In my single variable calculus book the derivation of the fundamental theorem of calculus only covered a special case and I would like help with generalisering the proof.

In my book the following was proven: If a function $f$ is continuous and differentiable on $(a, b)$ then it may be integrated over $[\alpha, \beta]$ satisfying $[\alpha, \beta] \subset (a,b)$ and its value is given by $F(\beta) - F(\alpha)$.

However, I believe the general case allows for $f$ to be integrated over $[a, b]$ given that $f$ is also continuous on $[a, b]$. This case was not proven. So I suppose that if i let $\alpha \to a^+$ and $\beta \to b^-$ and if I somehow show that the continuity of $f$ implies the continuity of $F$ on $[a, b]$ then perhaps that is enough to show that the integral of $f$ over $[a, b]$ is given by $F(b) - F(a)$ as well. How do I prove this?

  • $\begingroup$ In the case that $f$ is not defined at $a$ or $b$, what you are describing its called an improper integral. It is a limit, though, and as such may not necessarily be finite. $\endgroup$ Nov 18, 2016 at 18:27
  • $\begingroup$ In order for a function to be continuous on $[a,b]$, it must be defined at $a$ and at $b$. $\endgroup$
    – mrob
    Nov 18, 2016 at 18:29

1 Answer 1


Consider the counterexample $f(x) = \frac{1}{x}$, letting the interval be $(0,1)$. Certainly, $f$ is continuous and differentiable on that interval. And, indeed, if $\alpha = \epsilon > 0 = a$ and $\beta > \alpha$, we get its value as promised.

The problem is that $f$ is not defined on that closed interval $[0,1]$. In fact, for $f$ to be continuous, you need it to be defined on an open interval around $0$ (hence, the limit).

So, you could still sort-of apply that limit rule if you can prove $f$ is continuous and differentiable on $(a-\epsilon, b + \epsilon)$ for all reasonably small $\epsilon > 0$. Then you can then have your closed interval $[a, b]$ for your integral.

  • $\begingroup$ OP said $f$ is continuous on $[a,b]$, but your counterexample is not. $\endgroup$
    – mrob
    Nov 18, 2016 at 18:26
  • $\begingroup$ My argument is that it needs to be continuous around an open interval in order for this theorem to work on a closed interval. The only exception is if the "open interval" extends beyond the function's domain, in which case the intersection of the domain and the closed interval is topologically open wrt the function's domain. $\endgroup$
    – Larry B.
    Nov 18, 2016 at 18:32
  • $\begingroup$ I would like to see your argument then. If you could prove the OP's statement, or give a counterexample, I would appreciate that. $\endgroup$
    – mrob
    Nov 18, 2016 at 18:35

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