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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.


marked as duplicate by hardmath, Community Dec 3 '15 at 19:05

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  • $\begingroup$ 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. $\endgroup$ – hardmath Dec 3 '15 at 19:01
  • $\begingroup$ The proof of moving a constant in and out of the integral implicitly assumes the constant $c\neq 0$ $\endgroup$ – Brenton Dec 3 '15 at 19:02

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.

  • $\begingroup$ 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. $\endgroup$ – Mario Carneiro Dec 3 '15 at 19:06
  • $\begingroup$ Yes, I'm not making myself clear. I'll edit the answer. $\endgroup$ – Martin Argerami Dec 3 '15 at 19:07
  • $\begingroup$ That's fine .the original question seems satisfactory enough $\endgroup$ – theAnkor Dec 3 '15 at 19:11

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