Problem source

Find constants $a_{1}$, $a_{2}$, $u_{1}$ and $u_{2}$ such that, whenever ${\mathrm P}$ is a cubic polynomial,
\[\int_{-1}^{1}{\mathrm P}(t)\,{\mathrm d}t
=a_{1}{\mathrm P}(u_{1})+a_{2}{\mathrm P}(u_{2}).\]

Solution source

Since this is true for all cubic polynomials, it must be true in particular for $1, x, x^2, x^3$, therefore:

\begin{align*}
\int_{-1}^{1} 1 {\mathrm d}t
&=a_{1}+a_{2} &=2\\
\int_{-1}^{1} x {\mathrm d}t
&=a_{1}u_1+a_{2}u_2 &= 0 \\
\int_{-1}^{1} x^2 {\mathrm d}t
&=a_{1}u_1^2+a_{2}u_2^2 &= \frac23\\
\int_{-1}^{1} x^3 {\mathrm d}t
&=a_{1}u_1^3+a_{2}u_2^3 &= 0\\
\end{align*}

\begin{align*}
&& \begin{cases}
a_{1}+a_{2} &=2 \\
a_{1}u_1+a_{2}u_2 &= 0 \\
a_{1}u_1^2+a_{2}u_2^2 &= \frac23\\
a_{1}u_1^3+a_{2}u_2^3 &= 0\\
\end{cases} \\
\Rightarrow && \begin{cases}
a_{1}(u_1^2 - \frac13) + a_{2}(u_2^2 - \frac13) &= 0 \\
a_{1}u_1(u_1^2 - \frac13) + a_{2}u_2(u_2^2 - \frac13) &= 0
\end{cases} \\
\Rightarrow && \begin{cases} u_i = \pm \frac1{\sqrt{3}} \\ a_i = 1\end{cases}
\end{align*}

Therefore we have:

\[\int_{-1}^{1}{\mathrm P}(t)\,{\mathrm d}t
={\mathrm P} \l \frac1{\sqrt{3}} \r+{\mathrm P}\l -\frac1{\sqrt{3}} \r \]

[Note: this question is actually asking about Gauss-Legendre polynomials, and could be done directly by appealing to standard results]