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I am pretty sure that this statement is true, altough I don't find the correct proof for it. I am pretty sure it involves using the Fundamental Thorem of Calculus, but again I can't seem to picture how to use it to prove it.

Thanks

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  • $\begingroup$ Are you asking if every continuous function has an anti-derivative? $\endgroup$ – Ben Longo Nov 29 '15 at 4:12
  • $\begingroup$ At the end of the day, if $f$ is continuous on $[a,b]$, then $F(x)=\int_a^x f(y) dy$ is well-defined for $x \in [a,b]$, and $F'(x)=f(x)$ for each $x \in [a,b]$. There are some technicalities here that may or may not matter for your purposes, such as the proof that every continuous function is Riemann integrable. $\endgroup$ – Ian Nov 29 '15 at 4:14
  • $\begingroup$ Offered merely as a curiosity: Suppose one had studied derivatives but not integrals. With a bit of skill you can prove this: If $f$ is a bounded function on an interval $[a,b]$ then there exists a Lipschitz function $F$ such that $F'(x)=f(x)$ at every point $x$ at which $f$ is continuous. It is easier, of course, to rely on the Fundamental Theorem of the Calculus to do the work for you. $\endgroup$ – B. S. Thomson Nov 29 '15 at 16:44
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This is word for word the exact statement of the fundamental theorem of calculus.

The fundamental theorem of calculus states that if $f$ is continuous on $[a,b]$ then the function $F(x)=\int_a^x f(t) dt$ is differentiable and $F'(x)=f(x)$

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  • $\begingroup$ Very nice :). To be fair, though, in my experience what most students think of as the fundamental theorem is a (to an undergraduate calculus student, much more useful) corollary: $``$If $f$ is a real-valued continuous function on $[a,b]$ and $F$ is an antiderivative of $f$ in $[a,b]$, then $$\int _a^bf(t)\,\mathrm{d}t=F(b)-F(a)."$$ (Quoted from Wikipedia.) $\endgroup$ – A. Howells May 1 '17 at 17:48

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