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Homework Help: Integration Help!

  1. Aug 2, 2010 #1
    1. The problem statement, all variables and given/known data
    It's been a while since I've done some of these less common integrals. This is for a physics problem but I need to integrate....

    int from 0 to 2sec: (emf^2/R) (1-e^(-Rt/L))
    basically integrating voltage*current for an RL circuit.

    How do I do this?
     
  2. jcsd
  3. Aug 2, 2010 #2

    Mark44

    Staff: Mentor

    Is this the integral?
    [tex]\int_0^2 \frac{E^2}{R}(1 - e^{-Rt/L})dt[/tex]

    I'm assuming that the only variable is t. Split the integral above into two integrals, the first of which is very easy. For the second integral, use the substitution u = -Rt/L, and du = -R/L dt.
     
  4. Aug 2, 2010 #3
    ok so the first integral [tex]\frac{E^2}{R}[/tex] can be pulled out leaving:

    [tex]\frac{E^2}{R}\int_0^2 \frac{-E^2}{R}(e^{-Rt/L})dt[/tex]

    correct?

    can you pull out the other E^2/R and have just:

    [tex]-\frac{E^4}{R^2}\int_0^2 (e^{-Rt/L})dt[/tex]

    and then by u substitution get:

    [tex]-\frac{E^4}{R^2}\int_0^2 (e^{u})du[/tex] ?

    Sorry, haven't done these for a lonng while.
     
  5. Aug 2, 2010 #4

    eumyang

    User Avatar
    Homework Helper

    No. You need to split the integrals using
    [tex]\int (f(x) - g(x)) dx = \int f(x) dx - \int g(x) dx[/tex]

    Distribute the fraction, then split the integral.


    69
     
  6. Aug 2, 2010 #5
    Ok, but in my problem the E and R are constants so can't I take them out?
     
  7. Aug 2, 2010 #6

    Mark44

    Staff: Mentor

    If you take out E^2/R first, you will have two terms in the integrand. If you do it that way, I would still recommend splitting the integral into two parts.
     
  8. Aug 2, 2010 #7
    Could someone please go step by step and show me how to do this? It's such a small part of the original problem. Thank you
     
  9. Aug 2, 2010 #8

    rock.freak667

    User Avatar
    Homework Helper

    You have this right?

    [tex]\int_0^2 \frac{E^2}{R}(1 - e^{-Rt/L})dt[/tex]

    So let's do as Mark44 said and take out the E2/R. You will be left with


    [tex]\frac{E^2}{R} \int_0^2 (1 - e^{-Rt/L})dt[/tex]

    Are you able to evaluate this now?
     
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