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Thread: integrbility proof

  1. #1
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    integrbility proof

    % 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 $\displaystyle c\in(a,b)$ 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+$\displaystyle \delta$
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  2. #2
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    Re: integrbility proof

    Use TEX instead of tex!!!
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    Re: integrbility proof

    it works now
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    Re: integrbility proof

    Quote Originally Posted by transgalactic View Post
    % 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 $\displaystyle c\in(a,b)$ 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+$\displaystyle \delta$
    "why they chose such delta and x??"
    That is directly from the given: $\displaystyle f$ is continuous on the right at $\displaystyle a$.
    We want an open interval on which $\displaystyle f$ is negative, so the integral will be negative.
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