Integration (Velocity to Displacement or Position)

by kieran1black2
Tags: displacement, integration, position, velocity
kieran1black2 is offline
Jun1-07, 06:47 AM
P: 23
please help me with the integration in the word document.
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G01 is offline
Jun1-07, 10:30 AM
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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.?
dt19 is offline
Jun1-07, 11:13 AM
P: 47
This is the integration in LaTeX, if anyone else can't see it:

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

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

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

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.

G01 is offline
Jun1-07, 02:58 PM
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Integration (Velocity to Displacement or Position)

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.
kieran1black2 is offline
Jun1-07, 03:41 PM
P: 23
i have already tried that... it didnt seem to work... could you show it in latex? so i can see what im doin wrong?
jackiefrost is offline
Jun1-07, 05:21 PM
P: 137
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?
kieran1black2 is offline
Jun2-07, 12:06 AM
P: 23
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?

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