Let $f(x):[a,b]\to \mathbb R$, be differentiable on $[a,b]$ (and continuous) so that $f'(x)$ is integrable on $[a,b]$.

I need to show that: $$\int _a^b\left(f'\left(x\right)\right)\mathrm dx = f\left(b\right)-f\left(a\right)$$

my thought was to show that $f'$ is continuous, and then using the Fundamental theorem of calculus to complete the proof.

but, is $f '(x)$ continuous on $(a,b)$?


  • 1
    $\begingroup$ Do you think that every derivative is continuous ? $\endgroup$ – Tony Piccolo Jun 21 '14 at 15:29
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    $\begingroup$ $f'$ need not be continuous in general. $\endgroup$ – Olivier Bégassat Jun 21 '14 at 15:35
  • $\begingroup$ @TonyPiccolo That's what he's asking $\endgroup$ – leo Jun 21 '14 at 15:46
  • $\begingroup$ Your statement is true as long as $f'$ is Riemann integrable (see here: en.wikipedia.org/wiki/… ). If you use the Lebesgue integral, it is also true if $f'$ is Lebesgue integrable, see e.g. here: math.stackexchange.com/questions/813230/… (this is even in the case of Banach valued functions, but the proof works (even simpler) in the scalar case). $\endgroup$ – PhoemueX Jun 21 '14 at 15:54
  • $\begingroup$ @leo Thank you for the notice. I was wrong, I meant integrable instead of continuous. $\endgroup$ – Tony Piccolo Jun 21 '14 at 15:59

Let $f$ be differentiable on $[a,b]$.

It is a fact that the following statement is true:

for every $\varepsilon>0$ there is a $\delta(x)>0$ so that $$\left|\sum_{i=1}^n f'(\xi_i)(x_i-x_{i-1})-(f(b)-f(a))\right|<\varepsilon$$ whenever $$a=x_0<x_1<\dots<x_{n-1}<x_n=b$$ and $$\xi_i \in [x_i-x_{i-1}]$$ with $$|x_i-x_{i-1}|<\delta(\xi_i)$$

but the following is false:

for every $\varepsilon>0$ there is a $\delta>0$ (not depending on $x$) so that $$\left|\sum_{i=1}^n f'(\xi_i)(x_i-x_{i-1})-(f(b)-f(a))\right|<\varepsilon$$ whenever $$a=x_0<x_1<\dots<x_{n-1}<x_n=b$$ and $$\xi_i \in [x_i-x_{i-1}]$$ with $$|x_i-x_{i-1}|<\delta$$


In other words, one can always approximate $f(b)-f(a)$ by Riemann sums of $f'$ "pointwise" but cannot always do it "uniformly".

So your aim is hopeless if you just use the classical Riemann integral.

  • $\begingroup$ This is incorrect. The claim you state is false is in fact the definition of what it means for $f'$ to be Riemann integrable with integral equal to $f(b)-f(a)$. $\endgroup$ – Santiago Canez Jun 22 '14 at 16:38
  • $\begingroup$ I say that not every derivative is Riemann integrable. $\endgroup$ – Tony Piccolo Jun 22 '14 at 17:21
  • $\begingroup$ Yes, but the OP is assuming that $f'$ is integrable, so your answer isn't really an answer to the question which was asked. $\endgroup$ – Santiago Canez Jun 22 '14 at 17:56
  • $\begingroup$ Do you think that OP's "so that $f'(x)$ is integrable" is an assumption? It seems to me a consequence (if you are right, why doesn't OP use a simple and ?). $\endgroup$ – Tony Piccolo Jun 22 '14 at 18:52
  • $\begingroup$ Yes, it is ambiguous, but I think his/her comments make it clear this was meant as an assumption. But fair enough, I cannot remove my downvote unless the answer is edited. $\endgroup$ – Santiago Canez Jun 22 '14 at 18:54

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