prove with fundamental theorem of calculus how should i prove that $f(x) = x$, assume $f$  is continuous on $[0, \infty)$, $f(x)$ is not $0$ and $x$ is positive, also $[f(x)]^2 =2\int_0^x f(t) dt$;
 A: By the chain rule and the fundamental theorem of calculus, we can differentiate the equation $[f(x)]^2=2\int_0^xf(t)dt$:
$$
\frac{d}{dx}[f(x)]^2=\frac{d}{dx}2\int_0^x f(t)dt\\
\Rightarrow 2f(x)f^\prime(x)=2f(x)
$$
Thus after rearranging this we have $f(x)(1-f^\prime(x))=0$.  Since we are assuming $f(x)$ is not the zero function, we must have $f^\prime(x)=1$; this is a very simple differential equation that I think you should be able to solve (but be careful to address the constant of integration!)
Technical caveat: we should really be assuming that $f(x)$ is differentiable on $[0,\infty)$ for this to hold in the "basic calculus" setting.  Otherwise $f^\prime$ is meaningless.
A: Start with the integral and use the Second Fundamental Theorem of Calculus. I'm going to define the integral first as a function called F(x) and then relate it to f(x).
Using the First Fundamental Theorem of Calculus, we can write the integral as:
2(F(x)-F(0))=(f(x))^2. 
We then take the derivative of both sides. Because F(0) is just a constant, its derivative 
is zero. Remember to use the chain rule on (f(x))^2.
2F'(x) - 0 = 2(f(x))(f'(x))
Now, because F(x) is defined as the integral of f(x), then we know that F'(x)=f(x).
2(f(x))=2f(x))(f'(x))
We can then cancel out 2(f(x)) on both sides because it is stipulated that f(x) does not equal 0.
1=f'(x)
We integrate f'(x) to get f(x). The integral of 1 is x + C where C is a constant.
However, we then plug in x + C as f(x) into our equation to see what value C has.
(x+C)^2 = 2 times the integral from 0 to x of (x+C) dx
x^2 + 2Cx + C^2 = 2((.5)x^2)+Cx) - 0
x^2 +2Cx + C^2 = x^2 + 2Cx
C^2 = 0
C = 0
Therefore, f(x) = x + 0 = x
