0

It is a well known fact that

$$ \int f'(g(x))g'(x)dx=f(g(x))+C $$

This follows directly from the chain rule. However, sometimes it is easier to perform the substitution $u=g(x)$:

\begin{align} u&=g(x) \\ \frac{du}{dx}&=g'(x) \\ du&=g'(x)dx \\ \therefore \int f'(g(x))g'(x)dx&=\int f'(u)du\\&=f(u)+C\\&=f(g(x))+C\end{align}

Ordinarily, one cannot simply multiply both sides of an equation by $dx$. After all, '$dx$' on its own is not even formally defined. Here is my attempt to justify why this is allowed:

As well as knowing that $ \int f'(g(x))g'(x)dx=f(g(x))+C $, we also know that

$$ \int f'(u)du=f(u)+C $$

Note that this holds even when $u\neq g(x)$. However, the above fact becomes useful when we set $u=g(x)$, as $f(u)+C$ becomes $f(g(x))+C$. Knowing that $ \int f'(g(x))g'(x)dx$ and $\int f'(u)du$ evaluate to the same thing, we can say that

$$ \int f'(g(x))g'(x)dx=\int f'(u)du $$

when $u=g(x)$. Hence, $g'(x)dx$ can be replaced by $du$, without the value of the integral changing. Since differentiating both sides of $u=g(x)$ and multiplying by $dx$ yields the same result, it is safe to treat $\frac{du}{dx}$ as a fraction in this case.

I have two questions:

  1. Is this justification good (or, at the very least, valid)? If it can be improved upon, how so?
  2. Is there a more fundamental reason as to why $\frac{du}{dx}$ can be treated as a fraction? Is it something to do with how we can treat derivatives as fractions when using the chain rule?
Joe
  • 22,603
  • 1
    I think it's more intuitive to go directly from $u=g(x)$ to $\mathrm du=g'(x)\mathrm dx$. The differential wrt $u$ is the differential wrt $x$ scaled by the differential in $g(x)$, i.e. the derivative (since as you take the limit, functions becomes approximately linear). I might be wrong but I believe that's where the notation $\mathrm du/\mathrm dx$ comes from. – Andrew Li Jul 26 '20 at 18:31
  • "Is there a more fundamental reason as to why $\frac{du}{dx}$ can be treated as a fraction?" Well, no: it's rather the ingenuity of its inventor, Leibniz, that it behaves in like one. – Michael Hoppe Jul 26 '20 at 18:43
  • I don't think it is a consequence of the Chain Rule, it depends crucially on the Fundamental Theorem of Calculus. If you are to get any sort of proof you have to sort out the $x$s; the LHS is free of $x$, the RHS is a function of $x$. I think all you are doing is juggling with the notation. – ancient mathematician Jul 26 '20 at 18:46
  • As an aside, it is not always true the $\int_a^b f'(x),dx=f(b)-f(a)$. See This on Volterra's function – Mark Viola Jul 26 '20 at 18:52
  • Also, check out this question – Jackozee Hakkiuz Jul 26 '20 at 19:02
  • @MarkViola I'm sorry but I don't quite follow. What I wrote was that $\int f'(x)dx=f(x)+C$. I don't understand how definite integrals come into this. It seems like a lot of users have made similar comments, so I am wondering if that's part of my confusion. – Joe Jul 26 '20 at 19:04
  • 1
    Not all derivatives of functions are Riemann integrable. – Mark Viola Jul 26 '20 at 19:10
  • the phrase "Note that this holds even when $u\neq g(x)$" is not clear in the given context –  Jul 26 '20 at 19:19
  • Pherhaps this helps...I really enjoyed this video when first trying to understand substitution: https://www.youtube.com/watch?v=CoJbL_quLSU&list=LLl23_bzq5JqaZVpRetJlNjg&index=4&t=0s – joshuaronis Jul 31 '20 at 19:26

0 Answers0