I thought these were different words for the same thing, but it seems I am wrong. Help.
 3$\begingroup$ Not really a direct answer to your question, but note that there are examples of functions which are integrable but which don’t have an antiderivative, and examples of functions which have an antiderivative but are not integrable. Using “indefinite integral” to mean “antiderivative” (which is unfortunately common) obscures the fact that integration and antidifferentiation really are different things in general. $\endgroup$ Nov 30, 2013 at 1:37
 3$\begingroup$ Wolfram Mathworld says that an indefinite integral is “also called an antiderivative”. This MIT page says, “The more common name for the antiderivative is the indefinite integral.” One is free to define terms as you like, but it looks like at least some (and possibly most) credible sources define them to be exactly the same thing. $\endgroup$– user46234May 24, 2016 at 13:24
9 Answers
“Indefinite integral” and “antiderivative(s)” are the same thing, and are the same as “primitive(s)”.
(Integrals with one or more limits “infinity” are “improper”.)
Added: and, of course, usage varies. That is, it is possible to find examples of incompatible uses. And, quite seriously, $F(b)=\int_a^b f(t)\,dt$ is different from $F(x)=\int_a^x f(t)\,dt$ in what fundamental way? And from $\int_0^x f(t)\,dt$? And from the same expression when $f$ may not be as nice as we’d want?
I have no objection if people want to name these things differently, and/or insist that they are somewhat different, but I do not see them as fundamentally different.
So, the real point is just to be aware of the usage in whatever source…
(No, I’d not like to be in a classroom situation where grades hinged delicately on such supposed distinctions.)
 1$\begingroup$ www.ma.utexas.edu/users/sadun/F11/408N/FTC.pdf This defines them differently though. Is it incorrect? $\endgroup$ Nov 29, 2013 at 21:27
 6$\begingroup$ Well, this is not a question of “truth” but of mathematical usage. It surprises me to see that an integral with “variable” upper bound is called “indefinite”. (Not to mention the serious but potentiallynonsensical question of “what is a variable, as opposed to constant?) Or, for all I know, usage may have drifted over the years… One must always pay attention to context, which may entail usages different from what’s expected. As long as it’s upfront and not deceitful, it’s mostly harmless, even if jarring. $\endgroup$ Nov 29, 2013 at 21:31
 2$\begingroup$ @paulgarrett : many people call “the primitive” of a function the antiderivative that has the value $0$ at $0$. I thought this was universal, but I Googled it and found several people who used it for any antiderivative. $\endgroup$ Nov 30, 2013 at 1:26
 2$\begingroup$ Pugh’s “Real Mathematical Analysis” also uses “indefinite integral” to mean the function $F(x) = \int_a^x f(t)\,dt$. $\endgroup$ Nov 30, 2013 at 1:32
 2$\begingroup$ The difference is that the antiderivative has to be differentiable; the indefinite integral does not. $\endgroup$– epsilonJan 22, 2020 at 4:47
An antiderivative of a function $f$ is a function $F$ such that $F’=f$.
The indefinte integral $\int f(x)\,\mathrm dx$ of $f$ (that is, a function $F$ such that $\int_a^bf(x)\,\mathrm dx=F(b)F(a)$ for all $a<b$) is an antiderivative if $f$ is continuous, but need not be an antiderivative in the general case.
 5$\begingroup$ In which books is the definition of the indefinite integral stated? $\endgroup$– ArbujaOct 1, 2016 at 15:23
An antiderivative of a function $f$ is one function $F$ whose derivative is $f$. The indefinite integral of $f$ is the set of all antiderivatives of $f$. If $f$ and $F$ are as described just now, the indefinite integral of $f$ has the form $\{F+c \mid c\in \mathbb{R}\}$. Usually people don’t both with the setbuilder notation, and write things such as “$\int \cos(x)\,dx = \sin(x)+C$”.
This is what I was taught. One of the other answers here is completely different. I did some Googling, and, to my surprise, Wikipedia defines an improper integral as a single function. I found a link at http://people.hofstra.edu/stefan_waner/realworld/tutorials4/frames6_1.html that agrees with my answer. I don’t know if there is any consensus in the math community about which answer is correct.
(http://math.mit.edu/suppnotes/suppnotes0101a/01ft.pdf) p 1 sur 7
Antiderivative is an indefinite integral.
The answer that I have always seen: An integral usually has a defined limit where as an antiderivative is usually a general case and will most always have a $\mathcal{+C}$, the constant of integration, at the end of it. This is the only difference between the two other than that they are completely the same.
You will be safe in class though if you assume them to be identical if neither of them has a defined limit.
(J. Stewart. Calculus pp 391) I believe Stewart defines an antiderivative as an indefinite integral.
 2$\begingroup$ Nope. This is only true if $f$ is continuous. Stewart is assuming $f$ is continuous, as you can see in the paragraph above the paragraph that you quoted. $\endgroup$– epsilonJan 20, 2020 at 21:33
i think that indefinite integral and anti derivative are very much closely related things but definitely equal to each other. indefinite integral denoted by the symbol”∫” is the family of all the anti derivatives of the integrand f(x) and anti derivative is the many possible answers which may be evaluated from the indefinite integral. e.g: consider the indefinite integral: ∫2xdx in this indefinite integral, 2x is the differentiated function with respect to x and the integration results in the function x^2+c where “c” is the constant but what is the antiderivative of 2xdx? the answer is not one it can be many i.e x^2+10, x^2+4, x^2+0 etc and these all are the members of ∫2xdx. so the family i.e ∫2xdx is the integral and all its member are the antiderivatives.
I can’t believe I haven’t found a single answer that says this, but:
This is exactly what the fundamental theorem of calculus is about.

An integral is an area function. An integral of function $f:\mathbb R \to \mathbb R$ (usually assumed continuous I guess) is any $F:\mathbb R \to \mathbb R$ s.t. $F(x) – F(a) = \int_a^x f(t)dt$. $F$ tells you the area under $f$ from $a$ to $x$.

An antiderivative $G:\mathbb R \to \mathbb R$ is any differentiable function whose derivative is $f$.
The whole point of the fundamental theorem of calculus is to tell you
antiderivatives and integrals are identical.

Integrals are antiderivatives: Such an area function $F$ of $f$ is actually not only a continuous function but a differentiable function whose derivative is $f$ itself, i.e. $F'(x)=f(x)$.

Antiderivatives are integrals: Such an antiderivative $G$ of $f$ can actually be used to compute the area under $f$ like $\int_a^x f(x) dx = G(x)G(a)$.
Notes:

$F(x)F(a)=G(x)G(a)$ for any area functions / integrals / antiderivatives $F,G$ and for any $x,a \in \mathbb R$.

This is because the rule is any antiderivatives $F$ & $G$ will differ by a constant. Like for all $F,G$ there exists a $C$ s.t. for all $z \in \mathbb R$, $F(z)G(z)=C$. In particular, $F(x)G(x)=F(a)G(a)=C$.

It’s just occurring to me that the fundamental theorem of calculus’ saying that integrals and antiderivatives are identical is kinda like in complex analysis where they say that holomorphic and analytic are identical. (This isn’t the fundamental theorem of complex analysis though, which is apparently Cauchy’s Integral Theorem. Kinda anticlimactic, but ok I guess.)
The indefinite integral is a definite integral in which we ignore the limits of integration:
$$\int f(x) \,dx$$
The utility of the indefinite integral is in finding the antiderivative $F(x)$ of a complicated function. That is, the antiderivative is the result of solving the indefinite integral:
$$ \int f(x) \,dx = F(x) \longrightarrow F'(x)=f(x) $$