0

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?

Narasimham
  • 40,495
Aditya Agarwal
  • 4,595
  • 2
    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. – Gerry Myerson Apr 15 '15 at 07:24
  • 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. – Jared Apr 15 '15 at 08:13
  • Please explain the downvote. – Aditya Agarwal Aug 15 '15 at 13:31

1 Answers1

3

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.

nathan.j.mcdougall
  • 1,854
  • 16
  • 22
  • In the second line with the integrals, where does the fraction $\frac{1}{2}$ come from? – d56 Nov 25 '20 at 11:42
  • @d56, It comes from derivative $\frac{\mathrm{d}u}{\mathrm{d}x}=2$. That means that $\frac{\mathrm{d}u}{\mathrm{d}x}\frac{1}{2}=1$, and we can safely multiply anything by $1$ without changing the result. So we strategically multiply by this expression with the $\frac{1}{2}$ to introduce the derivative into the integral. – nathan.j.mcdougall Nov 25 '20 at 19:28
  • Thanks, understood! But what still bugs me is that we treat the differential $\frac{du}{dx}$ as a fraction by cancelling the denominator with $dx$. I cannot understand why we do that... – d56 Nov 26 '20 at 09:34
  • @d56, The Integral here is true because of the chain rule: $$\int\frac{\mathrm{d}y}{\mathrm{d}x}\mathrm{d}x=\int\frac{\mathrm{d}y}{\mathrm{d}u}\frac{\mathrm{d}u}{\mathrm{d}x}\mathrm{d}x$$ However, the next integral below is not true because we "cancel out" on the RHS of the above, but rather because $x$ and $u$ are dummy variables: we are effectively just changing notation: $$\int\frac{\mathrm{d}y}{\mathrm{d}x}\mathrm{d}x=\int\frac{\mathrm{d}y}{\mathrm{d}u}\mathrm{d}u$$ – nathan.j.mcdougall Nov 26 '20 at 19:10
  • So the final line of my answer is a tautology; it doesn't follow from the previous equation, but rather from a straightforward change in notation. The fact that this tautology is true, and also the chain rule is true, means that if you use both equations together, it looks like cancellation, even though algebraically the thing going on is not actually cancellation at all (well, at least not at this high-level view of things.) Hopefully that makes things clear. – nathan.j.mcdougall Nov 26 '20 at 19:12