Chain rule question

1. bitrex

195
I'm having some trouble following this equation:

$$\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)] \$$

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. CompuChip

4,296
There is a theorem in calculus that
$$\frac{d}{dx} \int_0^x f(y) \, dy = f(x)$$
(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 $x_C - w/2 < x_0 < x_C + w/2$ and write
$$\int_{x_C - w/2}^{x_C + w/2} dx B(x) = \int_{x_C - w/2}^{x_0} dx B(x) + \int_{x_0}^{x_C + w/2} dx B(x) = \int_{x_0}^{x_C + w/2} dx B(x) - \int_{x_0}^{x_C - w/2} dx B(x)$$

3. bitrex

195
Thank you! That's a good theorem to know!

4. HallsofIvy

41,265
Staff Emeritus
Yes, it's called the Fundamental Theorem of Calculus!

5. bitrex

195
Well, I suppose if I can't recognize that, this is all kind of a lost cause!

6. elibj123

240
actually a more general theorem is:

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