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We often write e.g. $$\int x^2 dx=\tfrac{1}{3}x^3+c$$ for any $c \in \mathbb{R}$, where $c$ is the constant of integration.

We can show (via limits) that, if $g(x)=\frac{1}{3}x^3+c$, then $\frac{dg}{dx}=x^2$ for any $c \in \mathbb{R}$. But this doesn't exclude the possibility that some function $f=f(x)$ that doesn't have the form $\tfrac{1}{3}x^3+c$ also has the derivative $\frac{df}{dx}=x^2$. So...

Q: Are indefinite integrals unique up to the constant of integration? If so, how do we know?

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They are unique in connected domains, in particular in intervals. Otherwise they are not unique. – Git Gud Jul 6 '14 at 11:51
up vote 8 down vote accepted

Hint: consider the difference of primitives, the derivative of which vanishes. now if the domain is connected then this implies that the difference must be a constant (this follows from the mean value theorem).

To expand on git guds fitting remark above: take the domain to be $\Omega:=(-1,0)\cup(0,1)$ then the characteristic function $\chi_{(0,1)}$ is differentiable on $\Omega$ and its derivative is zero. If $g$ is a primitive of $f$ on $\Omega$, then $g+\chi_{(0,1)}$ is another primitive but the difference $\chi_{(0,1)}$ is not constant.

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Thanks! (My interpretation: it simplifies to showing $\frac{df}{dx}=0$ only for constant functions $f$, and this follows from the mean value theorem since any non-constant function would have a non-zero derivative somewhere.) – Rebecca J. Stones Jul 6 '14 at 12:10
@Rebecca: this is exactly the point. but here it is best to give a direct proof: if $\Omega$ is a connected subset of $\mathbf{R}$ (which is equivalent to $\Omega$ being an intervall) and $[x,y]\subset\Omega$ then by the mean value theorem $(f(y)-f(x))/(x-y)=0$ i.e. $f(y)-f(x)=0$. Notice how the connectedness is used here: I gave above an example of a nonconstant function with zero derivative! – Zlatan der Zechpreller Jul 6 '14 at 14:25

Let $f$ and $g$ two antiderivative of $x^2$ then we have

$$\frac{d}{dx}(f(x)-g(x))=x^2-x^2=0$$ hence $$f(x)-g(x)=\text{constant}$$

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It was mentioned that the domain must be connected. Example: Sometimes people write $$ \int \frac{dx}{x} = \ln|x| + C . \tag{1}$$ But of course an antiderivative on the domain $(-\infty,0)\cup(0,+\infty)$ could have different values of $C$ on the positive part and on the negative part of the domain. Making (1) a questionable thing to write.

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My high school calculus teacher gave us a brain-teaser involving this particular problem (actually, I think it was about $\mathrm{arcsec}$), and ever since, I have considered this "formula" for $\int dx/x$ as being dangerously misleading. – Ryan Reich Jul 6 '14 at 20:03
When I want to be precise, I call $C$ a locally constant function instead of just a constant. – Per Manne Jul 11 '14 at 8:49

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