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Math Help - Trouble with Integration by Parts

  1. #1
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    Trouble with Integration by Parts

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

    my work:

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

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

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

    thus..
    <br />
\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 \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 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 = \int_0^1 \! xe^{-x^{2}} \, dx
    No need to use integration by parts; just substitute 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 y=xe^{-x^{2}} and the x axis over the interval [0,1]."

    my work:

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

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

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

    thus..
    <br />
\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 \int \! \frac{e^{-x^{2}}}{-2x}?

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

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