= Solution
Choose a <polynomial> $P$ with $\|f-P\|_\infty<\epsilon$, using part (i). For all sufficiently large $n$, its degree is at most $2n-1$, so $I_n(P)=\int_0^1P$. Positive weights summing to one give $|I_n(f-P)|\le\epsilon$, and $|\int_0^1(f-P)|\le\epsilon$. Hence
$$
\left|I_n(f)-\int_0^1f(t)\,dt\right|\le2\epsilon.
$$
Since $\epsilon$ is arbitrary, \b[the quadratures converge to the integral of every continuous $f$]. The positivity established in part (ii), not just <polynomial> exactness on its own, controls the approximation error uniformly in $n$.
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