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How did we come to know that treating the differential notation as a fraction will help us in finding the integral. And how do we know about its validity?
How can $\frac{dy}{dx}$ be treated as a fraction?
I want to know about how did u-substitution come about and why is the differential treated as a fraction in it?


marked as duplicate by 5xum, Claude Leibovici, hunter, user147263, Hans Lundmark Apr 15 '15 at 14:41

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    $\begingroup$ Integration by substitution is just the Chain Rule of differentiation put into practice. It works because the Chain Rule works. The differential is treated as a fraction because that works, too. $\endgroup$ – Gerry Myerson Apr 15 '15 at 7:24
  • $\begingroup$ I want to state that $\frac{d}{dx}$ is not a ratio and is never treated as such--$\frac{d}{dx}$ is an operator which means to differentiate with respect to $x$. If instead you mean $\frac{dy}{dx}$ then you should edit your question to be more clear. You can also realize the operator $d$, i.e. the differential of a function $f$: $df$ which might shed light on your problem. As an example, when I taught Calculus and we dealt with solids of revolution, I always had my students write out: $dV = \pi r^2 dh$ (for washer problems) and always illustrate what $r$ and $dh$ were in the picture. $\endgroup$ – Jared Apr 15 '15 at 8:13
  • $\begingroup$ Please explain the downvote. $\endgroup$ – Aditya Agarwal Aug 15 '15 at 13:31

It doesn't necessarily need to be.

Consider a simple equation $\frac{dy}{dx}=\sin(2x+5)$ and let $u=2x+5$. Then $$\frac{du}{dx}=2$$ Traditionally, you will complete the working by using $du=2\cdot dx$, but if we were to avoid this, you could instead continue with the integral: $$\int\frac{dy}{dx}dx=\int\sin(u)dx$$ $$\int\frac{dy}{dx}dx=\int\sin(u)\cdot\frac{du}{dx}\cdot\frac{1}{2}dx$$ $$\int\frac{dy}{dx}dx=\frac{1}{2}\int\sin(u)\cdot\frac{du}{dx}dx$$ $$y=c-\frac{1}{2}\cos(u)$$ $$y=c-\frac{1}{2}\cos(2x+5)$$

But why is this? Can we prove that the usefulness of the differentiatals' sepertation is justified? As Gerry Myerson has mentioned, it's a direct consequence of the chain rule:

$$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$$ $$\int\frac{dy}{dx}dx=\int\frac{dy}{du}\frac{du}{dx}dx$$ But then if you 'cancel', it becomes $$\int\frac{dy}{dx}dx=\int\frac{dy}{du}du$$ Which is what you desired.


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