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Integration is often said to be a global operation. I suppose that this refers to the fact that the integral $$ \int_{a}^{b} f(x) \, dx $$ depends on the value of $f(x)$ on an entire interval $[a,b]$. This is in contrast to the derivative, which simply depends on behaviour of a function around a given point.

However, in practice to compute the above integral I would simply find an antiderivative of $f$ and apply the fundamental theorem of calculus. And since antidifferentiation seems as local as differentiation, I don't understand how this explains why the mechanical process of integration (both indefinite and definite) is so much harder than differentiation.


For a reference, Terry Tao makes this statement in the following Math Overflow post: 'Why is differentiation mechanics and integration art?'.

Joe
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  • "I would simply find an antiderivative of $f$ and apply the fundamental theorem of calculus." You can't always do that, and in many cases finding an antiderivative is much more difficult than verifying that said antiderivative is correct. – DMcMor Jan 06 '21 at 21:05
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    Inverse operations are not always of the same difficulty. Which is easier - plugging values into a polynomial, or finding its roots? – Joshua Wang Jan 06 '21 at 21:09
  • @DMcMor I hadn't thought about that, thanks. However, even the integration of elementary functions is much harder than the differentiation of elementary functions. And I can't think of an example of an elementary function which isn't continuous (and all continuous functions are integrable). – Joe Jan 06 '21 at 21:12
  • @Joe Elementary functions do have antiderivatives, but the antiderivatives are not necessarily elementary. For example, $\exp(x^2)$ does not have an elementary antiderivative. This, I think, is what DMcMor means by "You can't always do that". – Robert Israel Jan 06 '21 at 21:56
  • @RobertIsrael Oh, I see. Thank you. Do you have any thoughts about this idea that integration is harder because it is global? – Joe Jan 06 '21 at 22:11
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    Compare the discrete version. Consider any function $f(n) $ given by a finite formula. It is easy to express $f(n) - f(n-1)$ as a finite formula, but usually difficult or impossible to express $f(1)+f(2)+\dots+f(n)$ as a finite formula. – Paramanand Singh Jan 07 '21 at 04:33

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