# Homework Help: Integral/continuety proof question

1. Feb 21, 2009

### transgalactic

function f(x) continues on [a,b]
suppose that for every sub part $$[\alpha ,\beta ]\subseteq [a,b]$$
we have$$\int_{\alpha}^{\beta}f(x)dx>0$$.
prove that f(x)>=0 for $$x\in[a,b]$$

if its wrong give a contradicting example??

i dont have a clue from here to start or how to go.
from the given i can conclude that if the sum of all subsections gives us a positive result then
the total sum from a to b has to positive too.

when we solve an integral we take the anti derivative of f(x)
then subtract (beta substituted by x) with alpha substituted by x.
so if we get a positive result, then the value beta substituted by x is bigger.

2. Feb 22, 2009

### Staff: Mentor

Haven't you ever run across any integrals that don't have antiderivatives that are elementary functions? There are lots of relatively simple functions that don't have nice, neat antiderivatives. One example is e^(-x^2).
That's assuming you actually have an antiderivative, which as I already mentioned, isn't guaranteed.
I would try a proof by contradiction. Suppose that for some x in [a, b], f(x) < 0. What does that imply as far as your assumption that for every subinterval $[\alpha, \beta] \subset [a, b]$ about your definite integral,
$$\int_{\alpha}^{\beta} f(x) dx ?$$

3. Feb 22, 2009

### HallsofIvy

Yes, it is wrong to give a contradicting example because there is NO contradicting example! The statement is, as you are told, true. You can't contradict it.

4. Feb 22, 2009

### transgalactic

how to prove it?

5. Feb 22, 2009

### Staff: Mentor

I gave you a suggestion in post 2. Did you read what I wrote? At some point you need to take some initiative and do your own work instead of always responding with
"how to prove it?"

6. Feb 24, 2009

### transgalactic

i tried to prove it by contradiction:
suppose
there is a section which has negative value
$$\int_{\alpha}^{\beta}f(x)dx<0$$
then
$$\int_{a}^{b}f(x)dx<0$$
that as far as i can think of

Last edited: Feb 24, 2009
7. Feb 24, 2009

### Staff: Mentor

What do you mean that a section has a negative value? Look again at what I wrote in post #2.

The approach here is a proof by contradiction. What conclusion are you trying to contradict?

8. Feb 24, 2009

### CompuChip

Do you have any idea why the statement should be true?
Can you explain in words why it should hold?

9. Feb 24, 2009

### transgalactic

i think that if every small part is possitive then the whole part is possitive to
its like saying that if
x>0
then 100x>0 too

how should i transform those word into a frormal prove??

10. Feb 24, 2009

### Staff: Mentor

OK, now how does this relate to the original problem? What I'm trying to get from you is some reasoning (but not yet a proof) as to why you think the statement in this problem is true.

After you do that, I would like to see exactly what you are going to try to prove by contradiction.
Let's not worry about that just yet.

11. Feb 25, 2009

### transgalactic

we are given that
$$\int_{\alpha}^{\beta}f(x)dx>0$$
so when we substitute beta into the anti derivative
then the value is bigger then the substitution of alpha into the anti derivative.
i know that the derivative says if the function is going up or down

so if the anti derivative is positive then the function is positive.

12. Feb 25, 2009

### Staff: Mentor

That definite integral is only part of what you're given. And how are you going to substitute alpha and beta into the antiderivative if you don't know what the antiderivative is? As I said in another thread, more often than not, you won't be able to get an antiderivative for an arbitrary function. All you know about the function are two things--that f is continuous on the interval [a, b] and one other thing. What's the other thing you are given about this function?

13. Feb 25, 2009

### HallsofIvy

Essentially then you are arguing that if
$$\int_\alpha^\beta f(x)dx> 0$$
for all $\alpha$ and $\beta$, $\beta> \alpha$, then
$$F(x)= \int_a^x f(t)dt$$
is an increasing function and so its derivative, f(x), is never negative.

Yes, that is a valid proof, and since it is your own, go with it! What the others have been thinking of is the indirect proof: if there were some point c in [a, b] such that f(x)< 0, then, because there is some neighborhood, $(\alph, \beta)$, in which f(x)< 0. The integral from that $\alpha$ to $\beta$ cannot be positive.

14. Feb 25, 2009

### transgalactic

if its continues [a,b]
then we can have a continues anti derivative to it on [a,b]

15. Feb 26, 2009

### transgalactic

what to do next?

16. Feb 26, 2009

### CompuChip

How do you mean, what to do next?

17. Feb 26, 2009

### HallsofIvy

Even if a function is NOT continous on an interval, its anti-derivative is.

18. Feb 27, 2009

### transgalactic

whats the second thing i was given?

19. Feb 27, 2009

### CompuChip

You're asking us what you were given?

Why are you still on this question anyway? Are you trying to do it Matt's way, proof by contradiction?

20. Feb 28, 2009

### transgalactic

i dont know how to prove it by contradiction..

21. Feb 28, 2009

### Staff: Mentor

You have asked for help with your proof, and have been given suggestions for two different approaches.

Pick one and follow it through.

22. Feb 28, 2009

### transgalactic

ok so if a functions derivative is positive then its anti derivative is increasing.
so here we were give that the anti derivative is increasing for every [a,b]
so the derivative is possitive.

do i need to add something else here
??

23. Feb 28, 2009

### HallsofIvy

Your logic is going the wrong way. "If A then B" does not imply "If B then A". You want to argue that since
$$\int_\alpha^\Beta f(x) dx> 0$$
it follows that
$$F(x)= \int_a^x f(t)dt$$
and so its derivative, f(x), is postive.

Saying the "because the derivative is positive (which you don't know, that's what you are trying to prove) the function is increasing" is unnecessary and irrelevant.

24. Mar 1, 2009

### transgalactic

but you said that my way of prove is correct
??

what part is correct in my way of thinking
and what to add so i will have a complete set.
?

25. Mar 1, 2009

### CompuChip

Yes but that was before you wanted to completely change your approach :tongue: