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Math Help - 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 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
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  2. #2
    Member anonimnystefy's Avatar
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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 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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