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Help with kinematics and integration.

  1. Oct 11, 2004 #1
    Yea, I've got a two problems which i can't really solve.

    1. if a particle's velocity, t seconds after leaving a fixed pt. A is,

    [tex] v=3-6e^-^0^.^5^t[/tex]

    state the value which v approaches as t becomes very large.

    2. Given that [tex] \int_{0}^{3} g(x) dx=4 [/tex] , evaluate

    [tex] \int_{0}^{2} 2 g(x) dx + 2 \int_{2}^{3} (g(x)+x) dx [/tex]

    The major problem i have is with the definite integral with range 3, 2.

    Thanks for any help.
  2. jcsd
  3. Oct 11, 2004 #2
    First off, please show your work here before asking for help.

    For part 1, plug in a few values and see how the exponent grows (or decays). It might be instructive to plot this graph and you'd get the right answer using that anyway :-D. There's also another way of doing this problem. Can you think of it?

    For part 2, do you know anything about the ADDITIVE property of integrals? Mathematically this breaks up an integral from the limits [a,b] to [a,c] and [c,b]:

    [tex]\int_{a}^{b} f(x)dx = \int_{a}^{c} f(x)dx + \int_{c}^{b} f(x)dx[/tex]
  4. Oct 11, 2004 #3
    Hey, ok, suddenly with the thing about drawing the curve, it somewhat fell into place. Since it is an exponential curve, but its negative, is the answer, in this case 0?

    as for the second one, do I actually?

    integrate x, with the range 3,2 so as to seperate it from the 2nd integral? But how do I take care of the 1st integral? is the integral of 2 g(x) the same as the integral for 2 times the integral of g(x). Thanks.
  5. Oct 12, 2004 #4


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    For the second part. All you need:
    [tex]\int_{a}^{b} f(x)dx = \int_{a}^{c} f(x)dx + \int_{c}^{b} f(x)dx[/tex]
    as Maverick said. And:
    [tex]\int_{a}^{b} Cf(x)dx = C\int_{a}^{b} f(x)dx[/tex]
    for some constant C.
    [tex]\int_{a}^{b} (f(x)+g(x))dx = \int_{a}^{b} f(x)dx + \int_{a}^{b} g(x)dx[/tex]

    I`m sure these are familiar to you.
  6. Oct 13, 2004 #5
    [tex]v=3-6e^{-0.5t}[/tex] is an exponentially increasing function of time.

    Lets rewrite it as

    [tex]v = -3 + 6(1-e^{-0.5t})[/tex]

    so that the y-intercept = -3 and the function thereafter can be thought of as a standard exponential buildup which levels off at +6-3 = +3 (precisely the limit of the function as t tends to infinity). Can you draw a graph now? If you still have trouble, have a look at the graphs of standard exponential functions first...check them using first and second derivatives, mathematica, etc.

    Last edited: Oct 13, 2004
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