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Math Help - Taking the Integral - did I do it right?

  1. #1
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    Taking the Integral - did I do it right?

    Hey everyone,

    I'm needing to take the integral of the following. However I don't get the correct answer (when I put in the bounds), so I'm wondering if I integrated it wrong somewhere?

    The equation is
    \int_2^3 \sqrt{1 + (x^2/8 -2/x^2)^2}


    So I started off by multiplying out the square,
    \int_2^3 \sqrt{1 + x^4/64 - 2x^2/8x^2 - 2 x^2/8x^2 + 4/x^4}

    =
    \int_2^3 \sqrt{1 + x^4/64 - 1/2 + 4/x^4}


    Square rooted everything:
    \int_2^3 {1 + x^3/8 - 1/\sqrt{2} + 2x^-3}


    Integrated it: x +x^4/32 - x/ \sqrt{2} - x^-2


    Did I do something wrong up there?

    Thanks in advance!
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  2. #2
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    Step 4. You can't just take the square root of everything inside the square root. It doesn't work.

    \sqrt{4+9} \ne \sqrt{4} + \sqrt{9}

    You will need to use integrating rules and finding the antiderivative to solve it.
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  3. #3
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    You can't take "square roots" term by term.

    I would do it this way...

    1 + \left(\frac{x^2}{8} - \frac{2}{x^2}\right)^2 = 1 + \left(\frac{x^4}{8x^2} - \frac{16}{8x^2}\right)

     = 1 + \left(\frac{x^4 - 16}{8x^2}\right)^2

     = 1 + \frac{x^8 - 32x^4 + 256}{64x^4}

     = \frac{64x^4}{64x^4} + \frac{x^8 - 32x^4 + 256}{64x^4}

     = \frac{x^8 + 32x^4 + 256}{64x^4}

     = \frac{(x^4 + 16)^2}{(8x^2)^2}

     = \left(\frac{x^4 + 16}{8x^2}\right)^2.



    Therefore \int{\sqrt{1 + \left(\frac{x^2}{8} - \frac{2}{x^2}\right)^2\,dx} = \int{\sqrt{\left(\frac{x^4 + 16}{8x^2}\right)^2}\,dx}

     = \int{\frac{x^4 + 16}{8x^2}\,dx}

     = \int{\frac{x^2}{8} + \frac{2}{x^2}\,dx}

     = \int{\frac{1}{8}x^2 + 2x^{-2}\,dx}

     = \frac{1}{8}\cdot \frac{x^3}{3} + 2\frac{x^{-1}}{-1} + C

     = \frac{x^3}{24} - \frac{2}{x} + C.


    Now substitute your terminals.
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  4. #4
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    Quote Originally Posted by funnytim View Post
    Hey everyone,

    I'm needing to take the integral of the following. However I don't get the correct answer (when I put in the bounds), so I'm wondering if I integrated it wrong somewhere?

    The equation is
    \int_2^3 \sqrt{1 + (x^2/8 -2/x^2)^2}\ dx


    So I started off by multiplying out the square,
    \int_2^3 \left(\sqrt{1 + x^4/64 - 2x^2/8x^2 - 2 x^2/8x^2 + 4/x^4}\right)\ dx

    =
    \int_2^3 \left(\sqrt{1 + x^4/64 - 1/2 + 4/x^4}\right)\ dx


    Square rooted everything:
    \int_2^3 \left({1 + x^3/8 - 1/\sqrt{2} + 2x^{-3}}\right)\ dx this is incorrect


    Integrated it: x +x^4/32 - x/ \sqrt{2} - x^{-2}


    Did I do something wrong up there?

    Thanks in advance!

    \displaystyle\sqrt{\frac{x^4}{64}+\frac{1}{2}+\fra  c{4}{x^4}}=\sqrt{\left(\frac{x^2}{8}+\frac{2}{x^2}  \right)^2}
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  5. #5
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    Right, it was the square root thingy that was wrong.

    Thanks so much for your help guys
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