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Chain rule question

  1. Nov 26, 2009 #1
    I'm having some trouble following this equation:

    [tex]\frac {d \Phi_B} {dt} = (-) \frac {d}{dx_C} \left[ \int_0^{\ell}dy \ \int_{x_C-w/2}^{x_C+w/2} dx B(x)\right] \frac {dx_C}{dt} = (-) v\ell [ B(x_C+w/2) - B(x_C-w/2)] \ [/tex]

    Shouldn't the differentiation of the bracketed terms "killed" the integration of B(x)? Why is it still evaluated between the limits in the final expression? Thanks for any advice.
  2. jcsd
  3. Nov 27, 2009 #2


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    Homework Helper

    There is a theorem in calculus that
    [tex]\frac{d}{dx} \int_0^x f(y) \, dy = f(x) [/tex]
    (i.e. differentiating with respect to the integral boundary)
    which you should be able to find in your calculus book.

    You can prove your identity by taking any fixed point [itex]x_C - w/2 < x_0 < x_C + w/2[/itex] and write
    [tex] \int_{x_C - w/2}^{x_C + w/2} dx B(x) = \int_{x_C - w/2}^{x_0} dx B(x) + \int_{x_0}^{x_C + w/2} dx B(x) = \int_{x_0}^{x_C + w/2} dx B(x) - \int_{x_0}^{x_C - w/2} dx B(x)[/tex]
  4. Nov 28, 2009 #3
    Thank you! That's a good theorem to know!
  5. Nov 28, 2009 #4


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    Yes, it's called the Fundamental Theorem of Calculus!
  6. Nov 28, 2009 #5
    Well, I suppose if I can't recognize that, this is all kind of a lost cause!
  7. Nov 29, 2009 #6
    actually a more general theorem is:

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