1. Integrability-Anti-Derrivatives Question

Let f be a function that has an anti-derivative function F (F'(x)=f(x)) in [a,b].
Prove / Disproof :

f is integrable in [a,b].

--

Well, I said it shouldn't be, and I gave the example:
$
F(x):= 0 , x=0$

$x\sin\frac{1}{x} , 0

$x\in[0,1]$

therefore,

$f:=\sin\frac{1}{x}-\frac{1}{x}cos\frac{1}{x} , x\in[0,1]$

f can't be integrable there, because it is not bounded for $x\in(0,1]$.

Is this a proper disproof?

Than you

Let f be a function that has an anti-derivative function F (F'(x)=f(x)) in [a,b].
Prove / Disproof :

f is integrable in [a,b].

--

Well, I said it shouldn't be, and I gave the example:
$
F(x):= 0 , x=0$

$x\sin\frac{1}{x} , 0

$x\in[0,1]$

therefore,

$f:=\sin\frac{1}{x}-\frac{1}{x}cos\frac{1}{x} , x\in[0,1]$
$f(0)= \lim_{h\to 0}\frac{f(0+h)- f(0)}{h}= \lim_{h\to 0}\frac{hsin(1/h)}{h}= \lim_{h\to 0} sin(1/h)$ does not exist so this function is not defined on all of [0, 1].

f can't be integrable there, because it is not bounded for $x\in(0,1]$.

Is this a proper disproof?

Than you

3. Originally Posted by HallsofIvy
$f(0)= \lim_{h\to 0}\frac{f(0+h)- f(0)}{h}= \lim_{h\to 0}\frac{hsin(1/h)}{h}= \lim_{h\to 0} sin(1/h)$ does not exist so this function is not defined on all of [0, 1].
That's right...

Well, does that mean my disproof is wrong?

Do I need to prove it?

4. I still don't understand - my original claim:

Let f be a function that has an anti-derivative function F (F'(x)=f(x)) in [a,b].
Prove / Disproof :

f is integrable in [a,b].
Is it true, or false?