} e^{2t} e^{-t} dt = \int_{0}^{1} e^{t} dt = [e^{t}]_{0}^{1} = e - 1\).

Substituting these results into the equation for \(C\):

\[ C = 1 - 2a C (e - 1). \]

### Step 5: Solve for \(C\)

Rearranging gives:

\[ C + 2a C (e - 1) = 1, \]

which can be factored as:

\[ C(1 + 2a(e - 1)) = 1. \]

Thus, we find:

\[ C = \frac{1}{1 + 2a(e - 1)}. \]

### Step 6: Substitute \(C\) back to find \(f(x)\)

Now, substituting \(C\) back into our expression for \(f(x)\):

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

### Final Expression for \(f(x)\)

Thus, the final expression for \(f(x)\) is:

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