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Lim from pos for an integral

  1. Dec 15, 2008 #1
    Prove: If f is integrable on [a,b] then
    lim [tex]\int[/tex](a,x) f=0

    I put If F is integrable for every E>0, partition,P, U(f,P) - L(f,P) <E
    This also means that f is integrable on [a,a+]

    If [tex]\int[/tex] (x,x) f(x) dx =0
    F(x)-F(x)=0
    Since f is integrable, if can be differentiable (he said that this was wrong)
    lim [tex]\int[/tex] (a,x) f(x) dx
    x[tex]\rightarrow[/tex]a+

    lim [tex]\int[/tex] (a,x) f(x) dx
    x[tex]\rightarrow[/tex]a

    lim [tex]\int[/tex] (a,x) f(x) dx
    x[tex]\rightarrow[/tex]a-

    are all equal then
    lim [tex]\int[/tex] (a,x) f(x) dx
    x[tex]\rightarrow[/tex]a+


    what am i doing wrong?
     
  2. jcsd
  3. Dec 15, 2008 #2
    "[a, a+]" is not an interval.

    The limits of integration should not depend on x.

    Integrability does not imply differentiability. For example, |x| is integral on [-1, 1] but not differentiable at 0.

    It's not clear how the parts of your proof are connected.
     
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