A quadrature formula, proving weights are equal to 1? I am given a quadrature formula on the interval [-1,1] of the form:
$$\int_{-1}^{1} f(x) dx \approx w_{0}f(- \alpha) + w_{1}f(\alpha)$$
which uses the quadrature points $x_{0}= - \alpha$ and $x_{1}= \alpha$ where $0 <\alpha \leq 1$. 
I would like to show that the weights, $$w_{0}=w_{1}=1$$ (which are independant of the value $\alpha$). 
I understand that the formula is required to be exact whenever f is a polynomial of degree 1, but I am unsure how to begin with this problem.
I would also like to show that there is one particular value of α, such that the
formula is exact also for all polynomials of degree 2. After this, I would then like to find this α, and show that, for this value, the formula is also exact for all polynomials of degree 3.
 A: All polynomials of degree 1 can be represented as $f(x) = ax+b$. Since
\begin{align*}
\int_{-1}^{1} ax + b \, \mathrm{d}x = 2b
\end{align*}
then
\begin{align*}
w_{0}(-a\alpha + b) + w_{1}(a\alpha + b) = (w_{1} - w_{0})a\alpha + (w_{0}+w_{0})b = 2b
\end{align*}
From this we see that $w_{1} = w_{0} = 1$.
That should be enough to help you get the next part of your question. (Hint: $\alpha = 1/\sqrt{3}$.)
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mc}[1]{\mathcal{#1}}
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Check with $\ds{\,\mrm{f}\pars{x} = x^{0},x^{1},x^{2}}$:

$$
\left.\begin{array}{rcl}
\ds{2} & \ds{=} & \ds{w_{0} + w_{1}}
\\[2mm]
\ds{0} & \ds{=} & \ds{w_{0}\pars{-\alpha} + w_{1}\alpha}
\\[2mm]
\ds{2 \over 3} & \ds{=} & \ds{w_{0}\pars{-\alpha}^{2} + w_{1}\alpha^{2}}
\end{array}\right\}
\implies w_{0} = w_{1} = 1\,,\quad \alpha = {\root{3} \over 3}
$$

$$
\int_{-1}^{1}\mrm{f}\pars{x}\,\dd x \approx \mrm{f}\pars{-\,{\root{3} \over 3}} +
\mrm{f}\pars{\root{3} \over 3}
$$
