
#1
Feb611, 03:40 PM

P: 20

Given an indefinite integral,
[tex] \int f(x) dx = F(x) + C, [/tex] 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? 



#2
Feb611, 03:50 PM

HW Helper
P: 805

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.




#3
Feb611, 04:34 PM

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P: 11,863

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).
[tex] \mbox{The indefinite integral of the function f(x)} = \{F(x) + C  F'(x) = f(x), C\in\mathbb{R}\} [/tex] 



#4
Feb711, 08:23 AM

Math
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PF Gold
P: 38,879

What _is_ an indefinite integral?
A better notation is
[tex]\int^x f(t)dt[/tex]. 



#5
Feb711, 08:52 AM

Mentor
P: 14,433

I have seen the indefinite integral written as [tex]\int dx\,f(x)[/tex] which makes [itex]\int dx[/itex] look more like an operator. 



#6
Feb911, 02:42 PM

P: 1

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. 



#7
Feb911, 02:54 PM

P: 4,513

[tex]\int f(x) dx = F(x) + C = g(x,C)[/tex]




#8
Feb1711, 12:28 AM

P: 367

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




#9
Feb1711, 12:39 AM

P: 737





#10
Feb1711, 12:48 AM

P: 367





#11
Feb1711, 06:05 AM

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P: 11,863

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.




#12
Feb1711, 10:15 AM

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PF Gold
P: 38,879

Here, use [ tex ] [/tex ] (without the spaces) or [ itex ] [/itex ] for "inline" LaTeX. [tex]\int2x\,dx=x^2[/tex] [itex]\int2x\,dx=x^2[/itex] 



#13
Feb1711, 10:53 AM

Sci Advisor
P: 1,686

[itex]\int f(x) dx[/itex] 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.



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