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Homework Help: Use Upper And Lower Sums To Evaluate An Integral

  1. Aug 27, 2010 #1
    1. The problem statement, all variables and given/known data

    The question is to use upper and lowers sums, Un and Ln, on a regular petition of the intervals to find the integral from 1 to 2 of f(x) = [[x]], where [[x]] is the greatest integer function.

    2. Relevant equations

    [tex]\Delta[/tex]x = [tex]\frac{b-a}{n}[/tex]

    3. The attempt at a solution

    [tex]\Delta[/tex]x = [tex]\frac{2-1}{n}[/tex] = [tex]\frac{1}{n}[/tex]

    The minimum and maximum of f(x) on every subinterval of [1,2] would be 1 except for the subinterval which includes x=2 where the maximum value of f(x)=2, so for f(x) there exists a unique number I such that Ln [tex]\leq[/tex] I [tex]\leq[/tex] Un

    I know that the area will be 1, as for [1,2) Un = Ln = [tex]\Delta[/tex]x[tex]\sum[/tex]1 = [tex]\frac{1}{n}[/tex] x 1n = 1

    However I am not sure how to include the subinterval that contains x=2 into my calculations. Any help on this would be great!

  2. jcsd
  3. Aug 27, 2010 #2


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    Staff Emeritus
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    Gold Member

    If you have n subintervals, evenly spaced say, then n-1 will have f(x) be just 1, and one will contain a point where f(x) is 2. So your lower sum, using the lowest value on each interval, will just give you the area of 1 as expected. For the upper sum, you have

    [tex](\sum_{i=1}^{n-1} \frac{1}{n})+\frac{2}{n}[/tex] adding up all the parts where the upper bound is 1, plus the one part where the upper bound is 2. Can you find the limit of that summation as n goes to infinity? (Of course we know it should be 1)
  4. Aug 27, 2010 #3
    That makes perfect sense! Thanks a bunch :D
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