2
$\begingroup$

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 $(a,b)$?

thanks

$\endgroup$
  • 1
    $\begingroup$ Do you think that every derivative is continuous ? $\endgroup$ – Tony Piccolo Jun 21 '14 at 15:29
  • 1
    $\begingroup$ $f'$ 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
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.