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Integration Help!

  • Thread starter Pete_01
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  • #1
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Homework Statement


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?
 

Answers and Replies

  • #2
33,644
5,311
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.
 
  • #3
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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.
 
  • #4
eumyang
Homework Helper
1,347
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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?
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
 
  • #5
51
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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
Ok, but in my problem the E and R are constants so can't I take them out?
 
  • #6
33,644
5,311
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.
 
  • #7
51
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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.
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
 
  • #8
rock.freak667
Homework Helper
6,230
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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
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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