# proving question with integral

• June 1st 2013, 11:43 PM
orir
proving question with integral
f is an integrable no-negative function in [a,b]. i need to prove that if exists point $c\in(a,b)$, so f is continous in $c$ and $f(c)>0$, so $\intop_{a}^{b}f(x)dx>0$
• June 1st 2013, 11:50 PM
orir
Re: proving question with integral
is it correct to prove that way? if exists point c, so maintains: $0
so - rather the integral is negative and the fracture, or the integral is positive and so is the fracture. and, obviously we know the fracture is positive, so we can conclude that the integral is also positive.

can i do that? isn't that correct only when the function is continous?! or maybe because it's given that "f is continous in $c$" it's fine?
• June 2nd 2013, 02:01 AM
ibdutt
Re: proving question with integral
please check if the question has been correctly posted. The reason is the if the function is non negative then the function will have positive values over the interval. That is to say for any c in the given interval f(c) will be positive i.e, f(c) > 0
• June 2nd 2013, 03:36 AM
Plato
Re: proving question with integral
Quote:

Originally Posted by orir
f is an integrable no-negative function in [a,b]. i need to prove that if exists point $c\in(a,b)$, so f is continous in $c$ and $f(c)>0$, so $\intop_{a}^{b}f(x)dx>0$

Because $f$ is continous at $x= c$ you know that $\exists (u,v)\subset (a,b)$ such that $\forall x\in (u,v)\left[ {f(x) > \frac{{f(c)}}{2}} \right]$. WHY?

Thus $\int_a^b f = \int_a^u f + \int_u^v f + \int_v^b f \geqslant 0 + \frac{{f(c)}}{2}\left( {v - u} \right) + 0 > 0$
• June 2nd 2013, 03:42 AM
Plato
Re: proving question with integral
Quote:

Originally Posted by ibdutt
please check if the question has been correctly posted. The reason is the if the function is non negative then the function will have positive values over the interval. That is to say for any c in the given interval f(c) will be positive i.e, f(c) > 0

That is in fact not true. If $\forall x\in [a,b]\left[f(x)=0\right]$ then $f$ is non-negative on $[a,b].$
• June 2nd 2013, 04:00 AM
orir
Re: proving question with integral
Quote:

Originally Posted by Plato
Because $f$ is continous at $x= c$ you know that $\exists (u,v)\subset (a,b)$ such that $\forall x\in (u,v)\left[ {f(x) > \frac{{f(c)}}{2}} \right]$. WHY?

Thus $\int_a^b f = \int_a^u f + \int_u^v f + \int_v^b f \geqslant 0 + \frac{{f(c)}}{2}\left( {v - u} \right) + 0 > 0$

i didn't understand why ${f(x) > \frac{{f(c)}}{2}}$, and acctualy all.. :S sorry..
• June 2nd 2013, 04:35 AM
Plato
Re: proving question with integral
Quote:

Originally Posted by orir
i didn't understand why ${f(x) > \frac{{f(c)}}{2}}$, and acctualy all.. :S sorry..

That is the key to this problem. Review your textbooks/notes on continuous functions.

Find a theorem that says: If $f$ is continuous at $x_0$ then there is an open interval about $x_0$ on which $f$ has the same sign as $f(x_0)$.

Now if you cannot find that theorem then prove it. I will not.
• June 2nd 2013, 04:45 AM
orir
Re: proving question with integral
but what about my way... is it not good at all?
• June 2nd 2013, 05:16 AM
Plato
Re: proving question with integral
Quote:

Originally Posted by orir
but what about my way... is it not good at all?

You are trying to use a mean value theorem for integrals.
However, the given conditions are not sufficient to use that.
• June 19th 2013, 11:56 AM
bkarpuz
Re: proving question with integral
The converse is also true, i.e. for a function $f\in C([a,b],[0,\infty))$,
There exists $c\in[a,b]$ such that $f(c)>0$ if and only if $\int_{a}^{b}f(x) \mathrm{d}x>0$.
Proof of sufficiency is given in Plato's Post #4,
and for the proof of necessity, you can use the method of contradiction.