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Is it always the case that if:

$$ \int f(x) dx = F_1(x) + C $$ and $$ \int f(x) dx = F_2(x) + C $$ then $$ F_1(x) = F_2(x) $$

and why?

Is it a legitimate way to prove the equivalence of two functions, namely $F_1$ and $F_2$ ?

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We always have $F_1-F_2=C$ instead. You need an initial condition to verify what you are looking for. –  B. S. Feb 7 '13 at 11:57
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1 Answer 1

up vote 0 down vote accepted

Basically yes. We need some continuity conditions on $f$ as well. And then, your first two lines mean $F_1'=f=F_2'$, and hence $(F_1-F_2)'=0$. Now the main theorem is that $g'=0 \Rightarrow g=$ constant $c$, that is, $F_1-F_2=c$ can be concluded.

If you also know any point $x_0$ such that $F_1(x_0)=F_2(x_0)$, or one endpoint of the integral is determined (and the other varies), and the two $C$'s mean the same, then $F_1=F_2$ follows.

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So, basically, in the case of different constants for each integral (as Babak Sorouh pointed out), the functions are equivalent under the appropriate translation? (continuity assumed) That makes sense, thank you! –  Schlomo Steinbergerstein Feb 7 '13 at 12:05
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