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Taking the integral of xe^ax^2

  1. Dec 21, 2011 #1
    1. The problem statement, all variables and given/known data
    dy/dx=x*e^(ax^2)

    solve the differential equation



    2. Relevant equations
    integral of e^x=e^x


    3. The attempt at a solution
    im not really sure how to do it when there are two variables in the exponent? i tried several things like u=x^2 1/2du=xdx then 1/2*int[e^a*u] results, but i can't take the integral without a different substitution since int[e^a*u] is not = to e^a*u.

    the answer is 1/2a * e^ax^2 + C.. any methods to integrate this? several methods would be best
     
  2. jcsd
  3. Dec 21, 2011 #2

    Mentallic

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    Think about how you compute [tex]\frac{d}{dx}e^{f(x)}[/tex]
     
  4. Dec 21, 2011 #3
    a is a constant, keep that in mind . . .
     
  5. Dec 21, 2011 #4
    I solved it with a as a constant and got the right answer, but . . .

    How do i know a is a constant? I can't just assume that can I? What is the proof for a being a constant
     
    Last edited: Dec 21, 2011
  6. Dec 21, 2011 #5

    Mentallic

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    Because a is assumed to be a constant in these situations, just as how [itex]\pi[/itex] nearly always represents the irrational number, but in some other cases, it can be mean something completely different, such as representing a product.

    And if it wasn't a constant, then you can't solve the problem. So which do you think is more likely? :tongue:
     
  7. Dec 22, 2011 #6

    HallsofIvy

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    Was the question just "integrate [itex]xe^{ax}[/itex]" or was it to find [itex]\int xe^{ax}dx[/itex]?
     
  8. Dec 22, 2011 #7
    That makes sense but it still bugs me to just assume it is a constant :(

    And to the other person, the question was

    dy/dx=x*e^(ax^2)

    solve the differential equation, so yes it would be integral of x*e^ax^2 dx (except you forgot the squared on top of the x in e's exponent)

    thanks for everybody's help
     
  9. Dec 22, 2011 #8

    epenguin

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    The situation is no different than if you were told to solve dy/dx = a .
     
  10. Dec 22, 2011 #9

    Mentallic

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    What's the difference?
     
  11. Dec 22, 2011 #10

    HallsofIvy

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    The second, [itex]\int xe^{ax^2} dx[/itex], makes it explicit that the 'variable of integration' is x ("dx") while the first, "integrate [itex]xe^{ax^2}[/itex]", is ambiguous. Of course, it is a standard convention that such things as "x", "y", "z" are used as variables while such things as "a", "b", "c", etc. are used to denote constants.

    If it were [itex]\int xe^{ax^2}da[/itex], it would be [itex](1/x)\int e^{ax^2}+ C[/itex].
     
  12. Dec 22, 2011 #11

    Mentallic

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    Yes, but without those standard conventions then we could just as well argue that a=f(x) in the integrand, just as the OP has done.
     
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