# 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 $[f(x)]^2 =2\int_0^x f(t) dt$;

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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.

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Depends how fussy the course is. Since the tag says calculus, possibly not very. But even if we assume that the OP is clarified to say $f(x)\ne 0$ for all positive $x$, there is the little issue of showing $f$ is differentiable. Since $f$ is continuous, and $f^2$ is an integral, certainly $f^2$ is differentiable. It can then be shown that $f$ must be, but that takes a line or so. – André Nicolas Nov 30 '12 at 5:24
Good point. I suppose I assumed it was a basic calculus class, in which case showing $f$ is differentiable from the given assumptions might be too hard for the student. – icurays1 Nov 30 '12 at 5:29

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

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You can use $\LaTeX$ markup to write formulas. See meta.math.stackexchange.com/questions/107/… for more details. – Pragabhava Nov 30 '12 at 5:38