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If integral f =0, prove that f(x)=0 for all x in [a,b]

  1. Feb 3, 2009 #1
    1. The problem statement, all variables and given/known data

    Suppose f is continuous on [a,b] with f(x) >= 0 for all x in [a,b].

    If [tex] \int_a^b \! f = 0 [/tex] , prove that f(x)=0 for all x in [a,b].

    2. Relevant equations

    3. The attempt at a solution

    Assume [tex] \int_a^b \! f = 0 [/tex].

    Then the lower and upper sums equal 0 also. I looked at the definitions of these and came up with,

    [tex] \sum_{i=1}^n ( inf\{ f(x) : x_{i-1} \leq x \leq x_i \} )( \bigtriangleup x_i )=0 [/tex].

    This looks obvious to me but that usually means I'm missing something because I don't know where to use that f is continuous.
  2. jcsd
  3. Feb 3, 2009 #2


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    HINT: It may be better to use the Fundamental Theorem of Calculus.
  4. Feb 3, 2009 #3
    I don't "know" that yet. I am supposed to prove this without that.
  5. Feb 3, 2009 #4


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    Suppose there is an x0 in [a,b] such that f(x0)=c>0. Can you show there is a lower sum that is greater than 0?
  6. Feb 3, 2009 #5
    Should I define a partition? If f(x_0)=c and c > 0 then the function must go up to atleast c, and since it is continuous it does not "jump" there.

    I was thinking of making a small partition of a,x_0, and b. I then know what the sup or inf values of the two rectangles (speaking of the hieghts) will be greater than or equal to x_0 for the upper sum.

    Would this do it? I figure that the upper sum and the lower are equal since the integral in the problem was defined as existing?
  7. Feb 3, 2009 #6


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    No. Knowing a particular upper sum is greater than zero doesn't get
    you anywhere all you know is that the integral is less or equal to
    any upper sum.

    But, yes. Make a partition. Use the epsilon-delta definition of continuity to show there is an interval around x0 such that f>c/2 on that interval. The use the endpoints of that interval to define the partition.
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