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Math Help - Integrate 1/e^-2y

  1. #1
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    Red face Integrate 1/e^-2y

    Should I have multiplied by -2 below?

    \int\displaystyle\frac{y}{e^-2y}dy

    =\displaystyle\frac{-2.y^2.ln e^-2y}{2}+C

    =-y^2.ln e^-2y+C
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ashura View Post
    Should I have multiplied by -2 below?

    \int\displaystyle\frac{y}{e^-2y}dy

    =\displaystyle\frac{-2.y^2.ln e^-2y}{2}+C

    =-y^2.ln e^-2y+C
    I'm confused about something here. Why aren't you writing this as:
    \int \, ye^{2y} dy

    That, at least, would get rid of the pesky negative sign. Also, I don't know where you are getting the "ln" from. This can be done using integration by parts:
    \int u dv = uv - \int v du

    So let u = y and dv = e^{2y}dy

    Then
    du = dy and v = \frac{1}{2}e^{2y}

    So
    \int \, ye^{2y} dy = \frac{1}{2}ye^{2y} - \int \frac{1}{2}e^{2y}dy

    and
    \int \frac{1}{2}e^{2y}dy = \frac{1}{4}e^{2y}

    Thus
    \int \, ye^{2y} dy = \frac{1}{2}ye^{2y} - \frac{1}{4}e^{2y} + C (Don't forget the integration constant!)

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ashura View Post
    Should I have multiplied by -2 below?

    \int\displaystyle\frac{y}{e^-2y}dy

    =\displaystyle\frac{-2.y^2.ln e^-2y}{2}+C

    =-y^2.ln e^-2y+C
    A couple of tips about the LaTeX coding:

    First, I don't know what the \displaystyle tag is supposed to do, but it doesn't seem to be doing anything on this Forum.

    Also, if you want to write an exponent with more than one character, enclose the exponent in { }. Example:
    e^2y (No { }.)

    e^{2y} (With the { }.)

    Hint: You can see how others have coded their LaTeX by quoting the post. You will see the coding in the quote box.

    -Dan
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  4. #4
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    I used the form,

    \int\frac{y}{e^{-2y}}dy

    because question was such. Also

    \int\frac{1}{e^{-2y}} =

    -2ln e^{-2y} + C.

    Agree that

    \int y.e^{2y}dy

    is easier to work with. Thanks for the formatting info.
    Last edited by ashura; November 8th 2006 at 04:57 PM.
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ashura View Post
    \int\frac{1}{e^{-2y}} =

    -2ln e^{-2y} + C.
    For the record:

    -2ln e^{-2y} + C = -2(-2y)ln(e) + C = 4y + C

    \int\frac{1}{e^{-2y}}dy = \int e^{2y}dy = \frac{1}{2}e^{2y} + C

    As the two of these are not different by a constant, there is something wrong with how you integrated.

    -Dan
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  6. #6
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    Smile

    \int \frac{1}{e^{-2y}}dy = \int e^{2y}dy,

    let u = 2y

    \frac{du}{dy} = 2

    dy = \frac{du}{2}

    substituting for u,

    \int e^{2y}dy =  \int e^u \frac{du}{2}

     = \frac{1}{2}\int e^u{du}

     = \frac{1}{2}e^u+C

    substituting for u,

     = \frac{1}{2}e^{2y}+C

    Thanks.
    Last edited by ashura; November 9th 2006 at 09:58 AM.
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