TITLE: Is a differentiable function on $(-2, 4)$ always integrable on $[-2, 4]$?
QUESTION [19 upvotes]: So my question is, say I have a function that is differentiable on $(-2, 4)$. Is it always integrable on $[-2, 4]$?
I know that if $f$ is diff on $(-2, 4)$, then it is continuous on $(-2, 4)$. And I also know that if $f$ is continuous on $[-2, 4]$ then it is integrable on $[-2, 4]$. However, I am wondering if there is such a function so that there would be a problem at the endpoints of the closed interval so that it is differentiable on the open interval, but not integrable on the closed interval.

REPLY [20 votes]: Generally no, but with additional assumption that it is bounded -- yes. For the unbounded case we have easy counterexamples, as $1/x$ or $1/\ln x$ on $(0,1)$.