# Thread: proving question with integral

1. ## proving question with integral

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

2. ## Re: proving question with integral

is it correct to prove that way? if exists point c, so maintains: $\displaystyle 0<f(c)=\frac{1}{b-a}\cdot\intop_{a}^{b}f(x)dx$
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 $\displaystyle c$" it's fine?

3. ## 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

4. ## Re: proving question with integral

Originally Posted by orir
f is an integrable no-negative function in [a,b]. i need to prove that if exists point $\displaystyle c\in(a,b)$, so f is continous in $\displaystyle c$ and $\displaystyle f(c)>0$, so $\displaystyle \intop_{a}^{b}f(x)dx>0$
Because $\displaystyle f$ is continous at $\displaystyle x= c$ you know that $\displaystyle \exists (u,v)\subset (a,b)$ such that $\displaystyle \forall x\in (u,v)\left[ {f(x) > \frac{{f(c)}}{2}} \right]$. WHY?

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

5. ## Re: proving question with integral

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 $\displaystyle \forall x\in [a,b]\left[f(x)=0\right]$ then $\displaystyle f$ is non-negative on $\displaystyle [a,b].$

6. ## Re: proving question with integral

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

Thus $\displaystyle \int_a^b f = \int_a^u f + \int_u^v f + \int_v^b f \geqslant 0 + \frac{{f(c)}}{2}\left( {v - u} \right) + 0 > 0$
i didn't understand why $\displaystyle {f(x) > \frac{{f(c)}}{2}}$, and acctualy all.. :S sorry..

7. ## Re: proving question with integral

Originally Posted by orir
i didn't understand why $\displaystyle {f(x) > \frac{{f(c)}}{2}}$, 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 $\displaystyle f$ is continuous at $\displaystyle x_0$ then there is an open interval about $\displaystyle x_0$ on which $\displaystyle f$ has the same sign as $\displaystyle f(x_0)$.

Now if you cannot find that theorem then prove it. I will not.

8. ## Re: proving question with integral

but what about my way... is it not good at all?

9. ## Re: proving question with integral

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.

10. ## Re: proving question with integral

The converse is also true, i.e. for a function $\displaystyle f\in C([a,b],[0,\infty))$,
There exists $\displaystyle c\in[a,b]$ such that $\displaystyle f(c)>0$ if and only if $\displaystyle \int_{a}^{b}f(x) \mathrm{d}x>0$.
Proof of sufficiency is given in Plato's Post #4,
and for the proof of necessity, you can use the method of contradiction.