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Homework Help: The Antiderivative

  1. Dec 28, 2007 #1
    There are several techniques for derivation that have proved that the derivative for x^n is n*x^(n-1). There is the method of incrementation, working extensively with limits, etc... I'm unsure of the antiderivative. Is there any mathematical proof for the antiderivative or did they simply say that it reverses differentiation and came up with it just by looking at the method for derivation?
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  3. Dec 28, 2007 #2


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    well by definition:

    [tex]\int x^N dx =\frac{x^{N+1}}{N+1} + C[/tex]

    so you can just use that definition.
  4. Dec 28, 2007 #3


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    That's not a definition. It's a theorem. The proof that integration corresponds to antidifferentiation is called the "Fundamental Theorem of Calculus".
  5. Dec 28, 2007 #4


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    ah well I learned something today
  6. Dec 28, 2007 #5
    differentiate it!
  7. Dec 28, 2007 #6


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    Throughout mathematics we come upon "direct" problems and "indirect" or "inverse" problems. Here's a simple algebraic example: if you are given the function f(x)= [itex]x^5- 7x^4+ 3x^3- 2x^2+ x- 5[/itex] and asked "what is f(1)?", that's easy. You are given the formula to use and just do the arithmetic: f(1)= 1- 7+ 3- 2+ 1- 5= -9. If, however, you are asked to solve the equation, f(x)= [itex]x^2- 7x^4+ 3x^3- 2x^2+ x- 5= -9[/itex], that's extremely difficult! There is no "formula" for solving general 5th degree equations. Here, because we had already done the evaluation, we know that one solution is x= 1. However, there may be up to 4 more solutions that we don't know.

    Differentiation is a "direct" problem because we are given a formula for it. Integration, or "anti-differentiation" is defined only as the inverse of differentiation. A result of that fact is that the anti-derivatives of most integrable functions cannot be given in terms of elementary functions.

    If we can "remember" or otherwise find a function whose derivative is the given function- for example, we know that the derivative of [itex](1/(n+1)) x^{n+1}[/itex] is [itex]x^n[/itex]- then we know that the new function is the "anti-derivative" of the original function- [itex]x^n[/itex] is the anti-derivative of [itex](1/(n+1))x^{n+1}[/itex].
    Of course, we also know that there are an infinite number of other functions having that same derivative. fortunately, we can prove, using the mean value theorem, that if two functions have the same derivative, they must differ only by a constant- we know that any anti-derivative of [itex]x^n[/itex] must be of the form [itex](1/(n+1))x^{n+1}+ C[/itex] where C is a constant.
  8. Dec 28, 2007 #7
    I see. A lot of information given, thanks.
  9. Dec 29, 2007 #8

    Gib Z

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    When you say "anti derivative" it implies, the process of differentiation reversed. From the rest of your sentence I think you may have instead wanted to use the word "integral". The Only reason we can use those words interchangeably is because of our knowledge of the Fundamental theorem of Calculus, but remember they were originally two different things.

    Now if that quote was asking if there is another way to prove [tex]\int^b_a x^n = \frac{b^{n+1} - a^{n+1}}{n+1}[/tex], then yes. Riemann sums.
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