$\int_0^x f(t)dt = [f(x)]^2$ and f is never 0 Problem says: 
Suppose that continuous $f(x)$ is such that:
$$\int_0^x f(t) dt= [f(x)]^2$$
and $f(x)$ is never zero. Then find $f(x)$.
I do the question thinking it say "$f(x)$ is not the zero function"  I find $f(x) = \frac12 x$ as solution.
But, I see from again reading again $f(x)$ can never be zero. Now my solution is not correct. $f(0) = 0$ for my solution.
From how I have came to solution I wonder:
Is this possible even to find?
In the way I have solved it seemed this was the only way possible solution.
 A: If you put $x=0$ then you immediately get $f(x) = 0$. I guess what they meant was "$f$ is not identically zero", since $f \equiv 0$ is another solution.
Also, if you follow J.M.'s advice then it's quite helpful to know that $f(x) \neq 0$, at least for $x \neq 0$. You have to assume that $f$ is differentiable, though.
A: EDIT: Expanding on J. M.'s advice.
From $\int_0^x {f(t)dt}  = f^2 (x)$, we get $f(x) = \frac{d}{{dx}}f^2 (x)$ (using that $f$ is continuous). From this, one wants to write $f(x)=2f(x)f'(x)$. In general, however, differentiability of $f^2$ does not imply that of $f$ (consider, for example, the functions $|x-1|$ and $(x-1)^2$, $x \geq 0$; the former is not differentiable at $x=1$), and we are not given that $f$ is differentiable. 
Nevertheless, in our case, for any $x > 0$ we have
$$
f(x) = \frac{d}{{dx}}f^2 (x) = 
\mathop {\lim }\limits_{h \to 0} \frac{{f^2 (x + h) - f^2 (x)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{[f(x + h) + f(x)][f(x + h) - f(x)]}}{h},
$$
leading to 
$$
1 = \mathop {\lim }\limits_{h \to 0} \bigg[\frac{{f(x + h) + f(x)}}{{f(x)}}\frac{{f(x + h) - f(x)}}{h}\bigg],
$$
where we have used the assumption $f(x)>0$.
Since $f$ is continuous at $x$, it thus follows that
$$
1 = 2\mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h},
$$
and hence the conclusion that 
$$
f'(x) = \frac{1}{2}.
$$
(The rest is straightforward.)
Original answer (see first comment below).
If $f$ is continuous on $[0,\infty)$, then, on the one hand,
$$
\mathop {\lim }\limits_{x \to 0^ +  } \int_0^x {f(t)dt}  = 0,
$$
and, on the other hand,
$$
\mathop {\lim }\limits_{x \to 0^ +  } f^2 (x) = f^2 (0).
$$
So the condition $\int_0^x {f(t)dt}  = f^2 (x)$ implies that $f^2 (0) = 0$, hence also $f(0)=0$.
