# A really basic integration question concerning differentials

I'm really, really confused with this. Please, please help me.  My Calculus teacher taught me that the integral symbol and the differential with respect to which we are integrating are like parenthesis. He told us to think of the integral sign as an “open parenthesis” and the $dt$ as a “close parenthesis”.
If we were to integrate any function of $t$, say $v(t)$, we have to put the integral sign on the left of $v(t)$, and the differential $dt$ on the right of $v(t)$ ie $\int v(t) dt$  In physics, while while deriving equations of motion, our physics teacher did this: since v is a linear function of t, $$dv(t)/dt=a(constant)$$ $$\Rightarrow dv= a dt$$ She then simply put an integral sign on LHS and RHS to integrate and then got $$\int dv= \int a dt$$ $$v=at+C$$ But this does not fit into what my Calculus teacher had taught us. As per what he has taught: $$dv= a dt$$ We now have to add an integral sign and another differential on either side of the expressions in LHS and RHS respectively. Only then can we integrate. Could somebody please explain this idea of the integral sign as an “open parenthesis” and the $dt$ as a “close parenthesis”? Please could you explain how this is applied to the physics example?

• Why do you have to add another differential? You already have a differential on both sides, you just need to add the integral sign. – Gregory Grant May 24 '15 at 14:13
• Sir, as per what he had taught, regardless of whether a differential exists, we have to put an integral sign on the left and a differential on the right if we are to integrate something. I don't think this is correct, but still I am a complete novice to Calculus. – User1234 May 24 '15 at 14:16
• Sir, could you please explain these steps? $$dv(t)/dt=a(constant)$$ $$\Rightarrow dv= a dt$$ $$\int dv= \int a dt$$ $$v=at+C$$ Shouldn't it be $$\Rightarrow dv(t)= a dt$$ – User1234 May 24 '15 at 14:26
• I'm not sure how to explain it any better, hopefully somebody else will chime in – Gregory Grant May 24 '15 at 14:28
• Sir, just one last doubt. If $v(t) = At^2+Bt+c$ $$\dfrac{dv(t)}{dt} = 2At+B$$ So how can we follow my physics teacher's method in this scenario? – User1234 May 24 '15 at 14:30

## 3 Answers

No doubt you already know the fact $$\int dy=y+c$$ where $y$ can be any (sensible) thing whatsoever. You probably also know that $$\int \frac{dy}{dt} dt=y+c$$ since integration undoes differentiation (almost), by the fundamental theorem of calculus (you might not recognise that name, even if you recognise the fact). Indeed, you probably also know the chain rule which states $$\frac{dy}{dt}=\frac{dy}{du}\frac{du}{dt}$$ where $y$ and $u$ are any sensible quantities. People who aren't pure mathematicians (and especially physicists) tend be abit risqué and like to treat differentials like they're numbers, and write things like $$dv=\frac{dv}{du}du$$ since we can "divide by $dt$" (or by any other differential) and get the chain rule. While what your teacher is doing is absolutely incorrect, it's safe because of these theorems working in the background, and lets your notation be a bit more elegant.

In the example you give, what is really happening is: $$\frac{dv}{dt}=a \implies \int \frac{dv}{dt}dt=\int a dt \implies v=at+c$$

Which your teacher has abbreviated to $$dv=adt \implies \int dv = \int adt \implies v=at+c$$

P.S. At its core this isn't a basic question, and you shouldn't be worried about being confused by it.

• NB: When I use the word incorrect, I don't mean that it gives the wrong answer. Just that it's completely unjustified. Lots of things in physics are unjustified, and that shouldn't alarm you. – preferred_anon May 24 '15 at 14:53
• Thanks Sir.This has really cleared up most of my doubts! – User1234 May 24 '15 at 14:57
• Sir, I couldn't understand what you meant by: "divide by $dt$". – User1234 May 24 '15 at 14:58
• This is an example of treating differentials as if they were ordinary real numbers: If we have the equation $3x=9$ then we divide by $3$ to get $x=3$. This is not defined for differentials, but if we pretend it is then we get $dv=adt \implies \frac{dv}{dt}=a\frac{dt}{dt}=a$, for example. – preferred_anon May 24 '15 at 15:32
• Sir, but then if division is not defined for differentials, multiplication too should not be defined (as per my understanding). Then how did we solve this step: $$\frac{dv}{dt}=a \implies \int \frac{dv}{dt}dt=\int a dt \implies v=at+c$$ – User1234 May 24 '15 at 15:45

Your question is one of notation only. Think of an indefinite integral as an antiderivative and a definite integral as a limit of Riemann sums. The Fundamental Thm of Calculus provides a link between them.

The $dx$ and the $\int$ really only remind us of where these integrals "come from", but this is not necessary for understanding what they are. In fact, it is quite common to see just the symbol $\int f$ and everyone knows what you mean.

Because of chain rule, you can treat $\frac{dy}{dx}$ as a fraction. Some mathematicians are annoyed by this.

We now have to add an integral sign and another differential on either side of the expressions in LHS and RHS respectively. Only then can we integrate.

You don't need to do that. If you have a differential on both sides, you can integrate.

Also, some physicists like to write integrals like this:

$$\int dx f (x)$$

Also some people do not even write the $dx$

• If $v(t) = At^2+Bt+c$ $$\dfrac{dv(t)}{dt} = 2At+B$$ Sir how can we follow my physics teacher's method in this scenario? How do we now get v(t) following her steps? – User1234 May 24 '15 at 14:33
• You multiply both sides by $dt$ $$dv=(2At+B)dt$$integrate$$\int dv=\int (2At+B)dt$$and get$$v=At^2+Bt+C$$ – grdgfgr May 24 '15 at 14:36
• @grdfgr Sir, when you multiplied both sides by $dt$, shouldn't you get $$dv(t)=(2At+B)dt$$ – User1234 May 24 '15 at 14:50
• I didn't multiply anything by 11, and I am unable to see a difference. – grdgfgr May 24 '15 at 14:51
• Oh, Sirry sir, I meant to write $dt$. Really sorry – User1234 May 24 '15 at 14:52