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Finding area under cosx using midpoint rule

  1. Dec 18, 2011 #1
    Use the midpoint rule to approximate the following integral:
    ∫sin(x) dx


    This is what I did:
    Δx = (1-0)/4 = 1/4

    1/4(f(1/8) + f(3/8) + f(5/8) + f(7/8) + f(9/8))

    But the answer i get is wrong. Is that the correct midpoint rule formula, and are the values I plugged in right? Any help is appreciated, thanks.
     
  2. jcsd
  3. Dec 18, 2011 #2

    LCKurtz

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    Are the limits on your integral 0 and 1?. How many points in your partition? Is 9/8 in your interval?
     
  4. Dec 18, 2011 #3
    Ye, its from 1 to 0, and n = 4.
     
  5. Dec 18, 2011 #4

    LCKurtz

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  6. Dec 18, 2011 #5
    I tried with and without 9/8, both don't give me the right answer.
     
  7. Dec 18, 2011 #6

    LCKurtz

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    Why would you try it with 9/8 in the first place??? Unless you show us your work how can we help you find what you are doing wrong? Maybe something simple like having your calculator in degree mode instead radians? Show us your calculations.
     
  8. Dec 18, 2011 #7
    ohh, i got the answer, u were right, the calculator was suppose to be in radians mode -_- oops. Thanks.
     
  9. Dec 19, 2011 #8

    NascentOxygen

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  10. Dec 19, 2011 #9

    NascentOxygen

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    But title says: Finding area under cosx using midpoint rule

    Better establish whether sine or cosine :confused:
     
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