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So, as the title states. My question is about mathematical notation. Whenever my lecturer in calculus one computes an integral they do some in my opinion very redundant steps.

When they attempt to compute for example $\int x^3e^{x^2}\,dx$, they will write down $u = x^2; du=2xdx$ and proceed to turn the integral into $\frac{1}{2}\int ue^u\,du$.

I don't see why it is required to hide this simple mathematical truth that $d(x^2)=2xdx$ behind a substitution, yet they tells me that not doing so would technically be incorrect.

By the way, I am not asking this to proof my lecturer wrong, since they are an epic person and when I told them that I found it easier without the substitution I was given permision to do so. I am merely asking why it would be "technically incorrect" to do so.

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    $\begingroup$ It's fine, just confusing, especially if you are teaching, where you want to be as straightforward as possible. $\endgroup$
    – user619894
    Oct 30, 2020 at 15:40
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    $\begingroup$ Some would say that $\int x^2e^{x^2}\,d(x^2)$ is a Riemann-Stieltjes integral, not a Riemann integral, and then argue that (in this case) you get the same answer. $\endgroup$
    – GEdgar
    Oct 30, 2020 at 16:04
  • $\begingroup$ It boils down to whether you think writing $f^\prime dx$ as $df$ is "correct", which I'd argue it is because substitution is a theorem we can summarise as "that works". $\endgroup$
    – J.G.
    Oct 30, 2020 at 16:09
  • $\begingroup$ @martinvanIJcken Hi Martin. I hope you're doing well. If you would, please let me know how I can improve my answer. I really want to give you the best answer I can. And feel free to up vote and accept an answer as you see fit. ;-) $\endgroup$
    – Mark Viola
    Nov 3, 2020 at 18:56

2 Answers 2

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The notation is proper. The Riemann-Sieltjes for a real-valued function $f(x)$ for $x\in [a,b]$ with respect to a real-valued function $g(x)$ is written

$$I=\int_a^b f(x) dg(x)$$

If $g$ is differentiable on $[a,b]$, then $I$ can be written as

$$I=\int_a^b f(x)\frac{dg(x)}{dx}\,dx$$

However, $g(x)$ need not be differentiable and, in fact, can even be discontinuous.

In the case of interest, $g(x)=x^2$ and $f(x)=x^2e^{x^2}$. We have, therefore

$$\int_a^b x^2e^{x^2}d(x^2)=2\int_a^b x^3 e^{x^2}\,dx$$

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Students can feel more comfortable when they see what is substituted as a single entity. But of course

$$\int x^3e^{x^2}dx=\frac12\int x^2e^{x^2}d(x^2)=\frac12\int ue^{u}du$$

are equivalent and technically correct.

You can even do the integral mentally if you want (but make sure your instructor believes you).


Extra trick:

Write down $x^2$ on a label and read it loud "$u$".

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