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Homework Help: Riemann sums

  1. Feb 10, 2006 #1
    Ok, so I have two questions regarding something I don't understand in my textbook (Adams)

    0 if 0<=x<1 or 1<x<=2
    f(x) = 1 if x=1

    (by "<=" i mean less than or equal)

    I'm supposed to show that it is Riemann integrable on that interval.
    They chose P to be: {0, 1-e/3, 1+e/3,2}
    L(f,P)=0 (I think I understand this, its just the sum of zero?)

    They then get U(f,p)= 0(1-e/3) +1(2e/3) +0(2-(1+e/3))=2e/3 <-- I have no idea how they get this!

    And how do they chose that specific partition?

    (I get that f is integrable since U(f,P)-L(f,P)=2e/3<e
    (I've used e as epsilon :P)

    Question 2:
    1 if x is rational
    f(x)= 0 if x is irrational

    I have no idea how they get L(f,P) to be 0 and U(f,P) to be 1!

  2. jcsd
  3. Feb 10, 2006 #2


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    Science Advisor

    They chose that specific partition so that the "odd point" x= 1 where f(1)= 1 is in a very small interval. The fact is that 0 to 1-e, 1-e to 1+e, 1+e to 2 (e= epsilon some very small number) would have worked just as well.
    The interval from 0 to 1- e/3 has length 1- e/3, the highest and lowest values of f are both 0 since f is identically equal to 0. That product (the area of the rectangle) is 0(1- e/3)= 0. The interval from 1- e/3 to 1+ e/3 has length 2e/3 and f has highest value 1, lowest value 0 in that interval. The products (areas of highest and lowest rectangles) are (2e/3)(1)= 2e/3 and (2e/3)(0)= 0, respectively. Finally, the interval from 1+ e/3 to 2 is 2- (1+e/3)= 1-e/3 and, again, f is identically equal to 0 so the "upper" and "lower" products are 0(1- e/3)= 0 again.
    Okay, using the lowest values we get L(f,P)= 0+ 0+ 0= 0 and using the highest values we get U(f,P)= 0+ 2e/3+ 0= 2e/3

    Yes, since by taking e going to 0, we can make those the same: the integral, which, in this case, is 0.

    Every interval contains both rational and irrational numbers. What is the lowest value of f in any interval? What is the highest value of f in any interval?
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