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Accumulation Functions
Applied Contexts

Master how definite integrals model real-world accumulation: from net change and displacement to total quantities built from rates of change.

AP Calculus BC

What Is an Accumulation Function?

An accumulation function captures the total quantity that builds up as we integrate a rate function from a fixed starting point \(a\) to a variable endpoint \(x\):

The variable upper limit \(x\) makes \(F\) a function, not just a number. As \(x\) moves, the accumulated area changes continuously.

Follow the salt through a working mixing rig

A stirred transparent tank has one upper inlet and one lower outlet with a valve. The exact diagram links its 100 L salt balance to a computed approach to equilibrium.Suppose a tank receives and releases 4 gal/min for 5 minutes. Its 200 gal inventory stays constant while 20 gal enters and 20 gal leaves: net change is zero and total throughput is 40 gal.

Net inventory versus throughput

Suppose a tank receives and releases 4 gal/min for 5 minutes. Its 200 gal inventory stays constant while 20 gal enters and 20 gal leaves: net change is zero and total throughput is 40 gal.Suppose a tank receives and releases 4 gal/min for 5 minutes. Its 200 gal inventory stays constant while 20 gal enters and 20 gal leaves: net change is zero and total throughput is 40 gal.

Visualizing Signed Area

t f(t) a x b + area - area F(x) = net signed area from a to x

Why Accumulation Matters in the Real World

Displacement
If \(v(t)\) is velocity, then \(\int_a^b v(t)\,dt\) gives net displacement over \([a, b]\).
Total Volume
If \(r(t)\) is flow rate (liters/min), then \(\int_0^T r(t)\,dt\) gives total liters pumped.
Population Change
If \(P'(t)\) is growth rate, then \(\int_0^t P'(s)\,ds\) gives net population change.

Every time you integrate a rate over an interval, you're computing an accumulation: the total net change in the underlying quantity.

FTC Part 1: The Derivative of an Accumulation

In words: differentiating an accumulation function gives back the original rate function. The integral and derivative are inverse operations.

FTC Part 2: The Net Change Theorem

F(a) initial value + \(\int_a^b f(t)\,dt\) net change F(b) F(b) = F(a) + net change

Net Change vs. Total Change

Net Change Total Accumulated
Formula \(\int_a^b f(t)\,dt\) \(\int_a^b |f(t)|\,dt\)
Signed? Yes: cancellation occurs No: always positive
Motion example Displacement Total distance traveled
When equal? Only when \(f(t) \geq 0\) on entire \([a,b]\)

Connecting Rate and Quantity

The Net Change Theorem lets us write the quantity at any time \(t\) using its initial value plus accumulated change:

Position
\(s(t) = s(0) + \int_0^t v(\tau)\,d\tau\)
Water Level
\(W(t) = W(0) + \int_0^t r(\tau)\,d\tau\)

How to Apply Accumulation in Context

1
Identify the rate function \(f(t)\) and its units (e.g., gallons/min, people/year).
2
Determine the interval \([a, b]\) over which accumulation occurs.
3
Decide what is asked: net change \(\int_a^b f(t)\,dt\), or quantity at time \(b\): \(Q(a) + \int_a^b f(t)\,dt\)?
4
Evaluate the integral analytically, or approximate numerically if given a table of values (trapezoidal sum, etc.).
5
Interpret and include units. The integral's units = (rate units) × (time units).

Example 1: Water Tank

Water flows into a tank at a rate of \(r(t) = 6t - t^2\) liters per minute for \(0 \leq t \leq 6\). The tank initially contains 10 liters. How much water is in the tank at \(t = 6\)?

Example 1: Graph of the Rate

t r(t) 0 1 2 3 4 5 6 5 9 Shaded area = 36 liters

Example 2: Velocity and Displacement

A particle moves along a line with velocity \(v(t) = t^2 - 4t + 3\) m/s for \(0 \leq t \leq 4\). Find the net displacement and the total distance traveled.

Example 3: Table-Based Approximation

A factory produces waste at rate \(w(t)\) kg/hr. Use the data below and a trapezoidal sum with 4 subintervals to approximate \(\int_0^8 w(t)\,dt\).

\(t\) (hr) 0 2 4 6 8
\(w(t)\) 3 5 8 6 4

Common Mistakes to Avoid

Forgetting Initial Value
The integral gives net change, not the final quantity. Always add \(Q(a)\) when finding quantity at time \(b\).
Confusing Net vs. Total
Displacement \(\neq\) distance. Use \(\int |v|\,dt\) for total distance, not \(\int v\,dt\).
Dropping Units
The integral of (kg/hr) over hours yields kg. Always state units in your final answer.

Summary of Key Formulas

Concept Formula Meaning
Accumulation Function \(F(x) = \int_a^x f(t)\,dt\) Running total of signed area
FTC Part 1 \(\frac{d}{dx}\int_a^x f(t)\,dt = f(x)\) Derivative undoes integral
Net Change \(F(b) - F(a) = \int_a^b f(t)\,dt\) Total signed change on \([a,b]\)
Quantity at \(b\) \(Q(b) = Q(a) + \int_a^b Q'(t)\,dt\) Initial + accumulated change
Total Accumulation \(\int_a^b |f(t)|\,dt\) Unsigned total (e.g., distance)

Worked practice: Practice Problems, Problem 4 — applied

Work through this original lesson problem, then reveal the solution and illustrations.

A factory produces a chemical at a rate of P(t) = 50e^(−0.1t) kilograms per hour, where t is measured in hours. Simultaneously, the chemical is consumed at a constant rate of 20 kilograms per hour. At t = 0, the factory has 100 kilograms of the chemical in storage. (a) Write, but do not evaluate, an integral expression for the net amount of chemical in storage at time t = T. (b) Evaluate the expression from part (a) for T = 10 to find the amount of chemical in storage at t = 10. Round to the nearest kilogram. (c) Find the time t at which the amount of chemical in storage is at a maximum. Justify your answer. (d) As t → ∞, will the storage eventually be depleted? Use a limit argument to justify.

Show answer and explanation

Point-by-point rubric: (a) [1 point] The net rate of change of storage is P(t) − 20 = 50e^(−0.1t) − 20. The accumulation equation is S(T) = 100 + ∫₀ᵀ (50e^(−0.1t) − 20) dt. Award 1 point for correct integrand with initial condition. (b) [1 point] ∫₀¹⁰ (50e^(−0.1t) − 20) dt = [−500e^(−0.1t) − 20t]₀¹⁰ = (−500e⁻¹ − 200) − (−500 − 0) = −500/e − 200 + 500 = 300 − 500/e ≈ 300 − 183.94 = 116.06. So S(10) = 100 + 116.06 ≈ 216 kilograms. Award 1 point for correct evaluation and rounding. (c) [2 points] S′(t) = 50e^(−0.1t) − 20. Set S′(t) = 0: 50e^(−0.1t) = 20, so e^(−0.1t) = 2/5, giving −0.1t = ln(2/5), hence t = −10 ln(2/5) = 10 ln(5/2) ≈ 9.163 hours. For t < 10 ln(5/2), S′(t) > 0 (production exceeds consumption); for t > 10 ln(5/2), S′(t) < 0 (consumption exceeds production). Since S′ changes from positive to negative, S has an absolute maximum at t = 10 ln(5/2). Award 1 point for finding the critical time, 1 point for the sign-change justification. (d) [1 point] As T → ∞, ∫₀^∞ (50e^(−0.1t) − 20) dt = [−500e^(−0.1t) − 20t]₀^∞ = (0 − ∞) − (−500) → −∞. Therefore S(T) = 100 + (−∞) → −∞. Since S is eventually strictly decreasing and unbounded below, the storage will eventually be depleted (S reaches 0 at some finite time). Award 1 point for recognizing that the integral diverges to −∞ and concluding depletion occurs.

Stock starts at 100 kg, rises to about 216.74 kg at 9.16 hours, then declines.
Inventory peaks when declining production first equals constant consumption; gross inlet and outlet totals remain distinct from net change. The vessel cutaway illustrates flow connections; its contents and stock level are unspecified.
Production rate 50e^(−0.1t) and constant consumption rate 20 kg/h intersect at about 9.16 hours.
Production rate 50e^(−0.1t) and constant consumption rate 20 kg/h intersect at about 9.16 hours.

Question 1 of 5

Water flows into a tank at \(r(t) = 4t\) liters/min. If the tank starts with 20 liters, how many liters are in the tank at \(t = 3\) minutes?

Question 2 of 5

If \(\displaystyle F(x) = \int_2^x (3t^2 + 1)\,dt\), what is \(F'(5)\)?

Question 3 of 5

A particle has velocity \(v(t) = t - 2\) on \([0, 4]\). What is the total distance traveled?

Question 4 of 5

If \(\displaystyle G(x) = \int_1^{x^2} \sin(t)\,dt\), what is \(G'(x)\)?

Question 5 of 5

A population grows at rate \(P'(t) = 100e^{0.02t}\) people/year. If \(P(0) = 5000\), which expression gives the population at \(t = 10\)?