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Does + C melt together ALL constants?

  1. Aug 19, 2009 #1
    Does "+ C" melt together ALL constants?

    I had this integral [tex]\int \frac{dy}{-2y + 6}[/tex].

    I realize you can let [itex]u = -2y + 6[/itex] so that [itex]du = -2dy[/itex], and adjusting: [itex]-0.5 du = dy[/itex].

    The integral becomes

    [tex]-0.5 \int \frac{du}{u}[/tex]

    [tex]= -0.5 \ln{|-2y + 6|} + C[/tex]

    However, there is another way. Factor out the half before the integration:

    [tex]\int \frac{dy}{-2y + 6}[/tex]

    [tex]= \int \frac{dy}{-2(y - 3)}[/tex]

    [tex]= -0.5 \int \frac{dy}{(y - 3)}[/tex]

    [tex]= -0.5 \ln{|y - 3|} + C[/tex]

    BUT!

    [tex]-0.5\ln{|-2y + 6|} + C \neq -0.5\ln{|y - 3|} + C [/tex] !

    INSTEAD,

    [tex]-0.5\ln{|-2y + 6|} + C[/tex] equals [tex]-0.5\ln{|y - 3|} - 0.5\ln{2} + C[/tex] !

    Can anybody explain this discrepancy? Does [itex]-0.5\ln{2} + C[/itex] turn simply into just [itex] + C[/itex]? I understand that there can be an infinite number of antiderivatives for a function. But this is just odd. Both integrations are correctly done, no? If, on a test, I gave one answer, where the other one was preferred, what would happen?
    And yeah, both derivatives go back to being the same thing. Strange!
     
    Last edited: Aug 19, 2009
  2. jcsd
  3. Aug 19, 2009 #2
    Re: Does "+ C" melt together ALL constants?

    The issue is that evaluating the integral will produce a family of expressions which differ by an arbitrary constant (the C).

    So even though your two answers are unequal, they do belong to the same family of antiderivatives.


    If it makes it easier, you can use subscripts like C1 and C2 in order to distinguish between them. You are correct that the C's you've listed are unequal and are really related they way you've stated.

    --Elucidus
     
  4. Aug 19, 2009 #3

    nicksauce

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    Re: Does "+ C" melt together ALL constants?

    The integral is unique up to a constant. Since -0.5ln2 is a constant, both answers are correct antiderivatives. Both integrations are correctly done. Both answers would be correct on a test.
     
  5. Aug 19, 2009 #4
    Re: Does "+ C" melt together ALL constants?

    The "plus C" is a bastardization.

    While differential functions have unique derivatives, their anti-derivatives are NOT unique. The reason for this is simple: constant functions have a zero derivative. That means, you can always add a constant function to the end of your differentiable function, differentiate it, and the result is the same derivative no matter what constant function you chose.

    Algebraic notation does not handle this very well. The values on either side of an equation must be a single value. But as we said, there is no unique anti-derivative. We could write out everything that we're doing in words, but that takes up more room on the page.

    The trick we use in introductory calc is that, most of the time, we end up taking the difference of a single anti-derivative evaluated at two different points.

    So let's do something easy. f(x) = x^2.

    What are some anti-derivatives?

    F(x) = 1/3 x^3
    F(x) = 1/3 x^3 + 1
    F(x) = 1/3 x^3 + 2
    F(x) = 1/3 x^3 + 3
    ....

    We have a whole infinity of them. For any number c, F(x) = 1/3 x^3 + c is an antiderivative.

    But check it out. What is F(1) - F(0)?

    If F(x) = 1/3 x^3, then F(1) - F(0) = 1/3
    If F(x) = 1/3 x^3 + 1, then F(1) - F(0) = 1/3
    If F(x) = 1/3 x^3 + 2, then F(1) - F(0) = 1/3
    If F(x) = 1/3 x^3 + 3, then F(1) - F(0) = 1/3
    ....

    So when you take the difference of an antiderivative evaluated at two different points, the resulting value does not care about which antiderivative you choose. "The C's cancel".

    So that's the story. Don't think of C as a number or an "arbitrary constant", because that's not what's really going on. If you treat C like you'd treat any other number, you can arrive at contradictions.

    It is a handy short cut, though. And there are other tricks that people like to do. For example, if you have multiple C's added together, you can often throw all but one of them away. Sometimes. That happens to be the case in this problem. But if you don't end up killing the C's off by the end of the problem, you'll probably get the wrong answer. When you're working with multiple integrals, be careful there as well.

    Good luck!
     
  6. Aug 19, 2009 #5
    Re: Does "+ C" melt together ALL constants?

    Yes yes guys I understand the idea that "+ C" is just a way to portray the infinite number of antiderivatives, and I appreciate all your replies. And I commend nicksauce for his brevity. :smile:

    This integral was part of a differential equation I was solving:

    [tex]\frac{dy}{dx} + 2y = 6; x = 0; y = 1[/tex]

    by the way, this leads to

    [tex]y = 3 - 2e^{-2x}[/tex].

    BOTH ANTIDERIVATIVES work out for me! :tongue:
    It's so intriguing! How a slight change in my method of integration would give two different functions as the solution, but still work when put together in a differential equation!

    Thanks a lot everyone :smile:

    (sorry for any plausible incoherence, I've had no sleep since nine AM yesterday)
     
  7. Aug 19, 2009 #6
    Re: Does "+ C" melt together ALL constants?

    So ... when somebody does an integral and says: "I got -sin^2 but the book says cos^2, what did I do wrong?" you will now know the answer!
     
  8. Aug 21, 2009 #7
    Re: Does "+ C" melt together ALL constants?

    What did you do wrong? You forgot the "1" :rolleyes:
     
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