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Chain rule question

by bitrex
Tags: chain, rule
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bitrex
#1
Nov26-09, 08:21 PM
P: 196
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.
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CompuChip
#2
Nov27-09, 09:16 AM
Sci Advisor
HW Helper
P: 4,300
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]
bitrex
#3
Nov28-09, 01:21 AM
P: 196
Thank you! That's a good theorem to know!

HallsofIvy
#4
Nov28-09, 06:20 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,346
Chain rule question

Yes, it's called the Fundamental Theorem of Calculus!
bitrex
#5
Nov28-09, 03:57 PM
P: 196
Well, I suppose if I can't recognize that, this is all kind of a lost cause!
elibj123
#6
Nov29-09, 01:30 PM
P: 240
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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