anti..inte

G'day, Lizzie.

Yes, an antiderivative is the same as an integral. (Perhaps "indefinite" integral is more precise.)

Integration is the inverse operation of differentiation.
 
Ok, so the technical term for the process is indefinate integral, and it is commonly called aniderivative, like a nick name? I get it. Thanks.
 
I'm not exactly positive on this, but I've always thought of it like this:

An antiderivative is just taking a given expression and doing opposite of derivation. In other words, the antiderivative of \(\displaystyle x^2\) is \(\displaystyle \frac{1}{3}x^3\), because the second expression's derivative is the first one.

However, an integral is more of an operation and so it has the integral sign and sometimes endpoints.

So yes, they are the same, but if my teacher asked for an antiderivate I probably wouldn't put an integral sign on it. Those of you who are teachers: is this right? It's actually something that I was mildly curious about myself.

Ted
 
I think that's basically what Unco said. A definite integral has endpoints, so it's not the same as an antiderivative. But, an indefinite integral doesn't have specific endpoints and is therfore the same as an antiderivative. At least, that's what I gathered.
 
If you look up "antiderivative" on the MathWorld site, you get no article; just a reference to the "indefinite integral" article.

According to Wikipedia, the "integral" is a number (that is, the value returned by a definite integral) while the "antiderivative" is a function (that is, the value returned by an indefinite integral). The Wikipedia article says that the point of the Fundamental Theorem of Calculus is the relationship between integrals and antiderivatives.

The Visual Calculus site appears to use the Wikipedia definition.

I was always of the understanding that antiderivatives and integrals were closely related, but not identical.

Eliz.
 
Thanks Stapel, you always post very informative messages full of links - I appreciate it.
 
I was always of the understanding that antiderivatives and integrals were closely related, but not identical.
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??? It all becomes less clear when one uses numerical techniques to evaluate an integral expression. Does the integral, then, have ANYthing to do with an anti-derivative?

Please note also that an anti-derivative cannot be the "opposite" or "inverse" of a derivative, barring a little luck. The derivative is destructive to constants, thus the constant of integration.
 
tkhunny said:
Does the integral, then, have ANYthing to do with an anti-derivative?
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I think that's the point of the Fundamental Theorems of Calculus.

tkhunny said:
Please note also that an anti-derivative cannot be the "opposite" or "inverse" of a derivative, barring a little luck.
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Maybe it could be looked at as a many-to-one or one-to-many relation? Any number of functions differentiate to one derivative, and thus one function antidifferentiates to any number of antiderivatives. So 3x<sup>2</sup> + 5x + 2 and 3x<sup>2</sup> + 5x - 47 both map to 6x + 5, and 6x + 5, because of the constant of integration, maps back to both 3x<sup>2</sup> + 5x + 2 and 3x<sup>2</sup> + 5x - 47...?

Just an idea...

Eliz.
 
Granted and Granted. I just wanted to poke at it a bit more.
 
The real problem is that there are functions that are themselves a derivative of another function yet the first function has no ordinary integral on the finite interval over which the derivative exists. Such functions were first were found in the 1890’s and set off a century of research into definitions of an integral for which all derivatives are integrable.
 
wow, This has turned out to be quite the interesting topic ;). I'm actually very interested in this conversation because it all makes sense to me now and I am following what's going on. I mean, now I understand the relatonship between anti-derivatives and indefinite integrals (as well as the lack there of) and the lack of relation between anti-derivatives and definite integrals. I would like to see this keep going, unless everyone has gotten sick of it, lol.
 
The fundamental theorem of calculus applies only to continuous functions. In general it is perfectly possible for a function to be integrable (its definite integral exists) yet have no anti-derivative. For example take f: R->R given by f(0) = 1 and f(x) = 0 for any x != 0. Then f has no anti-derivative (since the derivative of any function satisfies the intermediate value property - it is Darboux continuous) yet it is integrable on any interval (with integral zero).

In fact the converse fails as well in general. It is possible for a function to be the derivative of another function yet not be Riemann integrable. For example set f(x) = x^2 sin(1/x) for x != 0 and f(0) = 0. Then f' exists but is not Riemann integrable (since it's not bounded). On the other hand if f: [a,b]->R has an anti-derivative F: [a,b]->R _and_ f is Riemann integrable on [a,b] then the integral of f on the interval [a,b] is equal to F(b)-F(a).

[Edit: made a correction in last paragraph]
 
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