- 66

- 0

This seems easy enough, but now I'm second guessing myself...

Problem:

http://www.mcschell.com/probl.gif [Broken]

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 [Broken]

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

Problem:

http://www.mcschell.com/probl.gif [Broken]

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 [Broken]

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-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 by a moderator: