Proving f(x) = 0 from f(x) = ∫₀ˣ f(x) dx

[JG89]

Homework Statement

Prove that if f(x) is continuous and $$f(x) = \int_0^x f(x) dx$$, then f(x) = 0.

Homework Equations

The Attempt at a Solution

If $$f(x) = \int_0^x f(x) dx$$, then by integrating by the FTC we have f'(x) = f(x). Thus the only solution to this equation will have the form $$f(x) = ce^x$$ for some constant c. Now, $$f(x) = \int_0^x f(x) dx = f(x) - f(0)$$, implying that f(0 = 0. So since we know the solution to the equation will be $$f(x) = ce^x$$ then we have $$0 = f(0) = ce^0 = c$$, implying that c = 0. Thus f(x) = 0. QED

Is this correct?

 

[Mark44]

I'm confused by two things:

1. Your use of x in one of the limits of integration and as the dummy variable in the integral. It would be better to use different variables.
2. Your use of ' (as in dx'). Is this supposed to mean the derivative with respect to x of the definite integral?

Based on these points, I believe you are saying that
$$f(x) = \frac{d}{dx}\int_0^x f(t)dt$$

Now maybe I've missed something in how I've interpreted your problem, but the equation just above is true for every function f that is continuous on [0, b], and where 0 <= x <=b, per the FTC, so it does not follow that f(x) is identically 0.

 

[JG89]

Oops, let me retype the question:

Prove if f(x) is continuous and $$f(x) = \int_0^x f(x) dx$$ then f(x) = 0.

That's the question, and the mark beside the dx was only a comma, it wasn't meant to denote the derivative of the integral. So now that I fixed the question, isn't f'(x) = f(x)? And so f(x) = ce^x, but 0 f(0) = ce^0 = c and so c = 0 and f(x) = 0.

 

[Dick]

You still may be causing some confusion by having the dummy variable of integration be the same as the limit, but I get what you are saying. I think that proof is ok. As f is the integral of a continuous function it's differentiable.

 

[Mark44]

Same here.

 

[JG89]

Ugh, I meant to change the variable of integration to t but I forgot! Sorry, and thanks for checking my work.