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    evaluate the integral of: xe^-x^2 dx.


    u= x
    dv/dx=e^-x^2

    problem is, i dont know how to integrate the dv/dx term.

    help pleaseee.
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    (Original post by J_Alom)
    evaluate the integral of: xe^-x^2 dx.


    u= x
    dv/dx=e^-x^2

    problem is, i dont know how to integrate the dv/dx term.

    help pleaseee.
    The differential of e^(-x^2) is....
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    (Original post by Clarity Incognito)
    The differential of e^(-x^2) is....
    -2xe^-x^2
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    (Original post by J_Alom)
    -2xe^-x^2
    The differential of  -\dfrac{1}{2}e^{-x^2} is...
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    Or do a substitution with u = x^2 since the derivative of x^2 is (nearly) present.
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    (Original post by Clarity Incognito)
    The differential of  -\dfrac{1}{2}e^{-x^2} is...

    where is this going?
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    So to integrate you divide the e^... term by what you multiplied it by to differentiate.
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    (Original post by J_Alom)
    where is this going?
    Nowhere, apparently.

    Use the substitution tinyhobbit suggested.
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    (Original post by Kasc)
    So to integrate you divide the e^... term by what you multiplied it by to differentiate.
    for my final answer i have:

    - e^-x^2 (all over 2) +c

    which the integral calculators say is correct.

    but i dont understand the logic behind it; what integration method did i use?????
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    (Original post by J_Alom)
    x
    Going back to your original question of integrating  \displaystyle \int xe^{-x^2} dx , I think you were trying to do it by parts with  u = x and  \frac {dv}{dx} = e^{-x^2} ? The problem is that the integral of dv/dx can't be found out by our usual means.


    What I think Clarity Incognito was trying to get at is considering the function  f(x) = e^{-x^2} because this appears in the final solution as well (the argument never changes). If you consider f'(x) then you obtain something similar, but quite the same as, the integral. They differ by a factor, say k, that is not a function of x, so if you were to multiply f(x) by k then you obtain the original integrand. This is called reverse chain rule, which relies on recognition of what needs to be derived to obtain the integral, and then just reversing the process.

    Mr M among others were getting at Integration by substitution . Reverse chain rule is basically a type of integration by substitution.
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    (Original post by J_Alom)
    -2xe^-x^2
    So from that:
    \frac{d}{dx}[e^{-x^2}] = -2xe^{-x^2}
    Therefore \int -2xe^{-x^2}dx = e^{-x^2} + C since integration is the inverse operation of differentiation.
    This is almost what we want but we're trying to integrate xe^{-x^2}, which is out by a factor of '-2' to the above.
    Therefore, since you can pull constants to the front of integrals, we have:
    -2\int xe^{-x^2} dx = e^{-x^2}+C
    If we divide through both sides by minus two, you have calculated the required integral.
    Alternatively a substitution of u=-x^2 would have done it too.
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    I would suggest revising standard forms
 
 
 
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