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Homework Help: Small integration question

  1. Apr 11, 2010 #1
    I wanted to see the proof for the energy stored in an inductor equation was,
    But i had trouble understand how part of this integration works i.e my math sucks

    -

    given [tex] \frac{dw}{dt} = iv [/tex]

    and [tex] VL = L \frac{di}{dt} [/tex]

    -

    solving i get,
    subtituting the ldt/dt for v

    [tex] \frac{dw}{dt} = i L \frac{di}{dt} [/tex]

    the dt's cancel out

    [tex] dw = i L di [/tex]

    then to find the work done I integrate the equation,

    [tex] \int dw = \int i L di [/tex]

    and the integral of [tex] \int dw = w [/tex]

    but how do I integrate

    [tex] \int i L di [/tex] ?

    I know L, the inductance of the inductor is constant so can pull that outside the integral,
    and get
    [tex] L \int i di [/tex]

    But what do I do here?

    integrating i I get [tex] \frac{1}{2} i ^2 * i [/tex] according to the integration rules i've learnt

    but everyone knows that the energy inside an inductor equation is

    [tex] wL = \frac{1}{2} L i^2, [/tex]

    but my integration shows it is [tex] \frac{1}{2} L i^3 [/tex]

    can someone please explain it to me
     
  2. jcsd
  3. Apr 11, 2010 #2

    rock.freak667

    User Avatar
    Homework Helper

    First I think this equation

    [tex]
    VL = L \frac{di}{dt}
    [/tex]

    should be

    [tex]V=L\frac{di}{dt}[/tex]

    Well you got as far as W= L∫i di.


    Now remember the rules of integration

    [tex]\int x^n dx = \frac{x^{n+1}}{n+1}[/tex]

    so what does your integral become?
     
  4. Apr 11, 2010 #3
    when I wrote VL i ment the voltage across the inductor Vl,

    [tex] \int i^1 di = \frac{i^(1+1)}{1+1} [/tex]

    I just always assumed that the integral of something [tex] \int di = i [/tex]
    or [tex] \int dx = x [/tex]

    since when I integrate, [tex] \int dw = L \int i di [/tex]
    I get [tex] w = \frac{1}{2} L i^2 [/tex]

    -

    I can see now that the di disappears, and all i'm integrating is i, which I can do easily
     
  5. Apr 11, 2010 #4

    rock.freak667

    User Avatar
    Homework Helper

    yes those are correct, but you were integrating i w.r.t i. Not just 1.
     
  6. Apr 11, 2010 #5
    so when I integrate x w.r.t x, I get the integral of x only?

    [tex] \int x dx = \int x = \frac{1}{2}x^2 [/tex]

    but when I integrate dx,

    [tex] \int dx = x [/tex] ?

    so that dx after an equation means with respect to dx?
    like in an integral equation, [tex] \int x dx [/tex] is means integrating x with respect to x?
     
  7. Apr 11, 2010 #6

    rock.freak667

    User Avatar
    Homework Helper

    Yes. But the 'dx' does not just disappear.

    [tex]\int 1 dx = x+C[/tex]


    [tex]\int x dx = \frac{1}{2}x^2 +C[/tex]

    (constant C for indefinite integrals)
     
  8. Apr 11, 2010 #7
    ok thanks, I'll figure out how differential/integral equations are set up properly

    instead of just using my cheap way and plugging everything into

    [tex]
    \int x^n dx = \frac{x^{n+1}}{n+1}
    [/tex]
     
  9. Apr 11, 2010 #8

    Mark44

    Staff: Mentor

    The differential part in an integral tells you what the variable of integration is. Don't leave it out as you did here. Also, don't leave out the constant of integration.
    It should be
    [tex] \int x dx = \frac{1}{2}x^2 + C[/tex]

    In this integral, a is the variable of integration.
    [tex] \int a da = \frac{1}{2}a^2 + C[/tex]

    In this one, x is the variable of integration, and a is just a constant.
    [tex] \int a dx = a \int dx = ax + C[/tex]
    In the integral above I used a property of integrals - the integral of a constant times a function is the constant times the integral of the function. Here the function is 1.
     
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