Is a differentiable function on $(-2, 4)$ always integrable on $[-2, 4]$?

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.

Solution 1:

The function $f(x)=\frac 1x$ is differentiable on $(0,1)$, yet it is not integrable on $[0,1]$.

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However, if you have a function $f$ which is differentiable on $[0,1]$, then it is necessarily continuous on $[0,1]$, hence measurable. Moreover, a continuous function on a compact is bounded, hence $f$ is bounded measurable, therefore integrable.

Solution 2:

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 $(0,1)$.