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Proof that you don't need +c

  1. Sep 13, 2008 #1


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    When you do limited integration you don't need to put +c on the end. My maths teacher showed me a proof for this once but I can't rember it. Can any one show me the proof?
  2. jcsd
  3. Sep 13, 2008 #2


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    The way i like to think of it is like this:

    \int_a^b {f(x)\;dx}
    is evaluated by evaluating the indefinite integral of f(x) at b and then subtracting the indefinite integral evaluated at a.

    If i let the indefinite integral of f(x) be F(x)+c where c is the constant of integration, then:

    \int_a^b {f(x)\;dx} = [F(b) + c] - [F(a) + c] = F(b) - F(a)

    Note how the constant cancels out?
  4. Sep 13, 2008 #3
    In all of mathematics, nothing annoys me more than the +c convention in calculus ;-)

    Let's start by looking at where the hell +c comes from.

    Take a derivative of a few functions.

    f(x) = x^2, f'(x) = 2x
    f(x) = x^2 + 1, f'(x) = 2x
    f(x) = x^2 + 2, f'(x) = 2x
    f(x) = x^2 + 3, f'(x) = 2x
    f(x) = x^2 + n, f'(x) = 2x, for any real number n

    You can see that for all of these functions, the derivative is the same. In other words, the derivative operation is mapping different inputs to the same output. You are *losing* information about the input function, f, and if you're only given f', then you have no idea what value n was in the original equation.

    In mathematics, there is an idea of an injective operator. An injective operator is any operator which preserves all information on the input, and thus, it can be undone. To "undo" an injective operator, we apply another operator, called that operator's inverse. Operators which are *not* injective have *no* inverses.

    The derivative operation is *not* injective. It maps different functions to the same output. You are losing information about the input. And so, there is no way to undo it, and no inverse operation. If you know f'(x) = 2x, then f(x) could be any of f(x) = x^2, f(x) = x^2 + 1, f(x) = x^2 + 2, .... In other words, the inverse of derivation is a multi-valued operator. It has no single value, but rather, you can think of it as being the set of possible values whose derivative is 2x.

    Math teachers and practical scientists don't want to deal with functions and set theory and inverse operations. Actually, most of the mathematical language to describe this was invented later, so calculus textbooks still teach calculus similar to how Newton and Leibniz did things. Since you can get the right answer without using a strictly mathematically sound technique, this abuse of notation persists.

    Since the poster above me gave the short answer, the long answer is that it is an abuse of notation caused by the fact that the derivative has no inverse operator. But treating it as if it were an arbitrary constant gets you the correct answer.
  5. Sep 13, 2008 #4


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    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.
  6. Sep 13, 2008 #5
    Why only real n Tac Tics?
  7. Sep 13, 2008 #6
    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 If (I for "Integral"... and the sub f, because it's dependent on a choice of f). The set If 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 If 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 If, 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)
    = (A(f))(a) - (A(f))(b)

    (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? =-)

    You're right, I should have accounted for hyperreal, surreal, dual, and quaternion n as well =-P
  8. Sep 13, 2008 #7
    you forgot the definite integral methinks
  9. Sep 13, 2008 #8
    The one thing I kinda skimped over was the indefinite integral. Some people take the indefinite integral to mean the anti-derivative. Most other people take it to mean the anti-derivative "+c". I like to think of the indefinite integral as a multi-valued function (or a relation). An operation which produces the set If. But that's just a personal way of thinking about it to remind me that those details need to be accounted for.
  10. Sep 17, 2008 #9
    I was just being sarcastic and pointing out that you wrote "definite integral" twice :)
  11. Sep 17, 2008 #10
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