Results 1 to 3 of 3

Math Help - Quick puzzling question (antiderivatives)

  1. #1
    Member Mentia's Avatar
    Joined
    Dec 2008
    From
    Bellingham, WA
    Posts
    134

    Quick puzzling question (antiderivatives)

    I am having some trouble with antiderivatives. On occasion I see something like:

    The antiderivative of f(x)  = \int_{0}^{x} f(x')dx'

    I don't understand why this is used since it isn't true for something as simple as f(x) = e^x, in which case it is off by a constant. Can someone let me know where my understanding has gone wrong and why this notation is used?
    Last edited by Mentia; March 3rd 2009 at 04:45 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member mollymcf2009's Avatar
    Joined
    Jan 2009
    From
    Charleston, SC
    Posts
    490
    Awards
    1
    Quote Originally Posted by Mentia View Post
    I am having some trouble with antiderivatives. On occasion I see something like:

    The antiderivative of f(x)  = \int_{0}^{x} f(x')dx'

    I don't understand why this is used since it isn't true for something as simple as f(x) = e^x, in which case it is off by a constant. Can someone let me know where my understanding has gone wrong and why this notation is used?
    I have never seen it written that way. All an antiderivative does is "undo" the derivative. The way this is written is redundant because when f(x) is written as an integral it is already the derivative. So, the antiderivative is usually written as F(x) to show that it was originally in the form of a derivative.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member Mentia's Avatar
    Joined
    Dec 2008
    From
    Bellingham, WA
    Posts
    134
    Okay I think I understand what's going on in case anyone was wondering (probably not). It is helpful notation when you don't know what f(x) actually is and you are just writing the general case. For instance:

     \frac{d }{dx } g(x) = f(x)

    Then,
    <br />
g(x) =  \int_{ 0}^{ x} f(x')dx' + C1

    Notice this is slightly different from

    <br />
g(x) =  \int f(x)dx
    g(x)    = F(x) + C2

    where F(x) is the antiderivative of f(x). It should still be true in general. Notice it makes things slightly easier when you have boundary conditions at zero. For instance, if your condition is g(0) = 1 this immediately gives C1 = 1, whereas you have no knowledge about C2 without knowing f(x).

    Similarly if your conditions are at some other constant, K, you can integrate from K to x, and then your constant of integration is equal to whatever your boundary condition value is:

    g(x) =  \int_{ K}^{ x} f(x')dx' + C1, g(K)=Q --> C1 = Q
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Puzzling Ans
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 20th 2011, 02:12 PM
  2. puzzling problem! please help
    Posted in the Advanced Statistics Forum
    Replies: 11
    Last Post: July 9th 2008, 05:14 PM
  3. [SOLVED] Puzzling puzzle
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: February 13th 2008, 10:19 PM
  4. Puzzling Number Sequences
    Posted in the Math Challenge Problems Forum
    Replies: 0
    Last Post: February 16th 2007, 05:02 PM
  5. Puzzling Probability
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: January 29th 2006, 07:06 PM

Search Tags


/mathhelpforum @mathhelpforum