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 $-f(0)^2$.


Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 202 / 1 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-202/1/c/solution