When the primitive of a function is the function itself, then $f(x)=0$ (using the FTC)? Let f: R → R be a continuous function such that:
$f(x)=\int_{0}^{x}f(t)\,dt$
Show that $f(x)=0$; for all $x$.
I did it like this:
According to the statement, $ f (x) $ is already the primitive of $ f (x) $, so $ f (x) $ is equal to $ F (x) $. By the Fundamental Theorem of Calculus we already know that $ F´ (x) = f (x) $ just like $ f (x) = F (x) $ we have $ F´ (x) = f´ (x) $ and therefore $ f (x) = f´ (x) $. I don't know how to show that it's zero.
As $f$ is continuous, all its antiderivatives are differentiable and the differential equation
$$f'(x)=f(x)$$ holds.
We know that the general solution of this ODE is $f(x)=ce^x$ (including $c=0$, from $f'(x)=f(x)=0$).
Now as $f(0)=0$, only the solution with $c=0$ is possible.
If $f(x)=f'(x)$ then $\frac{f'(x)}{f(x)}=1$. Integrating, you get that
$\ln f(x)=x+K$ whence $f(x)=ce^x$ for some $c\in\Bbb R$.
But you also know that $f(x)=\int_0^xf(t)dt$, so $f(0)=\int_0^0f(t)dt=0$.
Putting altogether $0=f(0)=ce^0=c$ and you'll get that $f(x)=0$.
EDIT: Thh above argument works whenever $f\ne0$. Observe that, in fact, $c\ne0$ but $f(x)=0$ is a trivial solution of $f'(x)=f(x)$. So we can write that all solutions of $f'(x)=f(x)$ are $f(x)=ce^x$ with $c\in \Bbb R$.