Defennder said:
Your post hasn't explained why in the context of the definite integral there isn't any +c, unlike danago's. But of course there is some even deeper maths theory I have yet to learn as to how the definite integral came about.
I didn't explain it because danago did =-)
So we're working with four things. The derivative, the anti-derivative, the definite integral, and the definite integral. Let's work out the definitions for the rest in terms of the derivative.
It's true that the derivative throws away some information about your function. But it doesn't throw it all away! It actually keeps *most* of the information about your function except for the value of f(0). If you know the value for f(0) and f'(x), you can figure out your function. (Assuming of course, your function is "well-behaved"... it probably needs to be analytic or something is my guess).
The anti-derivative is another operation on functions. It doesn't have a standard notation, so let's call it A(f) for the anti-derivative of f. It is the unique operation which satisfies the property:
[tex]\frac{d}{dx} A(f) \vert^{x} = A(\frac{df}{dx}) \vert^x= f(x) - f(0)[/tex]
The nice thing about the anti-derivative is that it is the inverse of derivation when you only consider functions where f(0) = 0. It's also an operation, meaning applying it to any function f gives you a single value. No +c's or anything going on here!
However, the anti-derivative operator A is NOT the inverse of derivation in general. As I said in my last post, no such operation exists, because derivation is non-injective. There is no inverse operation for derivation.
But set and function theory in mathematics says that even though not every operation has an inverse operation. Given a function f, the best you can do is generate the set of functions whose derivative is f: {F | dF/dx = f}. Let's call this set I
f (I for "Integral"... and the sub f, because it's dependent on a choice of f). The set I
f contains the anti-derivative of f, A(f), and all other functions of the form F(x) = A(f) + c, for any real number c. So we can write I
f as {F | F(x) = A(f) + c, for any real n}.
What math and physics teachers like to do is pretend is to ignore the squiggly brackets { and } denoting it as a set and treat it like a single value! They use the notation F for A(f), and then claim the indefinite integral of f is F(x) + c. But the indefinite integral is really the set of possible values! An infinite set too!
But the teachers and physicists don't care, because of a nice trick. The definite integral from b to a is obtained by taking any function in I
f, calling it F, and then evaluating F(a) - F(b). Since you only take a function out of the set once, you get to "fix" c to a constant. So if F(x) = (A(f))(x) + c, then
[tex]F(a) - F(b) = (A(f))(a) + c - (A(f))(b) + c)<br />
= (A(f))(a) - (A(f))(b)[/tex]
(Note that A(f) is a function, so by (A(f))(x), I mean taking the antiderivative of f, and then inputting a value of x to it).
Taking the difference of F(a) and F(b) causes the c to cancel itself.
So, as you can tell by this post, the reason for the +c is because the actual formulation is a little complicated! I blame it on the notation, because things like (A(f))(x) seem very unnatural to most people after they finite algebra and trigonometry. But once you get used to it, the ideas behind calculus aren't terribly hard. The derivative and anti-derivative "operations" (as I've been calling them) can actually be treated in mathematics as regular functions! Not functions from reals to reals, of course, but functions from functions to functions.
In mathematics and computer science, we have a nice little notation for function types. If we let R be the set of reals, set can talk about the set R->R, or real valued functions, which take a real number as input and output another real number. What the derivative and anti-derivative operators are, are actually elements of the set (R->R)->(R->R). You give them a real function, (f) and you get back another (df/dx).
But you don't really have to care about the details. As long as you can (A) get the right answer and (B) acknowledge there are more details you haven't considered, then it doesn't really matter, does it? =-)
Why only real n Tac Tics?
You're right, I should have accounted for hyperreal, surreal, dual, and quaternion n as well =-P