# Why can we make this integral change of limits? Is it obvious?

When deriving the equation for the impulse-momentum theorem, the following occurs:

$$\cdots=\int\limits_{t_1}^{t_2}\frac{d\vec p}{dt}dt = \int\limits_{\vec p_1}^{\vec p_2}d\vec p=\cdots$$

I know the $dt$s aren't simply canceling each other out, so why can we get rid of them and then why do the limits of integration change? Is it because once the $dt$s are gone the limits just change automatically since we're only left with $d\vec p$ or is there something that needs to be done and it just happens that the limits change because of it?

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Compare with $\int_a^b f'(x)\, \mathrm dx=f(b)-f(a)=\int_{f(a)}^{f(b)}\mathrm dx$. – Hagen von Eitzen Jul 29 '14 at 21:32
It's just the substitution rule, really. You might also compare it notationally with a typical statement of the chain rule: $\dfrac{dp}{du}=\dfrac{dp}{dt}\dfrac{dt}{du}$. ('Toss out' the $du$ in there, and you've got your statement.) Note though that this is just shorthand for the math. – Semiclassical Jul 29 '14 at 21:36
@HagenvonEitzen and Semiclassical Thank you for your comments, I think that makes sense. – John Jul 29 '14 at 21:49
It might be pedagogically useful to use the term "differentiation by substitution" instead of "chain rule". – Michael Hardy Jul 29 '14 at 22:09

This is the chain rule. Sometimes the chain rule is stated as $\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot\dfrac{du}{dx}$, just as if $du$ literally refers to a number and cancels.
$$\int_a^b \frac{dp}{dt}\,dt = \int_{t=a}^{t=b} f'(p(t))p'(t)\,dt \tag 1$$ where $f'$ is a constant function equal to $1$. So $f(p)=p+\text{constant}$. So by the chain rule $(1)$ is $$\int_{t=a}^{t=b} (f\circ p)'(t)\,dt = f(p(b))-f(p(a)) = \int_{p=p(a)}^{p=p(b)} f'(p)\,dp = \int_{p=p(a)}^{p=p(b)} 1\,dp.$$