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$$\int{f(x) \cdot f'(x)dx}=\frac{f^2(x)}{2}+c$$ Does this rule always work, as it appears this way to me. $$\int{y\frac{dy}{dx}dx}=\int{ydy}=\frac{y^2}{2}+c$$ This is my working so far but I was wondering if this works for all functions or are there some limitations?

A.Γ.
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Henry Lee
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    (1) It is just a $u$-substitution and (2) you could check the answer by differentiation and notice that it is correct. – Randall May 23 '18 at 18:44
  • your second line is essentially a proof that it works. – The Count May 23 '18 at 18:48
  • certainly for the functions I have tried it is true and I cannot see why it would not be true for all functions $f(x)$ but not sure if anyone else knew better? – Henry Lee May 23 '18 at 18:54
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    No, it does not always work. There are functions $f(x)$, with derivatives $f'(x)$, that exist and are bounded everywhere, but are not Riemann integrable. See Volterra's Function. – Mark Viola May 23 '18 at 18:57
  • @Randall This is not generally true See Volterra's Function. One needs to ensure that $f'(x)$ is continuous almost everywhere. – Mark Viola May 23 '18 at 18:58
  • OK, fine. Assuming the integral exists, yes. – Randall May 23 '18 at 19:02
  • @MarkViola You bring up an interesting subtlety. Does an indefinite integral involve any concept of integrability, or is it just another way of stating that one function is the derivative of another? – Umberto P. May 23 '18 at 19:02
  • ah I see now, this was what I meant for if it always works or not. If this was to be used how would you know whether or not it works? – Henry Lee May 23 '18 at 19:03
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    @UmbertoP. I interpret the symbol $\int$ as an integral, and prefer to write $\int_a^x f(t)f'(t),dt$. If it merely means here a function that has as its derivative $f(x)f'(x)$, then yes, $\frac12 f^2(x)$+C $ is it. – Mark Viola May 23 '18 at 19:05
  • @MarkViola That is interesting. I happen to take the other point of view. – Umberto P. May 23 '18 at 19:08
  • @UmbertoP. In taking that "other point of view," the symbol $\int$ can no longer be interpreted as an integral. It becomes merely a pure symbol. – Mark Viola May 23 '18 at 19:12
  • Yeah, I know. It is a convenient abuse of notation. – Umberto P. May 23 '18 at 19:13

1 Answers1

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Let $u=f(x)$, then $du= f'(x)dx.$

Thus $$\int{f(x) \cdot f'(x)dx}= \int{udu} = \frac {u^2}{2}+c = \frac{f^2(x)}{2}+c $$

Mohammad Riazi-Kermani
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