Solution
= Solution
Put $\mu_f=\int_0^1f\,d\lambda$. Subtracting this constant does not affect <variance>, and <pairwise independent random variables> have zero pairwise covariance. Therefore
$$
\begin{aligned}
D(f)
&=\operatorname{Var}\left(\frac1n\sum_{i=1}^nf(Z_i)\right)\\
&=\frac1{n^2}\sum_{i=1}^n\operatorname{Var}(f(Z_i))
=\boxed{\frac{\operatorname{Var}(f)}n}.
\end{aligned}
$$
Only pairwise independence is needed for the off-diagonal covariance terms to vanish.