Is there a theorem that says

[mcah5]

$$\frac {d} {dt} \int_{a}^{b} f(x,t) dx = \int_{a}^{b} \frac {\delta} {\delta t} f(x,t) dx$$

From trial and error, it seems true but I can't find it in my textbook anywhere. Am I missing something obvious?

 

[LeonhardEuler]

Yes, this is true provided f and $\frac{\partial f}{\partial t}$ are continuous. This is an example of interchanging limits. Be careful about doing that because in general you can't. This borders on advanced calculus, but my regular calc textbook gives a proof in the appendix. Maybe yours does too.

 

[MalleusScientiarum]

There's a very thorough proof on page 245 of the Bartle if you have access to one of those. Also, note that as a notational thing
$$\frac{\delta}{\delta g}$$
usually denotes a functional derivative, while what you want is \partial, not \delta in LaTeX, like
$$\frac{\partial}{\partial t}$$

 

[mcah5]

Ok, thanks. Just needed to confirm that the statement was true (my physics textbook used it in a proof and never justified the statement, so I wanted to make sure)

 

[amcavoy]

This is only true though when the bounds (a & b) are held constant, correct? I mean, if they were a=u(x) and b=v(x), the limits couldn't be switched.

 

[Galileo]

If the limits themselves are also functions of t, the theorem (called Leibniz integral rule, IIRC) is:

$$\frac {d} {dt} \int_{a(t)}^{b(t)} f(x,t) dx = f(b(t),t)\frac{d b}{d t}-f(a(t),t)\frac{d a}{d t}+\int_{a(t)}^{b(t)} \frac {\partial} {\partial t} f(x,t) dx$$

I know a sufficient condition for this to hold is that f be uniformly continuous in the region of interest.