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I have an elementary question. In this video(Link), Sal explains why swapping the limits of integration changes the sign of the result; In his video, he reasons that "dx"s in the reverse direction must have opposite sign too (because (a-b)/n must become (b-a)/n), however, that's something I don't remember I saw anywhere when I learned Riemann integral, to indicate that there's a relation between "dx"s and the order(higher/lower) of integral limits.

I have no problem with the result of definite integrals becoming negative since, well..., that's what happens when your ceiling is at the bottom of your basement! :D HOWEVER, I think it makes sense only when f changes to -f... but about the order of limits, I think it mustn't be so. Series are just algebraic sums, so I can still divide from b to a to divisions of (a-b)/n and then instead, sum all of them FROM RIGHT TO LEFT... So for example, d1.f + d2.f + d3.f + d4.f + d5.f becomes d5.f + d4.f + d3.f + d2.f + d1.f which has a similar sign and also because it's an algebraic sum, they must have the same values as well(To not confuse anyone, by "di" I mean division number i, and let's say "di" with less "i", refers to a "dx" at a position more towards the left). So I know integrals are very old and I'm definitely wrong :D But where am I making a mistake? Thanks.

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Here is one way to see it. An integral over a range can be split into $2$ integrals, each covering part of the range. In particular, you get for any integral function $f(x)$ over the domain that $\int_{a}^{b}f(x)dx + \int_{b}^{c}f(x)dx = \int_{a}^{c}f(x)dx$. Now, as the comment says, this holds for $a \le b \le c$. However, if you consider extending/defining this to also be true for cases where $c \lt b$, e.g., if $c = a$, you get

$$\int_{a}^{b}f(x)dx + \int_{b}^{a}f(x)dx = \int_{a}^{a}f(x)dx = 0 \tag{1}\label{eq1A}$$

This results in

$$\int_{a}^{b}f(x)dx = -\int_{b}^{a}f(x)dx \tag{2}\label{eq2A}$$

Here is a second way to try to see this. Note the formal definition of the Riemann integral involves using a partition of the interval $[a,b]$ and taking the upper/lower limits of the sums of the functions at those partitions, times the partition widths. If $b \lt a$, then the partition values $x_i$ would be decreasing instead of increasing, so the differences $x_{i+1} - x_{i}$ would be negative instead of positive. This means the result would the negative of the integral of the same function $f(x)$ when going from the smaller limit to the larger limit.

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    $\begingroup$ This seems a bit logically backwards to me. The first identity holds for $a \le b \le c$ to begin with. Then we define (not derive) $\int_a^b = -\int_b^a$ in order to make that identity hold for all $a$, $b$ and $c$. $\endgroup$ Nov 12, 2019 at 4:28
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    $\begingroup$ @HansLundmark You have a valid point. What I show above is how I view & understand switching the limits, but you would need to extend the concept to get what I'm using. I'm demonstrating it to be a logical extension, but based on the original concept, it's more a definition than a derivation. I've added some more details to make this more clear. $\endgroup$ Nov 12, 2019 at 4:35
  • $\begingroup$ @aderchox I've also just added it can also be seen from the formal definition of the Riemann integral. I hope this helps to explain to explain it better. There are undoubtedly other explanations as well, perhaps ones which make more sense to you, but I can't think of any offhand. In general, basically similar to other areas of math, this makes the definition consistent when expanding other concepts, similar to things like adding negative numbers being the same as subtracting positive numbers, use of imaginary numbers in polynomials, etc. $\endgroup$ Nov 12, 2019 at 4:51
  • $\begingroup$ Now that I think more about it, it's much more convenient to be able to use that identity without knowing about the real values of the limits..., e.g. we might want to use variables there. So I think this is a good convention. Thanks. (A philosophical question: How do people know things they set as conventions won't be destructive anywhere else?! :D) $\endgroup$
    – aderchox
    Nov 12, 2019 at 4:56
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    $\begingroup$ @aderchox I just noticed your question is really a duplicate of Why change the sign of the integral when switching the limits of integration? (so I'm voting to close as dup.). The question basically states what I just wrote & the accepted answer basically states what I wrote initially. As for your philosophical question, people don't know for sure. However, there are usually only a fairly few cases to deal with, so those should be checked just to help make sure. $\endgroup$ Nov 12, 2019 at 5:00
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The point your explainer was trying to make is right. You're integrating a differential over an interval. Since this is an infinitesimal change in the function, the change can either be an increase or a decrease (when it is nontrivial), in each case having a corresponding positive or negative sign. Thus, reversing the direction of integration also changes the sign of the integral.

That is, first consider the finite differences in your interval $\Delta x.$ Then these differences are positive or negative according as you move in the direction of increasing or decreasing $x$ correspondingly. Going to the limit will not affect these signs.

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  • $\begingroup$ I don't say these changes can not be decreasing. I think this is not the right answer to this question because if this was the case, I would say one can also use these infinitesimal decreases when still the lower limit is smaller and the higher limit is bigger... . It makes more sense to me that it is a convention as the above answer suggests it. $\endgroup$
    – aderchox
    Nov 12, 2019 at 5:13
  • $\begingroup$ @aderchox The point is that changes are either positive or negative according as the quantity increases or decreasing. Thus the infinitesimal changes are accordingly positive or negative. $\endgroup$
    – Allawonder
    Nov 12, 2019 at 14:14

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