Chain rule question

  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. CompuChip

    CompuChip 4,299
    Science Advisor
    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. Thank you! That's a good theorem to know!
     
  5. HallsofIvy

    HallsofIvy 40,367
    Staff Emeritus
    Science Advisor

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

    [tex]\frac{d}{dx}\int^{b(x)}_{a(x)}f(s)ds=f(b(x))\frac{db}{dx}-f(a(x))\frac{da}{dx}[/tex]
     
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