1. ## 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]

2. ## Re: test

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

3. ## Re: test

$\displaystyle x^2\sqrt{x}$

4. ## 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$

$\displaystyle so\ \mu=1$

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

$\displaystyle So\ \lambda \neq 0$

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

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

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

$\displaystyle BUT x^2 - y^2 = 1???$

5. ## Re: test

Use \phantom{-} in front of the positive terms of your matrices and it will line everything up for you...

6. ## 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...