What _is_ an indefinite integral?

[monea83]

Given an indefinite integral,

<br /> \int f(x) dx = F(x) + C,<br />

I am having some problems in understanding what this indefinite integral "is". The RHS is clearly a function, but what is the LHS? Judging by the equals sign, it should also be a function, but seemingly it isn't because there's no place at which to plug in a function argument (aka "x"). There is an "x" appearing on the LHS, but this is obviously a bound variable...

Of course I know how to use antiderivatives, but what I am trying to grasp here is the "nature" of an indefinite integral - what kind of object is it?

 

[gb7nash]

Simply put, the definition of an indefinite integral of a function f is the function, call it F, who's derivative is f. So you can think of the antiderivative as being the opposite of the derivative. However, it is not the inverse since the constant can be any real number.

 

[dextercioby]

In the reals, the indefinite integral (aka the antiderivative) of a function f(x) is the infinite set of differentiable functions whose first derivative equals f(x).

\mbox{The indefinite integral of the function f(x)} = \{F(x) + C | F&#039;(x) = f(x), C\in\mathbb{R}\}

 

[HallsofIvy]

A better notation is
\int^x f(t)dt.

 

[D H]

A better notation is
\int^x f(t)dt

Perhaps, but I have yet to see that notation used anywhere.

I have seen the indefinite integral written as

\int dx\,f(x)

which makes \int dx look more like an operator.

 

[C.I.]

I think monea83 is right. There is something fundamentally wrong with the notion and the notation of 'indefinite integral'. If
\[
\int2x\,dx=x^2
\]
(with or without constant) you are entitled to plug in, say, $x=1$, which will lead you to all kinds of funny conclusions.

P.S. I'm new here and I was hoping my LaTeX code would somehow be automatically translated.

 

[Phrak]

\int f(x) dx = F(x) + C = g(x,C)

 

[wisvuze]

so is the indefinite integral one particular primitive, or the set of all of them? :S

 

[TylerH]

Given an indefinite integral,
The RHS is clearly a function, but what is the LHS? Judging by the equals sign, it should also be a function, but seemingly it isn't because there's no place at which to plug in a function argument (aka "x"). There is an "x" appearing on the LHS, but this is obviously a bound variable...

There is no way to say \int f(x)dx=F(x)+C and explicitly define what F(x) is. It's called an implicit function because there is no direct translation. The purpose is to specify the relation between the two, and leaves the leg work of finding what F(x), in explicit terms of x, is left up to the solver.

 

[wisvuze]

\int f(x) dx = F(x) + C = g(x,C)

this sounds reasonable to me, but a lot of books define the indefinite integral as the collection of ALL primitives. Also, according to the wikipedia article: http://en.wikipedia.org/wiki/Antiderivative, the indefinite integral indeed refers to one primitive at a time

 

[dextercioby]

I've been taught that an antiderivative of a functions is an element of the set I wrote in post #3. The set however defines the indefinite integral of the function f.

 

[HallsofIvy]

I think monea83 is right. There is something fundamentally wrong with the notion and the notation of 'indefinite integral'. If
\[
\int2x\,dx=x^2
\]
(with or without constant) you are entitled to plug in, say, $x=1$, which will lead you to all kinds of funny conclusions.

P.S. I'm new here and I was hoping my LaTeX code would somehow be automatically translated.

Unfortunately, the tags used to delimit LaTeX code varies from board to board.

Here, use [ tex ] [/tex ] (without the spaces) or [ itex ] [/itex ] for "inline" LaTeX.

\int2x\,dx=x^2

\int2x\,dx=x^2

 

[disregardthat]

\int f(x) dx is the set of functions (with domain equal to the domain of f) whose derivative wrt. x is is f(x). If F(x) is an antiderivative, then F(x) + C is as well for any constant C, and these functions exhaust the set of antiderivatives. You could say it is shorthand for set construction notation. Indefinite integration is not an inverse operation (as an inverse to an operation from functions to functions) to differentiation in the strict sense of the term, since differentiation is not an injective operator.