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Thread: integration by parts

  1. #1
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    integration by parts

    HI

    calculating integral sign ("S") b=2; a=0 (e^-x/(1-e^-2)) dx

    So if u = x then du = dx

    then I need a dv .... if dv = e^-1 then v = e^-1 doesn't it? and how does this help me if I now introduce V?

    how can I get all of these "e"s into "u"s?

    Thanks

    Calculus Beginner
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  2. #2
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    Quote Originally Posted by calcbeg View Post
    HI

    calculating integral sign ("S") b=2; a=0 (e^-x/(1-e^-2)) dx

    So if u = x then du = dx

    then I need a dv .... if dv = e^-1 then v = e^-1 doesn't it? and how does this help me if I now introduce V?

    how can I get all of these "e"s into "u"s?

    Thanks

    Calculus Beginner
    did you mean ...

    $\displaystyle \int_0^2 \frac{e^{-x}}{1-e^{-2x}} \, dx$

    or what you posted?
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  3. #3
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    what I posted - there is an x then the e stuff
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  4. #4
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    Quote Originally Posted by calcbeg View Post
    what I posted - there is an x then the e stuff
    your notation is a bit confusing ...

    ("S") b=2; a=0 (e^-x/(1-e^-2)) dx

    is the integrand

    $\displaystyle \frac{e^{-x}}{1-e^{-2}} $

    or is it

    $\displaystyle e^{\frac{-x}{1-e^{-2}}}$
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  5. #5
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    If it really is, as you said, ("S") b=2; a=0 (e^-x/(1-e^-2)) dx
    which is $\displaystyle \int\frac{e^{-x}}{1- e^{-2}}dx$, then that is the same as $\displaystyle \frac{1}{1- e^{-1}}\int e^{-x}dx$ and that should be easy to integrate.
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    Quote Originally Posted by HallsofIvy View Post
    If it really is, as you said, ("S") b=2; a=0 (e^-x/(1-e^-2)) dx
    which is $\displaystyle \int\frac{e^{-x}}{1- e^{-2}}dx$, then that is the same as $\displaystyle \frac{1}{1- e^{-1}}\int e^{-x}dx$ and that should be easy to integrate.
    Why did you take out $\displaystyle \frac{1}{1- e^{-1}}$? Wouldn't the e be to the negative 2 power-- $\displaystyle \frac{1}{1- e^{-2}}$ since all of it is a constant?
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