Saturday, 14 July 2018

integration - Why do we treat differential notation as a fraction in u-substitution method




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?


Answer



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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