Thank you for your solutions to Daniel and Ben (no schools given) and to Rajiv from the International School of Seychelles and Shaun from Nottingham High School.

The integral equation is: $$\int_0^x f(t)\,dt = 3f(x)+k,\quad
\quad(\star)$$ where $k$ is a constant. Differentiating both sides
of $(\star)$ gives $$f(x) = 3f'(x)$$ If there is a solution of
$(\star)$ it must be of the form $$f(x) = Ae^{x/3},$$ for some
constant $A$. We check to see whether or not this is a
solution.

For $f(x)=Ae^{x/3}$ we have $$\int_0^x Ae^{t/3}\,dt =
\Big[3Ae^{t/3}\Big]_0^x = 3Ae^{x/3}-3A.$$ Thus $f(x)=Ae^{x/3}$ is a
solution if and only if $A=-k/3$. The unique solution is $$f(x) =
{-k\over 3} e^{x/3}.$$