integrbility proof

**transgalactic** · August 28th 2011, 02:22 AM

% Preview source code from paragraph 0 to 12
there is integrabile function f in {[}a,b{]} so f(a)<0.
f is continues from the right of a.
prove that there is $c\in(a,b)$ so $\intop_{a}^{c}f(x)dx<0$
prove:
there is $\epsilon=\frac{{|f(a)|}}{2}>0$ from the given continuety
in a there is a delta
$b-a>\delta>0$ so for every x in which $a\leq x<x+\delta$ $|f(x)-f(a)|<\varepsilon$ .
so $f(x)-f(a)<-\frac{f(a)}{2}$ and the ({*}) $f(x)<f(a)/2<0$
f is integrabile in [a,b] and so in [a,a+ $\delta/2$ ]
from ({*}) and from the monotonicity we conclude
$\intop_{a}^{a+\frac{\delta}{2}}f(t)dt\leq\intop_{a }^{a+\frac{\delta}{2}}\frac{f(a)}{2}dt=\frac{\delt a}{2}\frac{f(a)}{2}<0$
.
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+ $\delta$

**anonimnystefy** · August 28th 2011, 02:25 AM

Use TEX instead of tex!!!

**transgalactic** · August 28th 2011, 03:45 AM

it works now

**Plato** · August 28th 2011, 03:55 AM

Originally Posted by transgalactic

% Preview source code from paragraph 0 to 12
there is integrabile function f in {[}a,b{]} so f(a)<0.
f is continues from the right of a.
prove that there is $c\in(a,b)$ so $\intop_{a}^{c}f(x)dx<0$
prove:
there is $\epsilon=\frac{{|f(a)|}}{2}>0$ from the given continuety
in a there is a delta
$b-a>\delta>0$ so for every x in which $a\leq x<x+\delta$ $|f(x)-f(a)|<\varepsilon$ .
so $f(x)-f(a)<-\frac{f(a)}{2}$ and the ({*}) $f(x)<f(a)/2<0$
f is integrabile in [a,b] and so in [a,a+ $\delta/2$ ]
from ({*}) and from the monotonicity we conclude
$\intop_{a}^{a+\frac{\delta}{2}}f(t)dt\leq\intop_{a }^{a+\frac{\delta}{2}}\frac{f(a)}{2}dt=\frac{\delt a}{2}\frac{f(a)}{2}<0$
.
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+ $\delta$

"why they chose such delta and x??"
That is directly from the given: $f$ is continuous on the right at $a$ .
We want an open interval on which $f$ is negative, so the integral will be negative.

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