Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Integration (Velocity to Displacement or Position)

  1. Jun 1, 2007 #1
    please help me with the integration in the word document.

    Attached Files:

    • Doc1.doc
      File size:
      16.5 KB
  2. jcsd
  3. Jun 1, 2007 #2


    User Avatar
    Homework Helper
    Gold Member

    EDIT: I realize you may have work in your document, but I can't yet see it. So, if you have work in the document, ignore my lecture below:)

    According to the forum rules, you must show some work to get help. What have you tried? Where are you stuck? Etc.?
  4. Jun 1, 2007 #3
    This is the integration in LaTeX, if anyone else can't see it:

    [itex]v = \frac{e^{\frac{t - 1205.525}{-100}}-142000}{30}[/itex]

    [itex]s = \int v dt[/itex]

    [itex]s = \int \frac{e^{\frac{t - 1205.525}{-100}}-142000}{30} dt[/itex]

    Do you know how to integrate exponential functions?

    EDIT: I don't know what is up with the LaTeX, but there should be only one expression on each line. So ignore the bit after the second equals sign on the first line. And there shouldn't be an 's' after the fraction on the first line. No idea why it is doing this, there's nothing wrong with the code as I put it in.
    Last edited: Jun 1, 2007
  5. Jun 1, 2007 #4


    User Avatar
    Homework Helper
    Gold Member

    First, I would sugest splitting the integral up into as many smaller integrals as possible. HINT:[tex]e^{a+b}=e^ae^b[/tex]

    Using this relationship, plus splitting the integral up, you should end up with two smaller, easier integrals of known forms.
    Last edited: Jun 1, 2007
  6. Jun 1, 2007 #5
    i have already tried that... it didnt seem to work... could you show it in latex? so i can see what im doin wrong?
  7. Jun 1, 2007 #6
    The 1/30 comes can "come out" of the integrand, right? And the subtraction (e to the blah minus 140000) just yields two integrals: Integral[e to the blah dt] minus Integral[142000 dt]. So that leaves the tricky part: Integral[e to the blah]. That exponent can be broken into two fairly simple expressions by going ahead and doing the division by -100... get it?
  8. Jun 2, 2007 #7
    what i ended up with is

    (e^-t/100)/(-30/100) x e^(-1205.525/-100)/t - 142000/30t

    that is after integration... is that what you meant?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook