Why can you cancel out the dx in u-substitution?

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The discussion centers on the justification for canceling out the dx in u-substitution during integration, specifically in the context of the integral ∫(-cosx)sinx dx. Participants clarify that while the notation du/dx resembles a fraction, it is fundamentally a limit and not a true fraction, which complicates the notion of cancellation. The conversation emphasizes the importance of the chain rule and the fundamental theorem of calculus in understanding this process, asserting that the cancellation is not merely a trick but a deeper mathematical principle. The validity of infinitesimals in calculus is also debated, highlighting the distinction between intuitive and rigorous mathematical approaches.

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  • #31
micromass said:
This is where choice is used. You can never give exact definition of the x_n. This is why it requires choice.
Hmm, color me skeptical. Why can't you explicitly construct a sequence satisfying x_n \in (x - \frac{1}{n}, x + \frac{1}{n})? For instance, x_n can be the point a third of the way into the interval (x - \frac{1}{n}, x + \frac{1}{n}).
 
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  • #32
lugita15 said:
Hmm, color me skeptical. Why can't you explicitly construct a sequence satisfying x_n \in (x - \frac{1}{n}, x + \frac{1}{n})? For instance, x_n can be the point a third of the way into the interval (x - \frac{1}{n}, x + \frac{1}{n}).

Note that you also want |f(x_n)-f(x)|\geq \varepsilon.
 

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