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I need to get a few things straight about the integration operation (as an intro calc student). I understand that integration is a process that takes a function and returns its antiderivative. We can think of it as an operator, where $\displaystyle \int...dx$ is kind of like an opening and a closing bracket for an input function. This is how I interpret integration: the summa and the differential (the closing "bracket") are inseparable, since they are part of the same notation. However, my professor confused me with his derivation of velocity under constant acceleration:

$$\displaystyle \frac{dv}{dt} = a$$ $$\displaystyle v\frac{dv}{dt} = va$$ $$\displaystyle v\frac{dv}{dt} = a\frac{dx}{dt}$$ Then he "cancels" the $dt$ $$vdv = adx$$ The next part is what confuses me: $$\int_{v_1}^{v_2}vdv = a\int_{x_1}^{x_2}dx$$ which comes out to be $$\frac{1}{2}v_2^2 - \frac{1}{2}v_1^2 = a(x_2 - x_1)$$

My primary question is, how did the two summas appear, if there were no corresponding differentials at the time of application? This is disconcerting because the same operation is supposed to be applied to both sides of the equation, and those look like two different operations in terms of two different variables. Shouldn't he have done something like $\displaystyle \int_{t_1}^{t_2}vdv~dt = \int_{t_1}^{t_2}adx~dt$?

If it's no trouble, I have two additional questions. Why does he use $\displaystyle \int_{v_1}^{v_2}...dv$ rather than $\displaystyle \int...dv$? Also, what justifies that he "cancels" the dt differentials?

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  • $\begingroup$ Sorry, I wrote my answer before actually reading all of the way through your question. When you ask "Why does he use $\displaystyle \int_{v_1}^{v_2}...dv$ rather than $\displaystyle \int...dv$?" do you mean that you don't know what a definite integral is or are you just asking why your professor chose to use definite integration as opposed to indefinite integration? $\endgroup$ – user137731 Dec 1 '15 at 2:04
  • $\begingroup$ I know what a definite integral is. I am just asking why he chose to use a definite integral rather than an anitderivative. I want to know why, since using the definite integral derives the equations correctly, while the indefinite integral does not. $\endgroup$ – Wesley Dec 1 '15 at 2:14
  • $\begingroup$ Applying an indefinite integral to both sides would also work, but indefinite integrals are by nature less definite. Meaning you'd have that pesky $+C$ hanging around and would have to use some initial conditions to get rid of it. $\endgroup$ – user137731 Dec 1 '15 at 2:19
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So you can get to here without trouble:

$$\displaystyle v\frac{dv}{dt} = a\frac{dx}{dt}$$

The next step is simply to integrate both sides with respect to $t$ from $t_1$ to $t_2$:

$$\int_{t_1}^{t_2} \displaystyle v\frac{dv}{dt}dt = \int_{t_1}^{t_2} a\frac{dx}{dt}dt$$


Now we just need to show that the $dt$'s "cancel". To do so we just need to use $u$-substition. Forgetting about the RHS for the moment, we start with

$$\int_{t_1}^{t_2} v(t)\frac{dv(t)}{dt}dt$$

Let $u=v(t)$, then $du = \frac{dv(t)}{dt}dt$. So

$$\int_{t_1}^{t_2} v(t)\frac{dv(t)}{dt}dt = \int_{u(t_1)}^{u(t_2)} udu$$ or changing the variable name back to $v$, we have $$\int_{t_1}^{t_2} v(t)\frac{dv(t)}{dt}dt = \int_{v(t_1)}^{v(t_2)} vdv$$

Then do the same thing for the RHS.

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  • $\begingroup$ So this is the most rigorous way to approach this scenario? Are what he does (cancel the dt's in the beginning, and integrate without having the dt's as closing brackets) just short-cuts that happen to lead to the same solution? $\endgroup$ – Wesley Dec 1 '15 at 2:12
  • $\begingroup$ This is absolutely not the most rigorous way, but it's more rigorous than simply saying "now we cancel the $dt$'s". To make it more rigorous I'd have to change the notation in a way that I'm not sure you'd understand and I think this way gets the point across. As for why physics (and even sometimes math) professors say that you can cancel the $dt$'s, most of the time you'll get the right answer and thus it lets them skip little steps like the above $u$-substitution. $\endgroup$ – user137731 Dec 1 '15 at 2:15
  • $\begingroup$ Alright, that makes sense. I have more question. We are using u-substitution to circumvent cancelling out the differentials. However, in doing u-substitution, we are moving the differentials around as if they were separate entities, which to my knowledge is not all too rigorous. So why do we get a free pass when do it in the u-substitution? $\endgroup$ – Wesley Dec 1 '15 at 2:32
  • $\begingroup$ $u$-substitution is always presented in a rather unsatisfactory way in Calc I. Wait until you get to Real Analysis (or find a book and start looking through it now if you want). That's the class where you'll see the rigorous statements of the wishy-washy stuff that you learned your first time through calculus. $\endgroup$ – user137731 Dec 1 '15 at 2:34

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