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Integration by parts r e^r/2 dr

  1. Dec 7, 2006 #1
    i know this is integration by parts so here is the problem I am currently confused on how the get an answer
    r e^r/2 dr

    I know I the formula is int udv= uv - int du v

    what i am not getting is what to use for this one. I figured

    e^r/2 = v and du = r

    so u= r^2/2 and what will that make dv? i have always had trouble with this part.
    so while i try another guys plz help... make me not go insane...
     
  2. jcsd
  3. Dec 7, 2006 #2

    D H

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    Post replaced.


    It appears you have a basic misunderstanding of integration by parts. It appears you have some things backwards and you are entirely missing the
    treatment of the differential as an operator.

    This is not quite rigorous, but if
    [tex] u=F(x)[/tex]
    then
    [tex] du=\frac{dF(x)}{dx}dx[/tex]

    Suppose [tex]F(x)[/tex] is the indefinite integral of [tex]f(x)[/tex]. Reversing the above,
    [tex] du=f(x) dx[/tex]
    then
    [tex] u=F(x)[/tex]

    Now suppose
    [tex] v=G(x)[/tex]
    so that
    [tex] dv=\frac{dG(x)}{dx}dx = g(x)dx[/tex]

    Integration by parts is simply the chain rule applied to differentiation:
    [tex]\frac d{dx}(F(x)G(x)) = f(x)G(x) + F(x)g(x)[/tex]
    Integrating wrt x,
    [tex]F(x)G(x) = \int f(x)G(x) dx + \int F(x)g(x) dx[/tex]
    Applying the definitions of u,v, du, and dv,
    [tex]uv = \int v\,du + \int u\,dv[/tex]
    or
    [tex]\int v\,du = uv - \int u\,dv[/tex]

    So now, applying to
    [tex]\int r e^{r/2} dr[/tex],
    you need to identify v and du such that
    [tex]v\,du = r e^{r/2} dr[/tex]

    For example (these are your choices),
    [tex] v = e^{r/2}[/tex]
    [tex]du = r dr[/tex]

    Solving for dv and u,
    [tex] dv = 1/2e^{r/2} dr[/tex]
    [tex]u = 1/2 r^2[/tex]

    Substituting on the right hand side of the integration by parts equation yields
    [tex]\int r e^{r/2} dr = 1/2 r^2 e^{r/2} - 1/4 \int r^2 e^{r/2} dr[/tex]

    and now we are worse off than before!

    The problem is an unwise choice of u and v. A better choice will reduce the order/complexity of the problem. An unwise choice make the problem more complex.

    Try looking for a choice of u and v that reduces the order.
     
    Last edited: Dec 7, 2006
  4. Dec 7, 2006 #3
    bah but like i siad i noit getting the right answere and i tried it several times...
    e^r/2 ' = e^r/2 *1/2

    so dv = (e^r/2)/2

    right?
     
  5. Dec 7, 2006 #4
    i understand the princple if integration by parts.. the formula abnd stuff is easy to me.. the one part i am having problem is whenI choice my dv.

    if dv=e^r/2 what the hell is v. that is my problem.. most often goiung back is the problem.
     
  6. Dec 7, 2006 #5

    D H

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    The fact that you are saying "dv=e^r/2" indicates to me that you don't understand the principle of integration by parts. The correct expression is "dv=e^r/2 dr". Integrating, [tex]v=\int dv = \int e^{r/2} dr[/tex].

    I am leaving the last part of this up to you.
     
  7. Dec 7, 2006 #6
    UGHHHH !!! that the problem I not getting what the integral of that. I know i gotta take the integral.. i just leave out the dr caue i know.. what the problem is that I have difficulty going in rever and thaking a integral of e^r/3
     
  8. Dec 7, 2006 #7

    D H

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    What is so hard about this?

    To calculate
    [tex]\int e^{r/2} dr[/tex]
    use a change of variables that converts this to a form you do know how to integrate. I assume you know how to calculate
    [tex]\int e^t dt[/tex]
    What change of variables is needed to convert the first form to this?
     
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