Integral/continuety proof question

[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 don't 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.

 

[Mark44]

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 don't 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)

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).

then subtract (beta substituted by x) with alpha substituted by x.

That's assuming you actually have an antiderivative, which as I already mentioned, isn't guaranteed.

so if we get a positive result, then the value beta substituted by x is bigger.

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,
$$\int_{\alpha}^{\beta} f(x) dx ?$$

 

[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.

 

[transgalactic]

how to prove it?

 

[Mark44]

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?"

 

[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

 

[Mark44]

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?

 

[CompuChip]

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

 

[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??

 

[Mark44]

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

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.

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

Let's not worry about that just yet.

 

[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.

 

[Mark44]

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.

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?

 

[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.

 

[transgalactic]

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?

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

 

[transgalactic]

what to do next?

 

[CompuChip]

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!

How do you mean, what to do next?

 

[HallsofIvy]

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

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

 

[transgalactic]

whats the second thing i was given?

 

[CompuChip]

whats the second thing i was given?

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?

 

[transgalactic]

i don't know how to prove it by contradiction..

 

[Mark44]

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

Pick one and follow it through.

 

[transgalactic]

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!

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
??

 

[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.

 

[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.
?

 

[CompuChip]

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

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

 

[transgalactic]

no i didnt change my approach
i think they are the same

can you tell what is the correct approach??