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simple question , if we define $\displaystyle F(x) = \int_a^x f(t) \, \mathrm{d}t$ does that mean $x>a$? or $x$ could be smaller than a even though that expressoin means it's an upper bound?, thanks

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That we write integrals with the upper bound greater than lower bound is just convention. It's perfectly acceptable for $x \leq a$. Note, of course, that $\int_{a}^{x}f(x)dx = -\int_{x}^{a}f(x)dx$. –  AWertheim Aug 8 '13 at 0:10
    
Previously asked at mathoverflow.net/q/138868/12357 and mathoverflow.net/q/138871/12357 –  Joel Reyes Noche Aug 8 '13 at 0:17
    
Also at mathoverflow.net/questions/138862 –  user28111 Aug 8 '13 at 0:18

3 Answers 3

Your function $F$ is defined (in principle, at least) for all $x$, not just $x>a$. If it fails to be defined, it is because $f$ is "badly behaved" in some way, not because $x$ is below the "lower bound" of the integral. This is why it makes sense to write $$ \int_a^bh(x)dx=-\int_b^ah(x)dx.$$

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thanks you everything is clear now –  Larson Aug 8 '13 at 0:22

First, it is better to write this a little more clearly, or you will get all your x's mixed up. Best state it as F(x) = ∫ax f(t)dt. Now, the answer is, "a" can be anything. It can be bigger than, smaller than, or equal to x. The only thing you want to be careful of is that the interval from "a" to "x" is all in the domain of f, or the integral doesn't mean anything. (This is perhaps not 100% true, but 99% true and you don't want to deal now with the exceptions).

If a = x, the integral will be 0. If it is larger than x, this is the same as integrating from x to a, but the sign on the answer is reversed.

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People seem to have missed that your expression doesn't actually define anything because you can't have the same variable in the $d$"variable" as in the limit. The function you want is $$ F(x)=\int_{a}^x f(t)\ dt $$ Now as the other answers say, this can be defined for $x<a$ (as long as $f$ is integrable there) by sticking a minus sign in front and integrating from $x$ to $a$.

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So, you answered to point out the use of the dummy variable is not correct, and other people didn't see this? You can just comment next time. –  Pedro Tamaroff Aug 8 '13 at 0:26
    
Thanks for the tip. –  James Aug 8 '13 at 0:33

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