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