= Solution
On each interval of width $h=1/N$, Taylor expansion about its midpoint shows that the local midpoint-rule error is $O(h^3)$. Summing over $N$ intervals gives the <composite midpoint rule error>
$$
\boxed{I(f)-I_N(f)=O(N^{-2})}.
$$
To remove the endpoint singularity, set $x=s^2$. Then
$$
I(f)=\int_0^1 2s f(s^2)\,ds
=\int_0^1 2g(s^2)\,ds,
$$
whose transformed integrand is analytic. Applying the composite midpoint rule in $s$ with $s_n=(n+\tfrac12)/N$ gives
$$
I(f)\approx\frac2N\sum_{n=0}^{N-1}s_nf(s_n^2).
$$
This has the required form when
$$
\boxed{y_n=s_n^2
=\left(\frac{n+\tfrac12}{N}\right)^2}.
$$
The <quadratic substitution for a square-root endpoint singularity> therefore restores second-order convergence.
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