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Math Help - Simple integral is being reversed. Why?

  1. #1
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    Simple integral is being reversed. Why?

    When I integrate this:

    \int \frac{3}{60 - t} dt


    I get:

    -3ln(60 - t) + c


    However, when I plug this into Wolfram Alpha, it spits out this:

    -3ln(t - 60) + c


    I have no idea why this is happening. A footnote in Wolfram Alpha says it's equivalent, but when I use the answer I get with the rest of the problem, I get an erroneous result - whereas the reversed version works fine.

    I thought I understood how to take an integral of a simple fraction like this, but clearly I don't.

    What am I missing?
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  2. #2
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    Re: Simple integral is being reversed. Why?

    Quote Originally Posted by Lancet View Post
    When I integrate this:

    \int \frac{3}{60 - t} dt


    I get:

    -3ln(60 - t) + c


    However, when I plug this into Wolfram Alpha, it spits out this:

    -3ln(t - 60) + c


    I have no idea why this is happening. A footnote in Wolfram Alpha says it's equivalent, but when I use the answer I get with the rest of the problem, I get an erroneous result - whereas the reversed version works fine.

    I thought I understood how to take an integral of a simple fraction like this, but clearly I don't.

    What am I missing?
    absolute value ...

    \int \frac{3}{60-t} \, dt = -3\ln|60-t| + C = -3\ln|t-60| + C
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  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Simple integral is being reversed. Why?

    Answer to this:

    What is integral of  \frac{1}{-x} ?
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    Re: Simple integral is being reversed. Why?

    This integration took place in an ODE problem as part of the integrating factor.

    U = e^{\int \frac{3}{60 - t} dt} = (60 - t)^{-3}

    I was under the impression that the absolute value bars could be dropped in that circumstance...


    Edit: And when I keep the absolute value bars in, it doesn't change the end result in any way. So that does not appear to be the root issue.
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    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Simple integral is being reversed. Why?

    Quote Originally Posted by Lancet View Post
    This integration took place in an ODE problem as part of the integrating factor.

    U = e^{\int \frac{3}{60 - t} dt} = (60 - t)^{-3}

    I was under the impression that the absolute value bars could be dropped in that circumstance...


    Edit: And when I keep the absolute value bars in, it doesn't change the end result in any way. So that does not appear to be the root issue.
    f(x):R-->R

    f(x)=ln(x) defined only when x>0.
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    Re: Simple integral is being reversed. Why?

    Quote Originally Posted by Also sprach Zarathustra View Post
    f(x):R-->R

    f(x)=ln(x) defined only when x>0.
    I'm afraid I don't understand. Or rather, I understand that ln(x) is defined only for positive numbers, but I don't understand how that explains why the terms appear to need to be reversed.
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  7. #7
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    Re: Simple integral is being reversed. Why?

    As a solution to the given integral, neither -3ln(60- t)+ C nor -3ln(t- 60)+ C is correct. What you should have is -3 ln|60-t|+ C or -3ln|t- 60|+ C. Those are equal, of course.

    Now, when you looking for an integrating factor, you can set "C" equal to 0 and take the exponential: e^{-3ln|60-t|}= |60- t|^{-3} and you still need the absolute value. Since you are looking for a single integrating factor out of the infinite possible integrating factors (which is why we can set C equal to 0), you can use either (60- t)^{-3} or (t- 60)^{-3}. Either will work because (60- t)^{-3}= -(t- 60)^{-3} and you will be multiplying both sides of the equation by that so the negatives would cancel.
    Last edited by HallsofIvy; June 16th 2011 at 07:41 AM.
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