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stunner5000pt
Jan26-04, 10:47 AM
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.

stunner5000pt
Jan26-04, 10:48 AM
A littel correction U(f,P) - L(f,P) must be strictly LESSSER than Epsilon.

Jupiter
Jan26-04, 03:28 PM
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 -7\leq \sup \{L(f,Q)|\textrm{Q a partition of [2,7]}\} and \inf \{U(f,R)| \textrm{R a partition of [2,7]}\}\leq -7+2a. You can use a simple epsilon argument to show that the two are equal, and thus integrable.

NateTG
Jan26-04, 03:49 PM
Given \epsilon find \delta so that if |t_n-t_{n+1}|<\delta \forall t_n \in P \rightarrow U(f,P)-L(f,P) < \epsilon

Hint: If you can find a monotone increasing function g so that g(\delta)<U(f,P)-L(f,P) then you have g^{-1}(\epsilon) 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 \delta for a couple of \epsilon I suggest trying 1, .1,.01,.001,.0001.

stunner5000pt
Jan26-04, 11:08 PM
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...