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Exploring
Accumulations of Change

Discover how definite integrals measure total accumulated quantities, connecting rate functions to net change through the power of integration.

AP Calculus BC

What Is Accumulation?

In calculus, accumulation refers to the total quantity that builds up over an interval when we know the rate of change. If \( f(t) \) tells us how fast something changes at each moment, the accumulation from \( t = a \) to \( t = b \) is the total amount of that change.

Rate Function
Tells you how fast a quantity changes at each instant: \( f(t) \)
Accumulation
Tells you how much total change occurs: \( \displaystyle\int_a^b f(t)\,dt \)

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.

Approximating with Riemann Sums

Before computing exact integrals, we approximate accumulation by slicing the interval into small rectangles. As the number of rectangles \( n \to \infty \), the sum approaches the exact integral.

t f(t) a b

Sum of rectangle areas \( \approx \displaystyle\int_a^b f(t)\,dt \)

The Definite Integral as Net Change

The definite integral \( \displaystyle\int_a^b f(t)\,dt \) gives the net accumulation of the quantity whose rate of change is \( f(t) \). This net change accounts for both positive and negative values of \( f \).

Above axis (\( f > 0 \))
Area counts as positive accumulation: quantity increases
Below axis (\( f < 0 \))
Area counts as negative accumulation: quantity decreases

The Fundamental Theorem: Part 1

The First Fundamental Theorem of Calculus defines an accumulation function and states that differentiation undoes integration:

t f(t) a x F(x) = area F'(x) = f(x) Derivative of accumulation = original rate function

The Fundamental Theorem: Part 2

The Second Fundamental Theorem (Evaluation Theorem) provides the computational tool for evaluating definite integrals using antiderivatives:

This bridges the gap between accumulation (integral) and antiderivatives: find any \( F \) with \( F' = f \), then subtract endpoint values. The total accumulated change equals the difference of the antiderivative at the boundaries.

Net Change vs. Total Accumulation

A crucial distinction in accumulation problems: the signed integral gives net change, while the integral of the absolute value gives total accumulation.

Net Change Total Accumulation
Formula \( \displaystyle\int_a^b f(t)\,dt \) \( \displaystyle\int_a^b |f(t)|\,dt \)
Sign Can be negative Always non-negative
Velocity example Displacement Distance traveled
Accounts for Cancellation of + and − All movement counted

Graphing an Accumulation Function

Given a rate graph \( f(t) \), we can sketch \( F(x) = \displaystyle\int_a^x f(t)\,dt \) by tracking how area accumulates:

f(t): rate function + − F(x): accumulation function f = 0 → F peaks t x

How to Solve Accumulation Problems

1
Identify the rate function. Determine what \( f(t) \) represents: velocity, flow rate, population growth rate, etc.
2
Set up the definite integral. Write \( \displaystyle\int_a^b f(t)\,dt \) with the correct bounds for the interval of interest.
3
Determine what is asked. Net change uses the signed integral. Total accumulation uses \( \displaystyle\int_a^b |f(t)|\,dt \).
4
Evaluate. Find the antiderivative \( F(t) \) and compute \( F(b) - F(a) \), or use geometry, tables, or numerical methods.
5
Interpret in context. Include units and state the meaning: "The particle moved 12 meters to the right over 5 seconds."

Example 1: Basic Accumulation

A tank is being filled at a rate of \( f(t) = 6t - t^2 \) liters per minute for \( 0 \leq t \leq 6 \). How many liters are added?

t f(t) 0 1 2 3 4 5 6 9 peak at t = 3 f(3) = 9

Example 2: Signed Accumulation

A particle has velocity \( v(t) = t^2 - 4t + 3 \) m/s. Find the net displacement and total distance traveled on \( [0, 4] \).

Example 3: FTC with Chain Rule

Find \( \displaystyle\frac{d}{dx}\int_1^{x^3} \sin(t^2)\,dt \).

Accumulation from Data Tables

When \( f(t) \) is given as a table of values (no formula), we approximate the integral using trapezoidal sums or Riemann sums.

\( t \) (hrs) 0 2 5 8 10
\( r(t) \) (gal/hr) 4 7 10 6 3

Key Properties of Definite Integrals

1
Additivity: \( \displaystyle\int_a^b f\,dt + \int_b^c f\,dt = \int_a^c f\,dt \)
2
Reversal: \( \displaystyle\int_a^b f\,dt = -\int_b^a f\,dt \)
3
Constant Multiple: \( \displaystyle\int_a^b c \cdot f\,dt = c\int_a^b f\,dt \)
4
Sum/Difference: \( \displaystyle\int_a^b [f \pm g]\,dt = \int_a^b f\,dt \pm \int_a^b g\,dt \)
5
Zero Width: \( \displaystyle\int_a^a f\,dt = 0 \)

Common Mistakes to Avoid

✕ Ignoring Sign Changes
For total distance, you must split the integral where \( v(t) = 0 \) and take absolute values. Integrating directly gives net displacement, not total distance.
✕ Forgetting Chain Rule
When differentiating \( \displaystyle\int_a^{g(x)} f(t)\,dt \), you must multiply by \( g'(x) \). The answer is \( f(g(x)) \cdot g'(x) \), not just \( f(g(x)) \).
✕ Wrong Initial Condition
If asked for \( F(b) \) and given \( F(a) \), remember: \( F(b) = F(a) + \displaystyle\int_a^b f(t)\,dt \). Do not forget to add the initial value.
✕ Misreading the Graph
Area below the x-axis contributes negative accumulation. Do not count all regions as positive when computing net change.

Summary of Accumulation Concepts

Concept Formula Interpretation
Net Change \( \displaystyle\int_a^b f(t)\,dt \) Signed total: + and − cancel
Total Accumulation \( \displaystyle\int_a^b |f(t)|\,dt \) All change counted positive
FTC Part 1 \( \frac{d}{dx}\displaystyle\int_a^x f(t)\,dt = f(x) \) Derivative undoes integral
FTC Part 2 \( \displaystyle\int_a^b f(t)\,dt = F(b) - F(a) \) Evaluate via antiderivative
Initial Value \( F(b) = F(a) + \displaystyle\int_a^b f(t)\,dt \) Start value + accumulated change

Question 1 of 5

If \( \displaystyle\int_0^5 f(t)\,dt = 12 \) and \( \displaystyle\int_5^8 f(t)\,dt = -4 \), what is \( \displaystyle\int_0^8 f(t)\,dt \)?

Question 2 of 5

What is \( \displaystyle\frac{d}{dx}\int_2^{x^2} \cos(t)\,dt \)?

Question 3 of 5

A particle's velocity is \( v(t) = t - 3 \) m/s on \( [0, 6] \). What is the total distance traveled?

Question 4 of 5

If \( F(2) = 5 \) and \( \displaystyle\int_2^7 f(t)\,dt = -3 \), what is \( F(7) \)?

Question 5 of 5

The graph of \( f \) is shown below as a semicircle of radius 2 above the x-axis on \( [0, 4] \). What is \( \displaystyle\int_0^4 f(x)\,dx \)?

0 4 2 2 semicircle