= Solution
For any partition of $[0,t]$,
$$
f(t)^2-f(0)^2
=2\sum_k f(t_{k-1})(f(t_k)-f(t_{k-1}))
+\sum_k(f(t_k)-f(t_{k-1}))^2.
$$
The final sum is at most the largest increment of $f$ times its total variation. It tends to zero because $f$ is uniformly continuous. Part b then gives the integration-by-parts identity
$$
2\int_0^tf(s)\,df(s)=f(t)^2-f(0)^2.
$$
Thus the formula stated in the question holds when $f(0)=0$; for a general initial value the necessary endpoint correction is $-f(0)^2$.
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