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Importance of adding the constant of integration.

  1. Oct 10, 2011 #1
    Hello experts!
    I have a question below.

    Why is it important to add the constant of integration immediately when the integration is performed?

    Thanks in advance.
     
  2. jcsd
  3. Oct 10, 2011 #2
    If you're wondering why there is a constant of integration, it is because when you have a function f(x) such that F(x) = f'(x), then the derivative of g(x) = f(x) + C for any constant also equals F(x) (i.e., g'(x) = f'(x)), so when you integrate F(x), you need to capture that constant in the solution.

    Now, if you are asking why it has to be done immediately I'm not sure what you're asking since adding the constant is done as the last step.
     
  4. Oct 10, 2011 #3

    dextercioby

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    To the OP: Assume you must do a double integration. For example:

    [tex] \frac{d^2 f(x)}{d x^2} = x^3 + 5 [/tex]

    What is then f(x) equal to ?
     
  5. Oct 11, 2011 #4
    f(x)=[itex]\frac{x^{5}}{5}[/itex] + [itex]\frac{5x^{2}}{2}[/itex] + c

    where "c" is the integration constant.

    Now what?? :s
     
  6. Oct 11, 2011 #5
    Ok daveb thank you. I think you have answered my question.
    Thank you very much.
     
  7. Oct 11, 2011 #6
    Thank you dextercioby. You also made me to think the answer of my question.
     
  8. Oct 11, 2011 #7
    Not quite, this is why the constant of integration is important. The first integration gives

    [tex] \frac{df}{dx}=\frac{x^4}{4}+5x+c_1 [/tex]

    and the second integration gives

    [tex] f(x)=\frac{x^5}{20}+\frac{5x^2}{2}+c_1x+c_2 [/tex]

    Which is why we need one constant of integration for each integral
     
  9. Oct 11, 2011 #8

    HallsofIvy

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    Note that dextercioby's example is a linear second order non-homogenous differential equation which means that the set of all solutions is a two dimensional "linear manifold". That is why you need two undetermined coefficients.
     
  10. Oct 12, 2011 #9
    Ok cpt_carrot I got you. Thanks man.
    Now I have understood that for each integration a constant is important, as Hallsofivy said.

    So help guys on Physics Forums
     
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