# If ∫baf(x)dx>0\int_{a}^{b}f(x)dx>0 for every a,ba, b such that b>a>x0>0b>a>x_0>0 then …

Usually in continuum mechanics when investigating the second law of thermodynamics one encounters a situation as follows

∫baf(x)dx>0\int_{a}^{b}f(x)dx>0 for every a,ba, b such that b>a>x0>0b>a>x_0>0 where x0x_0 is some fixed positive real number and f:R→Rf:\Bbb{R} \to \Bbb{R} is a continuous function.

My question are that

What can we say about f(x)f(x)?
What can we say about the f′(x)f'(x) whenever f(x)=0f(x)=0?

Intuition says that we may simply conclude f(x)≥0f(x) \ge 0 over R\Bbb{R}. And for the second question it seems that f′(x)=0f'(x)=0 or does not exist.

I want to make a rigorous analysis. For starting the argument I use the mean value theorem for integration. So there exist a cc such that a0f(c)(b-a)>0

what will be my next step?

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About the derivative, wouldn’t the first deduction imply that the derivative is zero at the zeros of the function?
– O. Von Seckendorff
Oct 20 at 20:30

Yes, intuitively but don’t how to write it down. 🙂
– H. R.
Oct 20 at 20:33

@MathematicsStudent1122 Did you check the edit queue? He edited the question. And yes, it might not be differentiable at the zeros, which could be relatable with some kind of phase change, but I’m guessing he has to tell if it is or isn’t assumed to be.
– O. Von Seckendorff
Oct 20 at 20:46

@O.VonSeckendorff I apologize for that rude comment. Rough day.
– MathematicsStudent1122
Oct 20 at 21:07

1

@MathematicsStudent1122 No worries, internet stranger. In the end we came to a conclusion in 2 comments. Hope it gets better for you. 😀
– O. Von Seckendorff
Oct 20 at 21:12

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I’d rather go with a proof by contradiction from the very beginning: if f(x)<0f(x)<0 for some xx, by continuity there is an interval (a,b)∋x(a,b)\ni x such that f(y)<0f(y)<0 for all y∈(a,b)y\in (a,b). But then ∫baf(y)dy<0\int_a^b f(y)\,dy<0. After edit: If ff is differentiable at xx and f′(x)>0f'(x)> 0, then for y(≠x)y(\ne x) in a small interval (x−ε,x+ε)(x-\varepsilon, x+\varepsilon) it holds 12f′(x)≤f(y)−f(x)y−x≤32f′(x)\frac12f'(x)\le\frac{f(y)-f(x)}{y-x}\le \frac32 f'(x). Especially, for yxy>x, f(y)≤12(y−x)f′(x)+f(x)1x>1. What can you say about f(−35)f(-35)?
– G. Sassatelli
Oct 21 at 3:58

Let us just talk about x>1x>1! 🙂 So what can we say about f(35)f(35)? 🙂
– H. R.
2 days ago