What is the difference between an indefinite integral and an antiderivative?

Question

I thought these were different words for the same thing, but it seems I am wrong. Help.

Answer 1

"Indefinite integral" and "anti-derivative(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.)

Answer 2

An anti-derivative 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.

Answer 3

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 set-builder 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.

Answer 4

(http://math.mit.edu/suppnotes/suppnotes01-01a/01ft.pdf) p 1 sur 7

Antiderivative is an indefinite integral.

Answer 5

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.

Answer 6

(J. Stewart. Calculus pp 391) I believe Stewart defines an antiderivative as an indefinite integral.

Answer 7

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.

Answer 8

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.

1. 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 $x$.

2. 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

anti-derivatives and integrals are identical.

1. Integrals are anti-derivatives: 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. $F'(x)=f(x)$.

2. Anti-derivatives 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:

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

2. This is because the rule is any anti-derivatives $F$ & $G$ 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, $F(x)-G(x)=F(a)-G(a)=C$.

3. It's just occurring to me that the fundamental theorem of calculus' saying that integrals and anti-derivatives 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 anti-climactic, but ok I guess.)

Answer 9

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:

$$ \int f(x) \,dx = F(x) \longrightarrow F'(x)=f(x) $$