= Solution
Let $W_n$ be the interpolation in part (a). The operation
$$
I:C[0,1]\longrightarrow\mathbb R,\qquad I(f)=\int_0^1f(t)\,dt
$$
is continuous, since $|I(f)-I(g)|\leq\|f-g\|_\infty$. By the <Donsker invariance principle> and the <continuous mapping theorem>,
$$
I(W_n)\xrightarrow{\ d\ }\int_0^1B_t\,dt.
$$
The exact trapezoidal integral of the linear interpolation is
$$
I(W_n)=\frac1{n^{3/2}}
\left(\sum_{k=1}^{n-1}S_k+\frac12S_n\right).
$$
Therefore the statistic in question differs from $I(W_n)$ by $S_n/(2n^{3/2})$. Using <independence>, zero means, and unit variances gives
$$
\mathbb E\left[\left(\frac{S_n}{2n^{3/2}}\right)^2\right]
=\frac{n}{4n^3}=\frac1{4n^2}\longrightarrow0.
$$
This error tends to zero in $L^2$, hence in probability. The <Slutsky theorem> now proves the <integrated random-walk limit>:
$$
\boxed{\frac1{n^{3/2}}\sum_{k=1}^nS_k
\xrightarrow{\ d\ }\int_0^1B_t\,dt.}
$$
The limiting law can also be made explicit. The time integral is a Gaussian <random variable>, as a mean-square limit of linear combinations of a <Gaussian process>. It is centered, and the <covariance> identity $\mathbb E(B_sB_t)=\min(s,t)$ gives
$$
\operatorname{Var}\left(\int_0^1B_t\,dt\right)
=\int_0^1\int_0^1\min(s,t)\,ds\,dt
=2\int_0^1\int_0^t s\,ds\,dt=\frac13.
$$
Thus the terminal value of <integrated Brownian motion> here has law \b[$N(0,1/3)$].
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