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Math Help - Integration by Substitution

  1. #1
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    Integration by Substitution

    Hi, I'm new to integration. Can anyone help me figure out where I went wrong?
    Many thanks.

    Q. Evaluate the following:

    Please see attachment.
    Attached Thumbnails Attached Thumbnails Integration by Substitution-photo-2-.jpg  
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  2. #2
    MHF Contributor
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    Re: Integration by Substitution

    Notice your u version of the integrand has no traces of the x in the numerator of the original. To bring that part with you, you can re-express it (x, that is) in terms of u, or else use integration by parts.
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  3. #3
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    e^(i*pi)'s Avatar
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    Re: Integration by Substitution

    Quote Originally Posted by GrigOrig99 View Post
    Hi, I'm new to integration. Can anyone help me figure out where I went wrong?
    Many thanks.

    Q. Evaluate the following:

    Please see attachment.
    Your numerator disappeared when changing the variable. Express x in terms of u.

    edit: drat, beaten to it
    Last edited by e^(i*pi); November 15th 2011 at 01:47 PM. Reason: too slow
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  4. #4
    Super Member Quacky's Avatar
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    Re: Integration by Substitution

    You haven't taken into account the x on the numerator.

    We have:

    \int^4_0{\frac{x}{\sqrt{2x+1}}}dx

    I'm going to let u^2=2x+1 so that 2u~du=2~dx and dx=u~du

    When x=4, u=3 and when x=0, u=1

    I have:

    \int^4_0{\frac{x}{\sqrt{2x+1}}}dx

    = \int^3_1{\frac{u^2-1}{2u}}\cdot{u}~du

    Take it!
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  5. #5
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    Re: Integration by Substitution

    The parts version...



    ... where (key in spoiler) ...

    Spoiler:


    ... is the product rule, straight continuous lines differentiating downwards with respect to x.



    ... is lazy integration by parts, doing without u and v.



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  6. #6
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    Re: Integration by Substitution

    Thanks guys.
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