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How to differentiare/integrate e^t and the like .

  1. Feb 9, 2007 #1
    How to differentiare/integrate e^t and the like.....

    I'm taking linear algebra this semester and completely have forgotten how to integrate and differentiate e functions.

    This is within the context of trying to learn how to integrate vector functions...

    So how would you integrate t*e^t ? what do you do when there is a multiplicative factor?

    How about if it were e^t2? or alternatively e^-2t??

    Thanks for the help!:redface:
  2. jcsd
  3. Feb 9, 2007 #2

    D H

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    Use the chain rule to integrate [itex]f(t)e^t[/itex]. You can choose either [itex] u = f(t), dv = e^t dt[/itex] or [itex] u = e^t, dv = f(t) dt[/itex]. With the former approach, [itex]\int f(t)e^t = f(t)e^t - \int f^\prime(t) e^t dt[/itex], and the latter, [itex]\int f(t)e^t = F(t)e^t - \int F(t) e^t dt[/itex]. Use whichever approach simplifies the result.

    There is no elementary function [itex]f(t)[/itex] such that [itex]f(t) = \int e^{t^2} dt[/itex]. You can always define a special function that satisfies this equation. The reason no one has done so is because this integral does not come up very often.

    On the other hand, [itex]\int e^{-t^2} dt[/itex] comes up all the time. A special function has been defined based on this integral, the error function [itex]\text{erf}(x)[/itex].
  4. Feb 9, 2007 #3
    DH, thanks a lot for your reply.

    Can you please explain further why you use the chain rule?

    Also, just to use an example - lets say you have to integrate t^2*e^t? How would you do that?
  5. Feb 9, 2007 #4

    D H

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    You use integration by parts, which is the chain rule rewritten for integrals.

    Choose [itex] u = t^2, dv = e^t dt[/itex]. Then

    [tex]\int t^2 e^t dt = t^2 e^t - 2\int t e^t dt[/tex]

    Use integration by parts again on the integral on the right hand side

    [tex]\int t e^t dt = t e^t - \int e^t dt = (t-1)e^t[/tex]

    Now apply this result to the first result,

    [tex]\int t^2 e^t dt = t^2 e^t - 2(t-1)e^t = (t^2-2t+2)e^t[/tex]

    Verify by differentiating:

    [tex]\frac d {dt}\left((t^2-2t+2)e^t\right) = (2t-2)e^t + (t^2-2t+2)e^t = t^2e^t[/tex]
  6. Feb 9, 2007 #5

    D H

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    My bad! Integration by parts is the product rule rewritten for integrals.
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