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Math Help - Integral of r^2 times square root of 4r^2 + 1 dr.

  1. #1
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    Integral of r^2 times square root of 4r^2 + 1 dr.

    Can anyone help me with this?

    Integral of r^2 times square root of 4r^2 + 1 dr.-integral.jpg

    ?
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  2. #2
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    Re: Integral of r^2 times square root of 4r^2 + 1 dr.

    Quote Originally Posted by Frankkair View Post
    Can anyone help me with this?

    Click image for larger version. 

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    Let u = 4r^2  + 1\;,\,du = 8rdr.\;\& \,\begin{array}{*{20}c}   r &\vline &  0 & 2  \\\hline   u &\vline &  1 & {17}  \\ \end{array}

    Thus \int_1^{17} {\left( {\frac{{u - 1}}{4}} \right)\sqrt u \left( {\frac{{du}}{8}} \right)} .
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    Re: Integral of r^2 times square root of 4r^2 + 1 dr.

    Thank you, but again I have a question (now a more basic one)

    I got this, but I can't solve (guess my algebra is not working today)

    Integral of r^2 times square root of 4r^2 + 1 dr.-problem.jpg


    Could u help me? Thanks!
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  4. #4
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    Re: Integral of r^2 times square root of 4r^2 + 1 dr.

    Seriously? Just put "17" and "1" in for u and subtract. You really can't do much with 17^{5/2} or 17^{3/2} but, of course, 1^{5/2}= 1^{3/2}= 1.
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