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is there such a function that is defined in [0,1], differentiable in (0,1) but not integrable? Thanks in advance.

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Robert
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    @Regret It can still be discontinuous on the boundaries. – N.U. Feb 16 '15 at 19:02
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    @N.U.: A clarification: functions differentiable on open interval $I$ are continuous on $I$, and functions continuous on $I$ are Riemann integrable on $I$. – Regret Feb 16 '15 at 19:06
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    @Regret That is true for closed bounded intervals. – N.U. Feb 16 '15 at 19:15
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    To be clear, I refer to your last statement, that the function is Riemann integrable. – N.U. Feb 16 '15 at 19:22

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It's important to know that if a function is continuous on a closed interval, it is integrable on that interval. This question seems to be getting at how strict that condition is.

If a function is differentiable on an open interval, then it is automatically continuous on that interval. The question is whether it can be extended to a continuous function on the enclosing closed interval. And one way this can fail to happen is if the function has an infinite limit at an endpoint.

So that is how we construct the counterexample. The function $$ f(x) = \begin{cases} 0 & x=0 \\ \frac{1}{x} & x > 0 \end{cases} $$ is defined on $[0,1]$, differentiable on $(0,1)$, but not integrable on $[0,1]$.

Matthew Leingang
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    It is a bit unclear to me whether @Robert meant not integrable on $(0,1)$ or $[0,1]$, but in the case of the latter, have a +1. – Regret Feb 16 '15 at 19:02
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    @Regret: It's not clear to me either. – Matthew Leingang Feb 16 '15 at 19:04
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    I was just reading some given in the past questions for my exam, it's not specified in which interval :( – Robert Feb 16 '15 at 19:15
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    @Robert The Riemann integral is defined on the closed interval, I'm sure you will see this if you look in your textbook. This is also discussed on this site, see for instance: http://math.stackexchange.com/questions/461459/is-there-any-notable-difference-between-studying-the-riemann-integral-over-open?lq=1 – N.U. Feb 16 '15 at 19:35