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Integration with constant of proportionality

  1. Jan 21, 2007 #1
    This seems easy enough, but now I'm second guessing myself...

    Problem:
    [​IMG]


    I solved it as follows:
    [tex]
    \frac{dN}{ds} = k(250-s)[/tex]
    [tex]dN = k(250-s) ds[/tex]
    [tex]\int dN = \int k(250-s) ds[/tex]
    [tex]N = k\int (250-s) ds [/tex]
    [tex]N = k(\int 250 ds - \int s ds)[/tex]
    [tex]N = k(250s - \frac{s^2}{2}) + C[/tex]

    But this is what the solution manual gives:
    [​IMG]

    I understand how they worked the problem, (substituting with u = (250 - s) and then integrating), but I didn't think this was the correct way to handle this problem. Why can't you just integrate each term seperately after pulling out the constant k[\b]?
    My reasoining was that if you distributed the k before integrating you'd end up with my anser, not the solution guide's...

    Thanks,
    -GeoMike-
     
    Last edited: Jan 21, 2007
  2. jcsd
  3. Jan 21, 2007 #2

    cristo

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    Staff Emeritus
    Science Advisor

    You can do it both ways. Your way is equivalent to the answer in the textbook (however, you need an integration constant!)

    If you expand the answer in the book, you will get k(250s-s2/2+2502/2) +C. However, note that the last term in the bracket is simply a constant, and so your integration constant (D, say) will be equal to C +(2502k)/2.
     
    Last edited: Jan 21, 2007
  4. Jan 21, 2007 #3
    ACK! My mistake! Thanks!

    -GeoMike-
     
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