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Does the expression:

$$ \int_a^b f(x) dx $$

make any sense when a > b? Consider the simple case:

$$ \int_1^0 x^2 dx = - 1/3 $$

But this of course makes no sense since area under a positive function like x^2 can not be negative! But if it doesn't make sense, why we cannot integrate reverse? P.S. I have never read anywhere that $ \int_a^b f(x) dx $ when $a > b$ doesn't make sense, but this simple example seem to prove that...

Giack_89
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    As noted under Conventions on the wikipedia page we define $\int _a^bf(x)\ dx = - \int_b^af(x)\ dx$. We view it as the signed area and if it is traversed in the reverse sense it becomes negative. – Ross Millikan Aug 02 '22 at 15:07
  • @RossMillikan yea i saw it and must say i never read something similar... it looks very strange actually, because in this way we completely loose the meaning of signed area under the curve. In fact can't say any longer that area under $\int_a^b x^2 dx$ with a > b is positive as with a < b it should, because it's exactly same area under the same positive function x^2! – Giack_89 Aug 02 '22 at 15:38
  • This may be absolute mathematical blasphemy I’m talking about, but you can intuitively think about the “dx” (or $\Delta x$ in case of a Riemann Sum) as being “negative” as you go from 1 to 0, hence the negative sign. – insipidintegrator Aug 02 '22 at 15:45
  • Of course you can have negative areas. A negative area is being added to the area of Earth covered by glaciers every year. – David H Aug 02 '22 at 15:48
  • @insipidintegrator i don't think it's so blasphemous as it seems because Δ in the definition of integral is (b-a)/n as n tends to infinity, so if we can treat Δ as a real number i think it should be positive. I've never got why Δ or especially dx is treated by mathematicians as a very misterious object and not just a real number actually.... – Giack_89 Aug 02 '22 at 15:52

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