0
$\begingroup$

Am I setting myself up for a fundamental misconception if I consider antidifferentiation denoted by $ \int $ sign as the operation that is the inverse of the differential of a function denoted by $ d $ and that also returns an arbitrary constant $ C $? To make things clear suppose I have a function $F$ where $\frac{d}{dx}F(x)=f(x)$ for all x in some interval. Then it follows $dF(x)=f(x)dx$ now if I have $\int (dF(x))=F(x)+C$ from my definition of antiderivative operation denoted by $\int$ then it also means $\int f(x)dx=F(x)+C$. What do you guys think?

  • 0
    No, it is indeed an inverse operation.2012-05-26
  • 0
    Oh thanks, I guess it's just the notation. For some time, I've wondered how could mathematicians come up with such brilliant notations. I just came around asking; maybe I got it all wrong in my head.2012-05-26
  • 0
    @Ilya, I thought the antiderivative was the inverse of the derivative, not the differential... I even posted a question precisely on this. Could you help please? http://math.stackexchange.com/questions/1035286/relationship-between-primitive-and-differential?noredirect=1#comment2112879_10352862014-11-24

1 Answers 1