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Show that f is integrable

  1. Jan 20, 2009 #1
    1. The problem statement, all variables and given/known data
    Let f be a function on [0,3] so that f(x) = 1, for x in [0,3); 17, for x = 3. Show that f is integrable on [0,3] and the value of the integral is 3.


    2. Relevant equations
    Upper and lower sums


    3. The attempt iat a solution
    I think this is a valid proof. Let e > 0. Define a partition P = {0, 3 - e/32, 3}.

    Then U(f,P) = 1*(3 - e/32) + 17*(e/32) = 3 + e/2. Clearly, L(f,P) = 3. Hence, U(f,P) - L(f,P) = e/2 < e. Hence f is integrable.

    Now [tex]L(f,P) \leq 3 \leq U(f,P)[/tex]. Since f is integrable, there is only one number between all the lower and upper sums, hence the value of the integral is 3.
     
  2. jcsd
  3. Jan 20, 2009 #2

    Dick

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    I think that works just fine.
     
  4. Jan 21, 2009 #3
    Hmmm, cool. The only real change I made to this in the actual proof is letting e' = min{1,e}, in case someone wanted to choose a morbidly large epsilon.
     
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