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Integrability of a step function

  1. Jan 26, 2004 #1
    Let f (x) = 1 if 2<=x<4
    2 if x =4
    -3, if 4<x<=7
    Prove that this function is integrable on [2,7], state its value and prove that it is what you say it is.
    Obviously integral of f from [2,7] is -7. but its proof and the integrability have me and my friends snagged.

    Suggestions anyone?

    SO far we have the idea that we have to prove that U(f,P) - L(f,P) , Epsilon
    We computed that U(f,P) - L(f,P) =5 (Tj - Tj-1) where Tj and Tj-1 is the subinterval in which Tj-1<2<Tj. However we are stuck from here.

    My notation is from a book called Calculus by Spivak. Basically U(f,P) is the upper sum and L(f,P) is th lwoer sum for a partition P of the interval. P = {A=T0,T1,T2,...,TJ-1,TJ,...TN=B}and A and B are the endpoints of hte interval.
     
  2. jcsd
  3. Jan 26, 2004 #2
    A littel correction U(f,P) - L(f,P) must be strictly LESSSER than Epsilon.
     
  4. Jan 26, 2004 #3
    Use the partition P={2,4-a,4,4+a,7}, where a>0 is small. You have L(f,P)=1(4-a-2)+1(4-4+a)+(-3)(4+a-4)+(-3)(7-4-a)=-7. You have U(f,P)=1(4-a-2)+2(4-4-a)+2(4+a-4)+(-3)(7-4-a)=-7+2a. You must have [tex]-7\leq \sup \{L(f,Q)|\textrm{Q a partition of [2,7]}\}[/tex] and [tex]\inf \{U(f,R)| \textrm{R a partition of [2,7]}\}\leq -7+2a[/tex]. You can use a simple epsilon argument to show that the two are equal, and thus integrable.
     
    Last edited: Jan 26, 2004
  5. Jan 26, 2004 #4

    NateTG

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    Given [tex]\epsilon[/tex] find [tex]\delta[/tex] so that if [tex]|t_n-t_{n+1}|<\delta \forall t_n \in P \rightarrow U(f,P)-L(f,P) < \epsilon[/tex]

    Hint: If you can find a monotone increasing function [tex]g[/tex] so that [tex]g(\delta)<U(f,P)-L(f,P)[/tex] then you have [tex]g^{-1}(\epsilon)[/tex] as a solution.

    In this particular situation, you can also try manually calculating the maximum and minimum with intevals of varying lenghts. Remeber to have intervals that span the discontinuities.

    You can also try to find [tex]\delta[/tex] for a couple of [tex]\epsilon[/tex] I suggest trying 1, .1,.01,.001,.0001.
     
    Last edited: Jan 26, 2004
  6. Jan 26, 2004 #5
    i think i'll use NateTG's given statement whichi think my prof would accept. I would think i could put delta = min (3,epsilon/5)
    It works out fine that way. Thanks for the help. Alas i found out this forum, and idea, a dya AFTER the assignemtn was due...
     
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