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there is integrabile function f in {[}a,b{]} so f(a)<0.

f is continues from the right of a.

prove that there is $\displaystyle c\in(a,b)$ so $\displaystyle \intop_{a}^{c}f(x)dx<0$

prove:

there is $\displaystyle \epsilon=\frac{{|f(a)|}}{2}>0$ from the given continuety

in a there is a delta

$\displaystyle b-a>\delta>0$ so for every x in which $\displaystyle a\leq x<x+\delta$ $\displaystyle |f(x)-f(a)|<\varepsilon$.

so $\displaystyle f(x)-f(a)<-\frac{f(a)}{2}$ and the ({*}) $\displaystyle f(x)<f(a)/2<0$

f is integrabile in [a,b] and so in [a,a+$\displaystyle \delta/2$]

from ({*}) and from the monotonicity we conclude

$\displaystyle \intop_{a}^{a+\frac{\delta}{2}}f(t)dt\leq\intop_{a }^{a+\frac{\delta}{2}}\frac{f(a)}{2}dt=\frac{\delt a}{2}\frac{f(a)}{2}<0$

.

__i got stuck on the first step:__

why they chose such delta and x??

from the defintion it should be there is delta>0 for which in every

x a<x<a+$\displaystyle \delta$