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    test

    [math]x^2\sqrt{x}[\math]

    [math] \left(\begin{array}{cc}1\\1\\1\end{array}\right) = \left(\begin{array}{cc}2x\\-2y\\0\end{array}\right)\lambda + \left(\begin{array}{cc}2\\0\\1\end{array}\right)\m u [\math]
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  2. #2
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    Re: test

    [tex]x^2\sqrt{x}[\tex]
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    Re: test

    x^2\sqrt{x}
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    Re: test

    \left(\begin{array}{cc}1\\1\\1\end{array}\right) = \left(\begin{array}{cc}2x\\-2y\\0\end{array}\right)\lambda + \left(\begin{array}{cc}2\\0\\1\end{array}\right) \mu

    so\ \mu=1

    \left(\begin{array}{cc}2x\\-2y\\0\end{array}\right)\lambda = \left(\begin{array}{cc}-1\\-1\\0\end{array}\right)


     So\ \lambda \neq 0

    \frac{1}{\lambda} = -2x

    \frac{1}{\lambda} = -2y

    -2x = -2y, and\  x = y

    BUT x^2 - y^2 = 1???
    Last edited by Convrgx; July 11th 2014 at 10:20 PM.
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  5. #5
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    Re: test

    Use \phantom{-} in front of the positive terms of your matrices and it will line everything up for you...
    Thanks from topsquark
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    Re: test

    u = -x^2

    du = -2xdx

    \frac{-1}{2}du = xdx

    Now, substitute in \frac{-1}{2}du\  for\  xdx.

    Pull that 5 out to get \frac{-5}{2} outside the integral.

    It's cake from there...
    Last edited by Convrgx; July 12th 2014 at 06:44 PM.
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