What does an accumulation function tell me as its endpoint moves?

Keep the starting point fixed and let the endpoint vary. Write F(x) = ∫ₐˣ v(t) dt. Here, a is the fixed starting point, x is the endpoint and input, and t is the variable of integration. Each choice of x gives a value for the integral, so the integral defines a function of x. For a continuous v, the Fundamental Theorem of Calculus gives F′(x) = v(x). That means the value of v at the endpoint is F’s instantaneous rate of change there. [S1]

In words, “F at x” means the amount accumulated from a to x. F(x) tells you the accumulated value; F′(x) tells you how that value is changing at the endpoint.

Take v(t) = 1 and fix the starting point at 0. For x ≥ 0,

F(x) = ∫₀ˣ 1 dt = x.

This is the simple example in the calculus source. [S1] The input is x, the moving endpoint; the output is the accumulated value between 0 and x. For example, F(3) = 3, and F(5) = 5. Moving the endpoint from 3 to 5 changes the accumulated value from 3 to 5. In this example, each additional unit of endpoint adds one unit to the accumulated value, and F′(x) = 1. [S1]

The integral is not a different kind of object at each endpoint: the rule F(x) = x assigns an output to every allowed input x.

Use the same example to separate them. At endpoint 5, F(5) = 5 is the accumulated value from 0 to 5. But F′(5) = 1 is the instantaneous rate at which F changes as the endpoint moves at 5. Those are different questions: “How much has accumulated?” asks for F(5); “How quickly is that amount changing now?” asks for F′(5).

For a continuous integrand, F′(x) = v(x), so the integrand’s value at the endpoint supplies that rate. In this example v(5) = 1, matching F′(5). [S1]

First identify the fixed starting point a and the endpoint x that will vary. Name the quantity being accumulated v(t), then write F(x) = ∫ₐˣ v(t) dt. Say what the output means at a chosen endpoint. If v is continuous, you can use F′(x) = v(x) to interpret the instantaneous rate of change there. [S1]

For instance, with the example above, the setup is F(x) = ∫₀ˣ 1 dt, and the output at endpoint 4 is F(4) = 4. The derivative rule gives F′(4) = 1.

The continuity condition matters: the cited result gives the derivative conclusion for continuous v, not every possible integrand. The source notes that step-function behavior can produce a corner where the derivative conclusion does not hold at that point. [S1] So check the condition before applying the rule, rather than assuming it always works.