I am pretty sure this question has already been asked several times, but it's quite general so when I try to search I cannot find the answer. I am trying to understand if, say, we have $\int_2^axdx$, and if $a>2$ I have no problem understanding what it is, if $a=2$ then the integral is equal to $0$, but if $a<2$ then it becomes negative? I am not really sure about this. Is it correct to say if $a=1$ then $\int_2^{1}xdx$ ? Would it just become $-\int_1^2xdx$ or it will not exist at all? Say, if my restriction on the parameter is $a<2$ how would I interpret it? Thanks!

  • 2
    $\begingroup$ Yep, you're correct. $\endgroup$
    – florence
    May 7, 2017 at 16:30

1 Answer 1


Yes, you are correct.

$$\int_a^b f(x)dx = -\int_b^a f(x)dx$$


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