I have been thinking that the symbol $$\int f(x) dx$$ is a variable as a whole, referring to $F(x)$, where $F$ is some antiderivative of $f$. It can refer to any function that is an antiderivative of $f$, evaluated at $x$. So $A =$ the symbol means $A$ is equal to some function evaluated at $x$, and that function is some unknown antiderivative of $f$.

Does this symbol mean above? Am I correct?


What is the precise definition of an antiderivative of a function? What is the definition that is most commonly accepted? What is the one that I write, others will know what I am referring to? Is there one?

I need to know it because the symbol is based on an “antiderivative”.

According to Wikipedia, an antiderivative of $f$ is $F$ such that $F' = f$. And two functions are equal iff their domains are the same and outputs are the same for each input from the domain.

By that definition, $F: \mathbb{R} \rightarrow \mathbb{R}$, $F(x) = C$ is not an antiderivative of $f: (0,1) \rightarrow \mathbb{R}$, $f(x) = 0.$

Is this also correct? Is it the commonly accepted definition of an antiderivative?

  • 1
    $\begingroup$ That's correct. $F$ is an antiderivative of $F' : \Bbb R \to \Bbb R$, $x \mapsto 0$, and in particular not to any function $(0, 1) \to \Bbb R$. It's still true, of course, that $\int_a^b f \,dx = F(b) - F(a)$ for any $a, b \in (0, 1)$, simply because $(F \vert_{(0, 1)})' = f$. $\endgroup$ Commented Feb 26, 2022 at 20:19

3 Answers 3


When we write$$\int f(x)\,\mathrm dx=F(x),$$what that means is that $F$ is an antiderivative of $f$, that is, it is a function with the same domain as $f$ such that $F'=f$. If it turns out that the domain of $f$ is an interval (with more than one point) then every antiderivative of $f$ will be equal to $F+K$, for some constant $K$.

And, indeed, if you have $f\colon(0,1)\longrightarrow\Bbb R$ defined by $f(x)=0$, then no constant function $F\colon\Bbb R\longrightarrow\Bbb R$ is an antiderivative of $f$, since they have distinct domains. However, the restriction of $F$ to $(0,1)$ is an antiderivative of $f$.

  • 1
    $\begingroup$ So this F' = f is the common definition of an antiderivative of a function? And F'(x)= f(x) for all x in the domain of f is not? When I say an antiderivative of a function, will most people know what I am talking about (referring to the first definition)? $\endgroup$
    – TFR
    Commented Feb 26, 2022 at 20:26
  • 4
    $\begingroup$ The assertion $F'=f$ means that $F'$ and $f$ have the same domain $D$ and that, for each $x\in D$, $F'(x)=f(x)$. $\endgroup$ Commented Feb 26, 2022 at 20:58
  • $\begingroup$ Now, what is F’? Is F’ a function whose domain is the set of values on which F’s derivative exist, and is equal to the derivative of F for all values in its domain. Or, is it a function whose domain is the same as F and equal to the derivative of F at every value (in which case F’ exist iff F is differentiable at every point). $\endgroup$
    – TFR
    Commented Feb 27, 2022 at 17:05
  • $\begingroup$ You did not clarify what is an antiderivative. $\endgroup$
    – TFR
    Commented Feb 27, 2022 at 17:07
  • $\begingroup$ Indeed I did not (but now I've added it). Since it is a standard mathematical concept and since you used it several times in your question without asking for a definition, I didn't think that that would be a problem. I also did not define the meaning of, say, “interval” or “domain”. Do you think that I should add those definitions too? $\endgroup$ Commented Feb 27, 2022 at 17:21

The statement $$\int f(x)\,\mathrm dx=F(x)$$ is a really interesting one. In my opinion, it's a good example of context-dependent notation: it doesn't mean the same type of thing in different scenarios.

Let's use a simple example for better intuition: $$\int 2x\,\mathrm dx=x^2+c$$

What is the right-hand side? Is it an expression? A family of expressions? Well, let's say my original question was:

Find $\int 2x\,\mathrm dx\ \ \ \ $ (1 mark)

Here, the meaning of the question is "find the functions ($\mathbb{R}\rightarrow \mathbb{R}$) that have derivative $2x$". And I've found them all: it's the set $\{f\in \mathbb{R}^{\mathbb{R}}:\exists c\in\mathbb{R}:\forall x\in\mathbb{R}:f(x)=x^2+c\}$ i.e. the set of functions $f:\mathbb{R}\rightarrow\mathbb{R}$ of the form $f(x)=x^2+c$ (for some $c\in\mathbb{R}$).

So $x^2+c$ is not an expression, but a set of functions. And so (for this to be "equal to" the left-hand side) the indefinite integral is a set of functions.

But what if my original question was:

Given that $\frac{\mathrm dy}{\mathrm dx}=2x$ and $y=f(x)$ and $f(0)=3$, find $f$.$\ \ \ \ \ \ $ (2 marks)

Now, when I write, $$f(x)=\int 2x\,\mathrm dx=x^2+c$$ it is the first step in a two-step solution (the second step is $0^2+c=3\implies c=3\implies f(x)=x^2+3$).

Here, the first step is the statement "for some specific and knowable value of $c$, and all real values $x$, the function $f$ satisfies $f(x)=x^2+c$".

It's not a statement about a set of functions, because I'm looking for one specific function. Suddenly, the indefinite integral is a single function: a specific antiderivative of $2x$ called $f$.

The indefinite integral sign, and surrounding presentation of calculations in calculus, is an example of abuse of notation: either there is no formal definition of a symbol, or the symbol is commonly used to indicate a very clear meaning, but one that violates the definition.

Moreover, functions are a common example of abuse of definitions: something isn't a function if it doesn't have both a domain and a codomain. So "$f(x)=x^2$ and $x>0$", which you might commonly see as a definition of the "function" $f$, is not a function, because I don't know whether the codomain of $f$ is $f(\mathbb{R}^+)=\mathbb{R}^+$, or $\mathbb{R}$, or even $\mathbb{C}$. Sometimes you don't even see the possible input values specified. And sometimes you see authors use definitions of functions where the image is irrelevant, and two functions are equal iff they have the same domain and the same value on every input in the domain.

Hopefully, the reason for all of this is clear: mathematical notation, like all forms of language, is primarily designed for easy communication. If there is no (reasonable) way to misinterpret the bunch of symbols you write on the page, then you are using them correctly.

  • $\begingroup$ So it seems like I was correct in the first part of my question. This symbol in a statement from some proof means some “anti-derivative” evaluated at x of a function. And in formula books or problems, it mean the set of “anti-derivatives” of functions. $\endgroup$
    – TFR
    Commented Feb 26, 2022 at 20:49
  • $\begingroup$ Then what is the precise definition of an “antiderivative”? Is it the same as I wrote in the question? $\endgroup$
    – TFR
    Commented Feb 26, 2022 at 20:50
  • $\begingroup$ @TFR an antiderivative of the function $f$ is exactly what you wrote, a function $F$ such that $F'=f$. This entails $F$ and $f$ having the same domain and range (depending on your definition of function). $F$ is never unique, as $x\mapsto F(x)+c$ is another antiderivative for any $c\in\mathbb{R}$. Or, in my language, an antiderivative is a member of the set of functions $\int f(x)\, \mathrm dx$. $\endgroup$
    – A.M.
    Commented Feb 26, 2022 at 20:53
  • $\begingroup$ What does F have the same domain as f? F could have a domain larger than f’s but not differentiable except on values in the domain of f, so that F’ = f. $\endgroup$
    – TFR
    Commented Feb 26, 2022 at 20:58
  • 1
    $\begingroup$ @TFR $F'=f$ is a statement that the functions are equal. They're not equal without the same domain. In a common abuse of notation, $F'=f$ might mean what you say, but this is not what $=$ formally means. $\endgroup$
    – A.M.
    Commented Feb 26, 2022 at 21:37

To directly answer your question, $F$ is an antiderivative of $f$ if and only if $F' = f$. In this case, we can write $$F(x) = \int f(x) \ \text dx$$ This is the most common (and arguably, the only reasonable) definition of the word.

However, I think the source of confusion here is not your understanding of antiderivatives but that of functions. We like to throw around symbols and write things like $f = g$ because it's concise and precise, but it's not exactly clear without some prior understanding. When we define functions like $$ f: X\to Y \\ f: x\mapsto y$$ we're saying a lot of different things. First, there are sets $X$ and $Y$, called the domain and codomain, respectively. Then, there is a rule that associates to each $x\in X$ a particular $y\in Y$ (usually written as an expression involving $x$). We can think of the function as a triple: (domain, codomain, rule). Finally, $f$ is simply the name given to that function.

The upshot of all this is that a statement like $f = g$ means that the whole triple of (domain, codomain, rule) is the same between $f$ and $g$. Importantly, if $f$ and $g$ have different domains, then they are not the same function, even if they agree everywhere their domains overlap. This is why the statement at the end of your question is correct; $F'(x) = f(x)$ for $x\in (0,1)$, but $F'$ and $f$ have different domains and so are different functions.

  • $\begingroup$ I understand the meaning of f = g. What I wanted to be sure about is whether that is the definition (which is according to many). However, someone said F’ = g implies F’s domain is the same as f because F’ is a function whose domain is the same as f. What I have been thinking is that F’ is a function defined whenever F is differentiable, so that their domains might be different. The reason for this is because I saw the derivative of F at x is often written as F’(x), whether F is a differentiable function (differentiable at every point in domain) or not. $\endgroup$
    – TFR
    Commented Feb 27, 2022 at 16:39
  • $\begingroup$ @TFR It's certainly true that $F$ and $F'$ may have different domains, in fact this is quite typical. However, the equation $F' = f$ makes no direct reference to the domain of $F$; regardless of whether $F'$ and $F$ share a domain, $F' = f$ requires that $F'$ and $f$ do. $\endgroup$
    – Alex Jones
    Commented Feb 27, 2022 at 17:45
  • $\begingroup$ This is what I thought, F can have F’ when F is not a differentiable function. However, this is different from the other two answers which require F’ has the same domain as F. $\endgroup$
    – TFR
    Commented Feb 27, 2022 at 20:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .