Solving an integral equation

1. Since $f$ is continuous, $x\mapsto\int_0^xf$ is differentiable with derivative $f(x)$. So $f$ is differentiable and $f'(x)=f(x)$. Setting $x=0$ in the equation gives $f(0)=1$. The function $g(x)=f(x)e^{-x}$ has $g'=0$, so $g=g(0)=1$ and
$$f(x)=e^x.$$
Conversely $\int_0^xe^t\,dt=e^x-1$, so $1+\int_0^xe^t\,dt=e^x$. The unique solution is $f(x)=e^x$.

2. The same reasoning gives $f'=f$, but now $f(0)=0$. So $g=f e^{-x}$ is constant equal to $g(0)=0$, and $f=0$.

Remark. This is the integral formulation of the differential equation $f'=f$. It makes the initial condition automatic, and it is the starting point of the *Picard iteration* $f_{k+1}(x)=1+\int_0^xf_k$, which produces the partial sums $1+x+\frac{x^2}2+\dots$ of the exponential series.