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I seem to think that it should be zero as well because being a constant zero can be taken outside the integral and whatever be the answer of the remaining constant integration it is finite. However my textbook implies that it is the arbitrary constant c.

  • The phrase "indefinite integration" means finding an antiderivative. What is the derivative of a constant? Thus any constant is an antiderivative (indefinite integral) of that. – hardmath Dec 03 '15 at 19:01
  • The proof of moving a constant in and out of the integral implicitly assumes the constant $c\neq 0$ – Brenton Dec 03 '15 at 19:02

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The antiderivative of zero will be any function whose derivative is zero. So, as your book says, any constant function will be an antiderivative of zero.

The notation $\int f(x) dx$ means nothing else but "an antiderivative of $f(x)$".

The example you mention, in particular, shows that you have to careful with your arithmetic manipulations: when you take the constant zero "out" in $\int 0\,dx$, you would get $0\,\int 1\,dx=0$, as Mario Carneiro says. But if you add a constant to an antiderivative, you still get an antiderivative, so any equality between antiderivatives is up to a constant.

So, if you show that $\int f(x)\,dx=0$, what you know that is $\int f(x)\,dx$ differs from $0$ by a constant: so $\int f(x)\,dx=c$ for some constant.

Martin Argerami
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  • It's not a meaningless manipulation; the right side is $0\int dx=0\int 1,dx=0x=0$. But it needs an extra $+c$ in the equality since it is an equality of indefinite integrals. – Mario Carneiro Dec 03 '15 at 19:06
  • Yes, I'm not making myself clear. I'll edit the answer. – Martin Argerami Dec 03 '15 at 19:07
  • That's fine .the original question seems satisfactory enough –  Dec 03 '15 at 19:11
  • @Martin Argerami: what you say makes sense, as in, even if you further add a constant to the zero you get when you integrate, it'd still be an antiderivative of 0, so that the 'antiderivative 0 is a constant, not 0' thing still holds. But this throws into doubt a definition of indefinite integrals I've seen- as the family of all antiderivatives of a function. Here, you take the indefinite integral (which is zero when you take 0 out) and then further add a constant. So now you've got an antiderivative which isn't really an indefinite integral of 0. So is the definition wrong? – harry Apr 07 '21 at 07:30
  • Not entirely sure what you mean. Any two antiderivatives differ by a constant. – Martin Argerami Apr 07 '21 at 07:56
  • I understand that. When you calculate the indefinite integral of 0, you get 0, but then no matter what constant you add to it, the resulting number will be an antiderivative; so generally, the antiderivative of 0 is 'any constant'. But here, you have to add the constant after indefinite integration to get all the antiderivatives. We usually add a constant of integration to get all the antiderivatives, but since the c that we initially got in integrating this is multiplied by 0, we have to add yet another constant to get all antiderivatives. So what I'm trying to ask is; you can't always get – harry Apr 07 '21 at 09:27
  • All the antiderivatives from indefinite integration; you might need to further add constants, is it so? – harry Apr 07 '21 at 09:27
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    Yes, that's why "$+c$". You cannot treat $\displaystyle\int f(x),dx$ as "all the antiderivatives" and do arithmetic with it. – Martin Argerami Apr 07 '21 at 12:23