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Computing Integrals

  1. Jan 22, 2005 #1
    I was having trouble starting up this question.

    Let
    ------| 2, if x<6
    f(x) ={ 7, if x=6
    ------|-3, if x>6.

    Compute (the integral from 3 to 11) f and prove that your right.
    The problem i was having was choosing a partition for this integral.
    For example, [3,x,y,11], what type of values could i use that are close enough to 6. Is 5 and 7 close enough? or something like +/- epsilon/2(this part is what confuses me, the epsilon)?
     
  2. jcsd
  3. Jan 23, 2005 #2

    ehild

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    No, 5 and 7 are not close enough to 6. You have to integrate across the open intervals [3,6) and (6,11]. This means integrating from 3 to (6-epsilon) and then from (6+epsilon) and taking the limit epsilon -->0.

    [tex]\int_3^{11} {f(x)dx}= \lim_{\epsilon\rightarrow 0}[\int_3^{6-\epsilon}{2dx} + \int_{6+\epsilon}^{11}{-3dx}}][/tex]


    ehild
     
  4. Jan 23, 2005 #3
    oh ok, i was workin on it but i used delta instead, tho i will change it to epsilon, but does it matter as much since delta is smaller than epsilon.
     
  5. Jan 23, 2005 #4
    btw just wanted to make sure, i got the -9 as the integral. Could some1 check and tell me if its right? Also if i wanted to prove that i am right(-9), what r the steps involved in proving this. So far i used like the definition of Riemann Integrability Criterion, is that enough, if yes/no, any tips of what i should mention exactly?
     
  6. Jan 24, 2005 #5

    ehild

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    It does not matter what notation you use for the "infinitesimal small quantity".

    ehild
     
  7. Jan 24, 2005 #6

    ehild

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    It is correct. The integral is -9. RIC is OK, but you should refer to the additivity of integration and the definition of improper integrals. You also should say that the integral under one single point is zero. You prove the statement that the integral is -9 when you derive it from the principles.

    ehild
     
  8. Jan 24, 2005 #7
    You have to pick an interval such that when you look at the integral the discontinuities are 'removable'. Such an interval would have a width of epsilon around the discontuities for the function

    In this case [tex] For \epsilon > 0, P_{\epsilon} = [3,6-\epsilon,6+\epsilon,11] [/tex]

    Hereafter use the Riemann sums to compute the integral. Thus you can show what the function is really.

    As an aside, this kind of question sounds familiar because a prof at my university worded questions like that. You wouldn't happen to be at York University in Toronto, Ontario would you?
     
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