Let f be a continuous function satisfying the equation `int_(0)^(x)f(t)dt+int_(0)^(x)tf(x-t)dt=e^(-x)-1`, then find the value of `e^(9)f(9)` is equal to…………………..

Explanation

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

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

### Step 1: Use the property of integrals
We can use the property of integrals which states that:

\[ \int_{0}^{a} f(t) dt = \int_{0}^{a} f(a - t) dt \]

Applying this property to the second integral, we rewrite it as:

\[ \int_{0}^{x} t f(x-t) dt = \int_{0}^{x} (x - u) f(u) du \]

where \( u = x - t \). Thus, we have:

\[ \int_{0}^{x} t f(x-t) dt = \int_{0}^{x} x f(u) du - \int_{0}^{x} u f(u) du \]

### Step 2: Substitute back into the original equation
Substituting this back into the original equation gives:

\[ \int_{0}^{x} f(t) dt + \left( x \int_{0}^{x} f(u) du - \int_{0}^{x} u f(u) du \right) = e^{-x} - 1 \]

### Step 3: Combine the integrals
Combining the terms, we get:

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

This simplifies to:

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

### Step 4: Differentiate both sides
Now, we differentiate both sides with respect to \( x \):

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

Using the product rule and the Fundamental Theorem of Calculus, we have:

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

This simplifies to:

\[ \int_{0}^{x} f(t) dt + f(x) = -e^{-x} \]

### Step 5: Isolate \( f(x) \)
Rearranging gives:

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

### Step 6: Differentiate again
Differentiating both sides again:

\[ f'(x) = e^{-x} - f(x) \]

### Step 7: Solve the differential equation
This is a first-order linear differential equation. The general solution can be found by solving:

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

The integrating factor is \( e^{x} \):

\[ e^{x} f'(x) + e^{x} f(x) = 1 \]

Integrating both sides:

\[ e^{x} f(x) = x + C \]

Thus,

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

### Step 8: Find the constant \( C \)
To find \( C \), we can use the initial condition. Setting \( x = 0 \):

\[ f(0) = e^{0}(0 + C) \Rightarrow f(0) = C \]

From the original equation, we can find \( f(0) \) by substituting \( x = 0 \):

\[ 0 + 0 = e^{0} - 1 \Rightarrow 0 = 0 \]

This does not provide information about \( C \). However, we can assume \( f(0) = -1 \) from the context of the problem, leading to \( C = -1 \).

### Step 9: Final expression for \( f(x) \)
Thus, we have:

\[ f(x) = e^{-x}(x - 1) \]

### Step 10: Calculate \( e^{9} f(9) \)
Now, substituting \( x = 9 \):

\[ f(9) = e^{-9}(9 - 1) = e^{-9} \cdot 8 \]

Thus:

\[ e^{9} f(9) = e^{9} \cdot e^{-9} \cdot 8 = 8 \]

### Final Answer
\[ e^{9} f(9) = 8 \]