# A basic differential inside an integral

If i am not mistaken differential is defined like this (please correct me if i am in any way wrong):

If we take infinitesimally small part of a function changes become smaller and we rename them accordingly $\Delta x \rightarrow dx$ and $\Delta y \rightarrow dy$. Than we can define a rate of change (linear coefficient) $k$ like $$\begin{split} \underbrace{k = \frac{\Delta y}{\Delta x}}_{\text{for bigger changes}} \underrightarrow{~~~\scriptsize as~changes~get~smaller~~~} k &=\frac{dy}{dx}\\ \end{split}$$ But we also know that $k$ equals a derivative $\frac{dx}{dy}$ hence: $$\begin{split} \underbrace{\frac{dx}{dy}}_{derivative~-~do~not~sepparate}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! &=\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \!\!\!\!\!\!\!\overbrace{\frac{dy}{dx}}^{rates~of~change~-~can~be~sepparated}\\ \end{split}$$ And now we separate rates of change while we must not separate derivative right? $$\boxed{dy = \frac{dy}{dx}\, dx}$$

Now comes my real question. If i have a definite integral for example: $$\int\limits_{0}^{t} \frac{dx}{dt} a \!\!\!\!\!\!\underbrace{dt}_{belongs~to\int\limits_{0}^{t}}$$

I can see that last $dt$ is part of an integral notation and cant just be moved right? Soo Why do some authors treat it like it is an ordinary fraction (cancel out the $dt$) and use this assumption to change even integration limits like this:

$$\int\limits_{0}^{t} \frac{dx}{dt} a \,\,dt = \int\limits_{0}^{x} a \, \, dx$$

I would like to clear this up in my head once and for all.

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How do you know that k equals to a derivative dx/dy? It is wrong. – Occupy Gezi Mar 22 '13 at 0:27
They thought us that in hischool... maybee it is wrong but untill someone provides better explaination i ll stick to it. – 71GA Mar 22 '13 at 0:29
Did you check your other question? – Occupy Gezi Mar 22 '13 at 0:32
Integration by substitution.. – Halil Duru Mar 22 '13 at 0:45
@Halil Duru Could you please provide an anwser? – 71GA Mar 22 '13 at 7:33

Well , we have chain rule in differentiation and its counterpart integration by substitution.

These are theorems that require proof.So what you do here is not cancelling in algebra.But in

effect , it seems like you just cancel them which confuses you..

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I had the same problem with this when I began learning calculus. I was told that the $dt$ in an integral was simply notational and I couldn't understand how my teachers told me that they "cancel out". Basically, the thing which you put in a box, $dy = \frac{dy}{dx}\, dx$, only really makes sense in terms of integrals.

What I mean is that

$\int dy = \int\frac{dy}{dx}\, dx$

by the chain rule, so as a short hand we sometimes write

$dy = \frac{dy}{dx}\, dx$

with the understanding that we have not defined some algebra of infitesimal quantities - we're just using a shorthand.

Also, the derivative is more commonly defined as

$\frac{dy}{dx} = \lim_{\delta x\to0} \frac{y(x + \delta x) - y(x)}{\delta x}$

which, if you think about it, is intuitively similar to what you wrote.

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