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 $

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- Jun 1st 2013, 11:43 PMorirproving 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 $

- Jun 1st 2013, 11:50 PMorirRe: 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? - Jun 2nd 2013, 02:01 AMibduttRe: 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

- Jun 2nd 2013, 03:36 AMPlatoRe: proving question with integral
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 $\displaystyle \int_a^b f = \int_a^u f + \int_u^v f + \int_v^b f \geqslant 0 + \frac{{f(c)}}{2}\left( {v - u} \right) + 0 > 0$ - Jun 2nd 2013, 03:42 AMPlatoRe: proving question with integral
- Jun 2nd 2013, 04:00 AMorirRe: proving question with integral
- Jun 2nd 2013, 04:35 AMPlatoRe: proving question with integral
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. - Jun 2nd 2013, 04:45 AMorirRe: proving question with integral
but what about my way... is it not good at all?

- Jun 2nd 2013, 05:16 AMPlatoRe: proving question with integral
- Jun 19th 2013, 11:56 AMbkarpuzRe: 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.