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Help with integration proof with epsilon-delta

  1. Nov 14, 2005 #1
    Prove that if f is continuous on [a,b] and

    [tex]\int_a^b |f(x)|\,dx = 0[/tex]

    then f(x) = 0 for all x in [a,b].

    so I'll have to use an epsilon delta proof by contradiction here. I'll have to assume that there exists a c such that f(c) != 0 and for all x = f(c)/2, there exists a delta such that |f(x)-f(c)|< epsilon for |x-c| < delta. and then I should make |f(x)| > epsilon /2. This would contradict the original hypothesis...

    But I'm getting confused here...

    Thanks!
     
    Last edited: Nov 15, 2005
  2. jcsd
  3. Nov 14, 2005 #2

    benorin

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    Are you using Riemann integration or Lebesgue integration?
     
  4. Nov 15, 2005 #3
    Riemann integration
     
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