- #1
GeoMike
- 67
- 0
This seems easy enough, but now I'm second guessing myself...
Problem:
http://www.mcschell.com/probl.gif
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:
http://www.mcschell.com/solu.gif
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-
Problem:
http://www.mcschell.com/probl.gif
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:
http://www.mcschell.com/solu.gif
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-
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