If $f(x)=\int_0^1e^{x+t}f(t)\,dt$, then find the value of $f(1)$ 
Problem: If $f(x)=\displaystyle\int_{0}^{1}e^{x+t}f(t)\,dt$, then find the value of $f(1)$.

My attempt: Following from the definition of $f(x)$ we can write: $f(0)=\int_{0}^{1}e^tf(t)dt$ and $f'(x)=f(x)$. From the latter integral we can state that $f(x)=e^x+C$ and therefore $f'(x)=e^x.$ Integrating the former integral by parts we get: $$f(0)=(f(1)e-f(0))-\int_{0}^{1} f'(t)e^tdt=ef(1)-f(0)-({e^2-1\over 2})$$ $$\Rightarrow 2f(0)=ef(1)-{e^2-1\over 2}.$$Now observe that $f(1)=ef(0).$ Substituting for $f(0)$ in the above equation we get: $$f(1)={e(1-e^2)\over2(2-e^2)}.$$ However the answer is $f(1)=0.$ Please identify the flaw in my reasoning. 
 A: Ian Miller answered well, so I will find a different solution. Let's find $f(1)$ without differential equation $f'(x)=f(x)$. Observe that
$$
f(1)=\int_0^1 e^{1+t}f(t)dt,
$$
then
$$
e^{1+x}f(x)=e^{1+x}\int_0^1 e^{x+t}f(t)dt=e^{1+2x} \int_0^1 e^t f(t)dt.
$$
Therefore,
\begin{align}
f(1)&=\int_0^1 e^{1+x}f(x)dx\\
&=\int_0^1 e^{1+2x} \left(\int_0^1 e^t f(t)dt\right)dx\\
&=\int_0^1 e^{2x} \left(\int_0^1 e^{t+1} f(t)dt\right)dx\\
&=\left(\int_0^1 e^{t+1} f(t)dt\right)\int_0^1 e^{2x}dx\\
&=\frac{1}{2}(e^2-1)f(1)
\end{align}
and so $f(1)=0$.
A: You have solved $f'(x)=f(x)$ incorrectly. The solution is $f(x)=c\cdot e^x$.
This leads to:
$$f(0)=\int_0^1e^t\cdot c\cdot e^t\ dt$$
$$c=\frac{c(e^2-1)}{2}$$
And the only way this can be true is if $c=0$.
Hence $f(x)=0$.
So $f(1)=0$.
A: Consider that
$$f(x)=\int_0^1 e^{x+t}f(t)dt=e^{x-1}\int_0^1 e^{1+t}f(t)dt=e^{x-1}f(1)$$
Therefore
$$f(1)=\int_0^1 e^{1+t}f(t)dt=\int_0^1 e^{1+t} e^{t-1}f(1) dt = f(1)\int_0^1 e^{2t}dt$$
This holds only if $f(1)=0$ because $e^{2t}>1$ in $(0,1)$ implies $\int_0^1 e^{2t}dt>1$ (it is easy, of course, to give the exact value of this integral, but proving that it's value is not $1$ is enough).
This answer is similar to the previous one but I think it might be nicer. As said, the flaw in the reasoning in the question was in the solution to the differential equation..
