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Antiderivatives x^n

  1. May 1, 2006 #1
    I'm learning about antiderivatives now, and I just learned a formula to find the antiderivative of xn:
    if f(x) = xn, and F'(x) = f(x), then
    F(x) = (xn+1)/(n+1) + C, where C is a constant
    but if n < 0, then
    F(x) = (xn+1)/(n+1) + C1, if x > 0
    F(x) = (xn+1)/(n+1) + C2, if x < 0

    My question is why are there two different C's?

    The only example my book gives is of finding the antiderivative of f(x) = x-3. Its answer is:
    F(x) = -1/(2x2) + C1, if x > 0
    F(x) = -1/(2x2) + C2, if x < 0

    EDITED: f(x) should be x-3
     
    Last edited: May 1, 2006
  2. jcsd
  3. May 1, 2006 #2

    berkeman

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    The C value is determined by the initial conditions of the system, or other information that you are given. You need more information to solve for the C in any of those antiderivatives, so maybe the book is just trying to keep you from thinking that it will always be the same C?
     
  4. May 1, 2006 #3
    What would be different if you take a derivative of a constant ? If you don't like C1, C2, you can make your own C100 or C10000, who makes it hard for you ?
     
  5. May 1, 2006 #4

    HallsofIvy

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    There aren't two different C's- there are an infinite number of different C's!!
     
  6. May 1, 2006 #5

    VietDao29

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    Nah, I don't know what your book is doing either, C is just a constant, any constant will work.
    If [tex]f(x) = x ^ n , \ n \neq -1 \Rightarrow F(x) = \frac{x ^ {n + 1}}{n + 1} + C[/tex].

    No this example is wrong!!!! f(x) = x3, then
    F(x) = x4 / 4 + C
    C is any constant, and the derivative of a constant is 0, so if you differentiate F(x) with respect to x, you will get f(x):
    F'(x) = (x4 / 4 + C)' = (x4 / 4)' + C' = 4 . x3 / 4 + 0 = x3 = f(x)
    Can you get this? :)
     
  7. May 1, 2006 #6

    Curious3141

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    C is completely arbitrary, and there's no need to separate the cases based on the value of n (except for the special case where n = -1

    Sure it wasn't f(x) = x^(-3) ?
     
  8. May 1, 2006 #7

    VietDao29

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    Even if it was x-3, it's still wrong!!! :smile:
     
  9. May 1, 2006 #8

    Curious3141

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    Exactly how is it wrong ?

    [tex]\int{x^{-3}}dx = \frac{x^{-2}}{-2} + C = -\frac{1}{2x^2} + C[/tex]
     
  10. May 1, 2006 #9

    HallsofIvy

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    The reason for having two "different" constants when n< -1 is because they are separated by x= 0 for which xn+1 is not defined. Since the function is not continuous at x= 0, there is no reason to think the values, as you approach x=0 from each side, are the same.
     
  11. May 1, 2006 #10

    Curious3141

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    I don't get you. :confused:

    All the constant of integration says is that there's a whole family of curves y = F(x)(which can be translated vertically to superimpose on one another) that share the common property of having a derivative that varies in that fixed way (F'(x) = x^(-3) ) for every value of x where the function and its derivative are well defined. The value of the constant should have nothing to do with the domain of the function or its derivative.

    If they wanted to specify the domain properly they could just have stated :

    [tex]f(x) = x^{-3}, x \in[/tex] R\{0}

    which means the answer would be [tex]F(x) = -\frac{1}{2x^2} + C, x \in[/tex] R\{0}, [tex]C \in[/tex] R
     
    Last edited: May 1, 2006
  12. May 1, 2006 #11

    0rthodontist

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    :eek: !!!!!!
    I never thought of that before.
     
    Last edited: May 1, 2006
  13. May 1, 2006 #12

    Curious3141

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    Are you saying you understand or agree with Halls ?
     
  14. May 1, 2006 #13

    0rthodontist

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    Yes. Halls' point is that the antiderivative has an asymptote such that an integral spanning the asymptote doesn't make sense. So the things on either side of the asymptote can be shifted up or down independently of one another, and every definite integral that does not span the asymptote will not be affected.
     
  15. May 1, 2006 #14

    Curious3141

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    Ah, I think I see now. What he's saying then is that a solution like :

    F(x) = -(1/2)x^(-2) + 7 for all x>0
    AND
    F(x) = -(1/2)x^(-2) + 3 for all x<0

    is perfectly admissible as a particular solution to the diff. equation F'(x) = x^(-3) for all x in R\{0} ?

    I completely agree, that IS a new way to look at it, and I hadn't thought of it before. :approve:
     
    Last edited: May 1, 2006
  16. May 2, 2006 #15
    Thanks! I think I needed to visualize a graph that is discontinuous at x=0. But if it was continuous everywhere, then there would only be one constant. Since it's discontinuous, there are two constants which can be the same or could be (and probably are) different.:tongue:
     
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