BTW the above is taken from B. P. Demidovich: Problems in mathematical analysis. I am just mentioning this in case it helps somebody to understand the terminology - what is meant by $f(x)$ is infinitesimal as $x\to1$.

Are calculus of variations and functons of bounded variations the same

@BAYMAX Those are two entirely different topics.

oh...

I thought they were same...

ok

Integral of function b/w a nd b

is 0

and f is non-negative then without using geometric interpretation how do i show f = 0?

any hintts

I would say that even after brief look of Wikipedia articles Calculus of variations and Bounded variation you can see that those are rather different question.

@BAYMAX I do not really understand what this says?

What is b/w?

ok

Do you meant $\int_a^b f(x) dx=0$?

$\int_{a}^{b}f(x) dx = 0,f(x) \geq 0 \forall x \in [a,b] $ then $f = 0$..any hints

Actually asked this a 2min ago in main chatroom

Without further assumptions on $f$ this is not true.

Is $f$ continuous, by any chance?

Yes, differentiable too

Quick search returns several related posts: approach0.xyz/search/…

For example: Prove the integral of $f$ is positive if $f ≥ 0$, $f$ continuous at $x_0$ and $f(x_0)>0$

Although it would be nice to find a post with a picture - that would be probably simplest explanation.

i don't want to procced by geometrical idea

Why?

The picture leads you directly to a rigorous proof, if you want (need) to write such proof.

I know that definite integral represents area and area is zero implying f is 0

Yes.

All you need to do is to think a bit why this is true.

And then write it down formally.

But perhaps some of the posts on the main site will give you what you are looking for.

there is something always tickling my brain! .... sorry to disturb you even these are in the main...

I see that this discussion is going in two rooms simultaneously.

yyeah...

$$\int_a^b {f\left( x \right)dx} \geqslant \int_{{x_0} - \delta }^{{x_0} + \delta } {f\left( x \right)dx} \geqslant \int_{{x_0} - \delta }^{{x_0} + \delta } {\frac{{f\left( {{x_0}} \right)}}{2}dx} $$

I think the first inequality is because $2x_{0} < b -a$

The first inequality is because $[x_0-\delta,x_0+\delta]\subseteq [a,b]$ and $f\ge0$.

Yeah some of the measure theory concepts collision going on i think!

yes @MartinSleziak

That last answer you gave the link really helped..thanks@MartinSleziak!

ok

I'll have to go. Good luck!

thank you @MartinSleziak ! same to you!