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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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    Re: test

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

    $\displaystyle x^2\sqrt{x}$
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    Re: test

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

    $\displaystyle so\ \mu=1$

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


    $\displaystyle So\ \lambda \neq 0$

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

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

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

    $\displaystyle BUT x^2 - y^2 = 1???$
    Last edited by Convrgx; Jul 11th 2014 at 09:20 PM.
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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...
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    Re: test

    $\displaystyle u = -x^2$

    $\displaystyle du = -2xdx$

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

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

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

    It's cake from there...
    Last edited by Convrgx; Jul 12th 2014 at 05:44 PM.
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