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Integrate dy/dx=0 and arbitrary constant?

  1. Apr 17, 2012 #1

    sharks

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    Gold Member

    The problem statement, all variables and given/known data
    Integrate:
    [tex]\frac{dy}{dx}=0[/tex]

    The attempt at a solution
    [tex]\int \frac{dy}{dx}=\int 0\,.dx[/tex]
    The answer is: [itex]y=0[/itex] or [itex]y=0+A[/itex]?
    This is the part which is confusing, as i know that integral of 0 is 0, but do i have to add a constant of integration??
     
  2. jcsd
  3. Apr 17, 2012 #2
    How about a dy/dx which is as positive as negative over the interval, that could be zero. fix it was e.g. sin(x)
    But yeah, a constant.
     
  4. Apr 17, 2012 #3

    sharks

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    Now that i think about it, here is a simple example:
    Let [itex]y=5[/itex]
    [tex]\frac{dy}{dx}=0[/tex]
    Now, if i integrate 0, then i should normally get 5. So, the constant of integration has to be involved.
     
  5. Apr 17, 2012 #4

    Mark44

    Staff: Mentor

    The above should be
    [tex]\int \frac{dy}{dx}~dx=\int 0\,.dx[/tex]
    You are integrating both sides of the original equation, with respect to x.

    Zero is merely one antiderivative of 0.

    A simpler approach is as follows.
    Since dy/dx = 0, then y must be a constant. IOW y ##\equiv## C.

    Also, when you integrate both sides of an equation, there is a constant of integration for each side. So the integration that you did would look like this:
    y + C1 = 0 + C2

    Of course, you can subtract C1 from both sides to end up with y = C, where C = C2 - C1.
     
  6. Apr 19, 2012 #5

    sharks

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    Thank you for the clarification, Mark44. :smile:
     
  7. Apr 19, 2012 #6

    HallsofIvy

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    Science Advisor

    Most teachers will mark off on homework or a test if you do not include the "constant of integration". If F(x) is an anti- derivative of f(x)- that is, if F'(x)= f(x), then [itex]\int f(x)dx= F(x)+ C where C can be any number.
     
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