Sunday, February 24, 2019

calculus - Differential Notation Magic in Integration by u-Substitution





I'm really confused now. I always thought that the differential notation $\frac{df}{dx}$ was just that, a notation.



But somehow when doing integration by u-substitution I'm told that you can turn something like this $\frac{du}{dx} = 2x\;$ into this $\;du = 2x\ dx$.



But how is that even possible? I understand that the notation comes from the fact that $\frac{du}{dx}$ actually means the limit of the difference in $u$ over the difference in $x$, with $\Delta x$ approaching $0$.




$$u'(x) = \frac{du}{dx} = \frac{du(x)}{dx} = \lim_{\Delta x\to 0} \frac{u(x+\Delta x)\ -\ u(x)}{(x+\Delta x) - x} = \lim_{\Delta x\to 0} \frac{u(x+\Delta x)\ -\ u(x)}{\Delta x}$$



So if $\frac{df}{dx}$ is just a notation for the limit mentioned above, then what is the underlying argument to say that you can treat $\frac{du}{dx}$ as if it were an actual fraction?



Appreciate the help =)


Answer



It is really just a notation. And the trick with the substitution e.g. $du = 2xdx$ does not have any mathematical meaning, it is just a convenient way of memorizing the integration by substitution rule/law/theorem:



$$\int_a^b f(\phi(t)) \phi'(t) dt = \int_{\phi(a)}^{\phi(b)} f(x)dx $$




Going from left to right you might want to make the substitution $x=\phi(t)$. Our mnemonic tells us to $\frac{dx}{dt} = \phi'(t)$ or in other words that you have to replace $\phi'(t)dt$ with $dx$ if you replace $\phi(t)$ with $x$. If you look again at the equation above you see that this mnemonic does a nice job, so we do not have to memorize this whole equation.



I do use the mnemonic but still I always keep this equation in mind when doing so.


No comments:

Post a Comment

analysis - Injection, making bijection

I have injection $f \colon A \rightarrow B$ and I want to get bijection. Can I just resting codomain to $f(A)$? I know that every function i...