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Thread: Trouble with Integration by Parts

  1. #1
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    Trouble with Integration by Parts

    "Determine the area between the graph of $\displaystyle y=xe^{-x^{2}}$ and the x axis over the interval [0,1]."

    my work:

    Area = $\displaystyle \int_0^1 \! xe^{-x^{2}} \, dx$
    using the integral by parts rule " $\displaystyle \int \! udv = uv - \int \! vdu$ "

    u = x | v = $\displaystyle \frac{e^{-x^{2}}}{-2x}$

    du = 1 | dv= $\displaystyle e^{-x^{2}}$

    thus..
    $\displaystyle
    \int_0^1 \! xe^{-x^{2}} \, dx = x(\frac{e^{-x^{2}}}{-2x}) - \int \! \frac{e^{-x^{2}}}{-2x} (1)$

    This is where I'm like, am I even on the right track?

    and if so, how do I integrate $\displaystyle \int \! \frac{e^{-x^{2}}}{-2x}$?

    Thanks for the help.
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  2. #2
    Super Member girdav's Avatar
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    You don't need to use an integration by parts. What is the derivative of $\displaystyle x\mapsto \exp (-x^2)$ ?
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  3. #3
    Senior Member Sambit's Avatar
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    Quote Originally Posted by Savior_Self View Post

    Area = $\displaystyle \int_0^1 \! xe^{-x^{2}} \, dx$
    No need to use integration by parts; just substitute $\displaystyle y=x^2$
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  4. #4
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    Alright, thanks guys, but can we just assume I have to use integration of parts? It's kind of the point of the lesson.
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  5. #5
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    Quote Originally Posted by Savior_Self View Post
    Alright, thanks guys, but can we just assume I have to use integration of parts? It's kind of the point of the lesson.
    That is an old trick of most seasoned calculus teachers.
    In a section on integration by parts always throw in some that do not require parts.
    It is part of standard teaching practices.
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  6. #6
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    Quote Originally Posted by Savior_Self View Post
    "Determine the area between the graph of $\displaystyle y=xe^{-x^{2}}$ and the x axis over the interval [0,1]."

    my work:

    Area = $\displaystyle \int_0^1 \! xe^{-x^{2}} \, dx$
    using the integral by parts rule " $\displaystyle \int \! udv = uv - \int \! vdu$ "

    u = x | v = $\displaystyle \frac{e^{-x^{2}}}{-2x}$
    This is incorrect. With $\displaystyle dv= e^{-x^2}$, v is NOT $\displaystyle \frac{e^{-x^2}}{2x}$
    In fact, the anti-derivative of $\displaystyle e^{-x^2}$ is not an elementary function.

    du = 1 | dv= $\displaystyle e^{-x^{2}}$

    thus..
    $\displaystyle
    \int_0^1 \! xe^{-x^{2}} \, dx = x(\frac{e^{-x^{2}}}{-2x}) - \int \! \frac{e^{-x^{2}}}{-2x} (1)$

    This is where I'm like, am I even on the right track?

    and if so, how do I integrate $\displaystyle \int \! \frac{e^{-x^{2}}}{-2x}$?

    Thanks for the help.
    Follow Math Help Forum on Facebook and Google+

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