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Limit of a Reimann Sum

  1. Nov 11, 2013 #1

    Qube

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    1. The problem statement, all variables and given/known data

    https://scontent-b-mia.xx.fbcdn.net/hphotos-prn2/v/1419845_10201044047645089_1286462043_n.jpg?oh=adc74f67f112c0697cbfba79b4fa81fc&oe=5283F9AB

    2. Relevant equations

    delta x = (b-a)/n

    3. The attempt at a solution

    Well, from the delta x formula I can figure out the limits of integration. They're 4 and 0. That leaves us with three possible answer choices. I'm suspecting that the 4i/n term goes away and the answer is B, but I really don't know and I'm not even sure where to begin.
     
  2. jcsd
  3. Nov 11, 2013 #2

    Mark44

    Staff: Mentor

    You have the interval [0, 4] that you will divide into n subintervals of equal length. How would you write xi, the x value in the i-th subinterval? The x value could be at the left or right end of a given subinterval, or somewhere in the middle of it.
     
  4. Nov 11, 2013 #3

    Qube

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    I'm not sure what terms to write xi in terms of. I guess, (x/n) would give me the width of each subinterval and I'm not sure what else.
     
  5. Nov 11, 2013 #4

    Mark44

    Staff: Mentor

    No, the width of each subinterval would be 4/n. Since the summation has cos(2 + ...), that's going to show up in the integral as well.
     
  6. Nov 12, 2013 #5

    Qube

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    So the integral would have cos (2+x) as the integrand?
     
  7. Nov 12, 2013 #6

    Mark44

    Staff: Mentor

    Yes. Do you see how it works? Since i is running from 1 to n, 4i/n represents the x value at the right side of each subinterval, and cos(2 + 4i/n) is the function value associated with that x value.
     
  8. Nov 12, 2013 #7

    Qube

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    Alright, I see :)! 4/n is the width of each sub interval. The i represents each sub interval.
     
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