Question:

Step 1: Express the given integral equation for $f(x)$ by factoring out $e^x$. The given equation is $f(x) = \int_0^1 e^{x+t} f(t)\, dt$. We can rewrite the exponential term and factor $e^x$ out of the integral since the integration is with respect to $t$: $$f(x) = \int_0^1 e^x e^t f(t)\, dt = e^x \int_0^1 e^t f(t)\, dt$$ Step 2: Define the integral as a constant. Let the definite integral be a constant, as it does not depend on $x$: $$k = \int_0^1 e^t f(t)\, dt$$ Step 3: Express $f(x)$ in terms of the constant $k$. Substitute the constant $k$ back into the expression for $f(x)$: $$f(x) = k e^x$$ This equation holds for all $x \in [0, 1]$. Step 4: Substitute the expression for $f(t)$ back into the definition of $k$. Now, substitute $f(t) = k e^t$ into the definition of $k$: $$k = \int_0^1 e^t (k e^t)\, dt$$ $$k = \int_0^1 k e^{2t}\, dt$$ Step 5: Evaluate the definite integral to find a relationship for $k$. We can factor $k$ out of the integral: $$k = k \int_0^1 e^{2t}\, dt$$ Now, evaluate the integral: $$k = k \left[ \frac{e^{2t}}{2} \right]_0^1$$ $$k = k \left( \frac{e^{2(1)}}{2} - \frac{e^{2(0)}}{2} \right)$$ $$k = k \left( \frac{e^2}{2} - \frac{1}{2} \right)$$ $$k = k \frac{e^2 - 1}{2}$$ Step 6: Solve the resulting equation for the constant $k$. Rearrange the equation to solve for $k$: $$k - k \frac{e^2 - 1}{2} = 0$$ Factor out $k$: $$k \left( 1 - \frac{e^2 - 1}{2} \right) = 0$$ Simplify the term inside the parenthesis: $$k \left( \frac{2 - (e^2 - 1)}{2} \right) = 0$$ $$k \left( \frac{2 - e^2 + 1}{2} \right) = 0$$ $$k \left( \frac{3 - e^2}{2} \right) = 0$$ Since $e \approx 2.718$, $e^2 \approx 7.389$. Thus, $3 - e^2 \neq 0$. For the product to be zero, $k$ must be zero: $$k = 0$$ Step 7: Determine the value of $f(1)$. Since $k=0$, substitute this value back into the expression for $f(x)$: $$f(x) = 0 \cdot e^x$$ $$f(x) = 0$$ Therefore, for any value of $x$ in the domain, $f(x)$ is $0$. For $x=1$, we have: $$f(1) = 0$$ The final answer is $\boxed{\text{0}}$.