= Solution
{c}
Write $p=a+bx$. <orthogonality> of the residual to $1,x$ gives
$$
\frac a2+\frac b3=\int_0^1x\sqrt{1-x^2}dx=\frac13,
$$
$$
\frac a3+\frac b4=\int_0^1x^2\sqrt{1-x^2}dx=\frac\pi{16}.
$$
The first integral follows from $t=1-x^2$; for the second use $x=\sin\theta$, giving $\int_0^{\pi/2}\sin^2\theta\cos^2\theta\,d\theta=\pi/16$. The coefficient <matrix> has determinant $1/72$, so solving yields
$$
\boxed{p(x)=6-\frac{3\pi}{2}
+\left(\frac{9\pi}{4}-8\right)x.}
$$
Part (a) proves this is the unique best weighted $L^2$ approximation. The weight $x$ is essential; using ordinary unweighted <normal equations> would give different coefficients.
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