Accumulation Functions

You’ll be able to

Interpret an integral with a variable upper limit as an accumulation function
Differentiate an accumulation function with the Fundamental Theorem (Part 1)
Use accumulation to find a quantity from its rate of change

An integral as a function of its upper limit

An accumulation function has the form g(x) = ∫[a to x] f(t) dt: the integral runs from a fixed lower limit a to a variable upper limit x. As x grows, g accumulates more signed area under f. The variable t is just a dummy for integrating; x is what g depends on. Accumulation functions model any quantity built up from a rate — distance from velocity, volume from a flow rate.

The Fundamental Theorem, Part 1

The first part of the Fundamental Theorem says differentiation undoes accumulation: d/dx ∫[a to x] f(t) dt = f(x). The derivative of the accumulation function is simply the integrand evaluated at the upper limit. If the upper limit is itself a function like u(x), the chain rule attaches a factor: d/dx ∫[a to u(x)] f(t) dt = f(u(x))·u′(x).

Net change from a rate

The Net Change Theorem is accumulation in action: ∫[a to b] f′(t) dt = f(b) − f(a). Integrating a rate of change over an interval gives the total accumulated change in the quantity. So if you know a starting value and a rate, the value at time b is f(a) + ∫[a to b] f′(t) dt — the initial amount plus everything accumulated since.

Fundamental Theorem of Calculus (Part 1)

d/dx ∫[a to x] f(t) dt = f(x); with a variable limit, d/dx ∫[a to u(x)] f(t) dt = f(u(x))·u′(x)

Differentiating an accumulation function returns the integrand at the top limit, times the derivative of that limit.

Worked example

Let g(x) = ∫[2 to x] (t² + 1) dt. Find g′(x) and g′(3).

1.By the Fundamental Theorem Part 1, differentiating cancels the integral and evaluates the integrand at the upper limit.
2.The upper limit is simply x, so g′(x) = x² + 1.
3.Evaluate at x = 3: g′(3) = 3² + 1 = 9 + 1.
4.That gives 10.

Answer: g′(x) = x² + 1 and g′(3) = 10 — no antidifferentiation needed, just the integrand at the top limit.

Tip

To differentiate an accumulation function you do not integrate first — just plug the upper limit into the integrand (and multiply by the limit’s derivative if it is not a bare x). This shortcut is the whole point of Part 1 of the Fundamental Theorem.

Checkpoint

If g(x) = ∫[0 to x] sin(t) dt, what is g′(x)?

Checkpoint

A tank holds 50 L at t = 0, and water flows in at rate r(t) L/min. How much water is in the tank at t = 10?

On the exam

For "how much is there at time b" problems, use final value = initial value + ∫[a to b] rate dt. Integrating the rate gives only the change; you must add the starting amount to get the total. This structure appears on nearly every AP integral free-response.

Answer the 2 checkpoints as you read.