Accumulation, Limits, and Stationary Action

The big idea

A definite integral accumulates many small contributions. Each contribution must include both the rate being measured and the interval over which it acts. A Riemann sum approximates that accumulation with finitely many slices; a limit assigns a value independent of a particular slicing. If a parameter changes an entire candidate function, the integral becomes a new function of that parameter, and a derivative can locate a stationary candidate.

Worked example

For \(v(t)=2t\) on \([0,1]\), use two equal intervals. Right-endpoint speeds are 1 and 2, and each duration is \(1/2\), so

\[ S_2=1^2\left(\frac12\right)+2^2\left(\frac12\right)=\frac52. \]

The continuous accumulation is

\[ S=\int_0^1(2t)^2\,dt=\frac43. \]

The finite sum is an approximation, not a different underlying quantity. Now suppose a family produces \(S(a)=4+a^2\). Then \(S'(a)=2a\), so \(a=0\) is stationary; since \(a^2\ge0\), it is also the absolute minimum in that family.

How children may show it

A learner may use a table of interval rates and durations, shaded rectangles under a rate graph, a sequence of finer sums, or a graph of the parameter function \(S(a)\). Units are a useful check: rate squared times time has different units from rate squared alone.

Common mix-up

One common error is to add sampled rates without multiplying by interval width. Another is to differentiate with respect to time when the task asks which parameter selects a whole history. Ask, “What variable indexes the slices, and what variable labels the candidates?”

Try it together

Approximate \(\int_0^1(1+t)^2\,dt\) with one, two, and four right-endpoint rectangles. Then compute the integral exactly and discuss what changes as the partition becomes finer and what stays attached to the same continuous function.