
#1
Nov2609, 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_Cw/2}^{x_C+w/2} dx B(x)\right] \frac {dx_C}{dt} = () v\ell [ B(x_C+w/2)  B(x_Cw/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
Nov2709, 09:16 AM

Sci Advisor
HW Helper
P: 4,301

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] 



#3
Nov2809, 01:21 AM

P: 196

Thank you! That's a good theorem to know!




#4
Nov2809, 06:20 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,879

Chain rule question
Yes, it's called the Fundamental Theorem of Calculus!




#5
Nov2809, 03:57 PM

P: 196

Well, I suppose if I can't recognize that, this is all kind of a lost cause!




#6
Nov2909, 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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