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Integral Problem

  1. Mar 15, 2009 #1
    Please forgive any formatting errors.

    [tex]\int x[/tex]exdx = ?

    Is this correct?

    [tex]\int x[/tex]exdx

    = [tex]\int e[/tex]ln(x)exdx
    = [tex]\int e[/tex]exln(x)dx

    u = ex
    x = ln(u)
    du = exdx
    [tex]du/u[/tex] = dx

    = ∫ ln(u)u*(1/u)*du

    a = ln(u)u
    EDIT: da = ln(u)u-1du
    dv = 1/u du
    v = ln(u)

    = ln(u)u+1 - ∫ [ln(u)]udu

    u = ex

    = ln(ex)ex+1 - ∫ [ln(ex)]exdx

    ∫xexdx = xex+1-∫xexdx

    2∫xexdx = xex+1
    ∫xexdx = xex+1*(1/2)+C
     
    Last edited: Mar 16, 2009
  2. jcsd
  3. Mar 15, 2009 #2

    gabbagabbahey

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    :eek: eeks, that's not really how you differentiate that function is it?


    Have you tried differentiating this result?

    More to the point, does this integral actually exist/converge for all values of x?....Surely you are given integration limits?
     
  4. Mar 15, 2009 #3
    Well...I skipped some steps.

    a = ln(u)^u
    a = 1/u (chain rule) * u (bring down the exponent) * ln(u)^u-1 (subtract from the exponent)
    a = ln(u)^(u-1) (simplify)
     
  5. Mar 15, 2009 #4
    You can't do that when the exponent is the variable.
     
  6. Mar 15, 2009 #5
    Oh...Hm...what would you do then?
     
  7. Mar 15, 2009 #6
    make the function e^(something). Then you can use chain rule.
     
  8. Mar 15, 2009 #7

    gabbagabbahey

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    Again, there is an even bigger issue here than you incorrectly differentiating 'a':

    To illustrate the problem, I've plotted the integrand for x=[0,3]:

    http://img16.imageshack.us/img16/2144/hmmmp.th.jpg [Broken]

    Does it look as though the area under the curve is bounded?
     
    Last edited by a moderator: May 4, 2017
  9. Mar 16, 2009 #8
    Wait...what's wrong with the way I did it? I'm differentiating with respect to u.
     
  10. Mar 16, 2009 #9
    (d/dx)3^x is not x3^(x-1).
     
  11. Mar 16, 2009 #10
    what is the integral of x^(e^x) then?
     
  12. Mar 16, 2009 #11
    I really doubt you can integrate it indefinitely.
     
  13. Mar 16, 2009 #12

    gabbagabbahey

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    It is unreasonable to expect an antiderivative to exist, when the area under the curve of the integrand becomes unbounded at around x=5!...That means for almost half of the real number line, the integral does not exist.

    There is no antiderivative for this function, but if you are given specific integration limits, and the integral is bounded on that interval, you may still be able to evaluate analytically by using a trick or two, and if that fails numerical integration is always available.

    So, I ask for the 3rd time; Are you given limits of integration?
     
  14. Mar 16, 2009 #13

    gabbagabbahey

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    [tex]\frac{d}{dx} f(x)^{g(x)}= \frac{d}{dx} e^{\ln(f(x)^{g(x)})}=\frac{d}{dx} e^{g(x)\ln(f(x))}=e^{g(x)\ln(f(x))}\left(g'(x)\ln(f(x))+\frac{g(x)}{f(x)}f'(x)\right)[/tex]

    [tex]=f(x)^{g(x)}\left(g'(x)\ln(f(x))+\frac{g(x)}{f(x)}f'(x)\right)\neq g'(x)f(x)^{g(x)-1}[/tex]
     
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