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Integration where am I going wrong?

  1. Jan 15, 2016 #1
    Hi,
    I have an example in my studies as follows:
    If v = 200sin(100pi*t + 0.2) then evaluate Integral of v^2 dt between limits of 0.005 and 0.

    I have integrated it and used the double compound angle formula sin^2 A=1-cos2A and come up with the following as per the solution to the example in the home work studies:
    20000 [t - (sin(200pi*t + 0.4)/200pi) ] ... so far so good I think.

    However when I put the limits in it I get

    20000 [0.005 - (sin(200pi*0.005 + 0.4)/200pi)) - (0 - (sin(200pi*0 + 0.4)/200pi)) ]

    This I calculate as 20000 (0.005 + 0.0061) - (0 -0.0061) = 344

    According to the solution and online examples the answer should be 124.8

    The solution with the limits calculated shows 20000 [ (0.005+0.00062) - (0 - 0.00062)] but I can't get these figures from what I have above. Please can anyone see where I am going wrong? It is driving me mad!
     
  2. jcsd
  3. Jan 15, 2016 #2

    Samy_A

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    Homework Helper

    You miss a term in ##20000 [t - (\sin(200\pi t + 0.4)/200\pi) ]##
    Carefully redo the integral.
    Post your full calculation, so that we can help finding a possible error.
     
  4. Jan 15, 2016 #3
    Hi Thank you very much for your reply.

    I have just twigged my error in my calculation - I've literally been at it 3 hours!

    I don't believe I have missed a term the error was when dividing by 200pi on the calculator. I was dividing [sin(200pi(0.005)+0.4)] by 200pi not by (200pi).

    Glad that is sorted now it was giving me a real head ache and I thought dividing the sum by 200 (pi symbol) on the calculator would work fine, but it appears i needed to divide the sum by [200(pi symbol)]

    Thanks again.
     
  5. Jan 15, 2016 #4

    HallsofIvy

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    If you enter, on your calculator , "A/B*C" the calculator will interpret that as (A/B)*C. If you want A/(BC) you need to use parentheses.
     
  6. Jan 17, 2016 #5
    That's indeed how I was going wrong. Thanks for the clarification on how it works. It confirmed my suspicions.
     
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