Net Change Theorem

6 min read

Reference: Stewart §4.4

The Integral of a Rate = Net Change

Here’s one of the most powerful ideas in calculus, stated simply:

$$\int_a^b F'(x)\, dx = F(b) - F(a)$$

In words: The integral of a rate of change is the net change.

This is really just FTC2 written in a way that emphasizes its meaning. If $F'(x)$ tells you how fast something is changing, then integrating $F'$ from $a$ to $b$ tells you how much it changed overall.

Prerequisite Skills

Before You Start

Prerequisite Check: Can you answer these?

From FTC Part 2:

State the Fundamental Theorem of Calculus (Part 2): If $F$ is an antiderivative of $f$, what is $\int_a^b f(x)\, dx$?
Evaluate $\int_1^3 2x\, dx$ using FTC2.

From Computing Indefinite Integrals:

Find $\int (4t - 3)\, dt$.
Find an antiderivative of $e^x$.

Check Your Answers

$\int_a^b f(x)\, dx = F(b) - F(a)$
Antiderivative of $2x$ is $x^2$, so $\int_1^3 2x\, dx = [x^2]_1^3 = 9 - 1 = 8$
$\int (4t - 3)\, dt = 2t^2 - 3t + C$
$F(x) = e^x$ (since $\frac{d}{dx}[e^x] = e^x$)

If these feel unfamiliar, review the prerequisite pages before continuing.

Quick Reference

Property	Value
Concept	Indefinite Integrals & Net Change
Chapter	Chapter 4, Section 4
Difficulty	Intermediate
Time	~15 minutes

Key Concepts

The Net Change Theorem

Theorem: If $F'$ is continuous on $[a, b]$, then: $$\int_a^b F'(x)\, dx = F(b) - F(a)$$ The integral of the rate of change of a quantity gives the net change in that quantity.

Why “Net” Change?

The word “net” is crucial. If a quantity goes up and then down (or vice versa), the integral captures the overall difference, not the total movement.

Example: If water flows into and out of a tank:

Water in: +500 gallons
Water out: -300 gallons
Net change: +200 gallons

The integral would give 200, not 800 (the total volume that moved).

Applications Across Disciplines

If $F(x)$ represents...	Then $F'(x)$ is...	And $\int_a^b F'(x)\, dx$ gives...
Position $s(t)$	Velocity $v(t)$	Displacement (net position change)
Velocity $v(t)$	Acceleration $a(t)$	Change in velocity
Volume $V(t)$	Flow rate	Net volume change
Population $P(t)$	Growth rate	Net population change
Cost $C(x)$	Marginal cost $C'(x)$	Total additional cost
Charge $Q(t)$	Current $I(t)$	Net charge transferred

Setting Up Net Change Problems

Step 1: Identify what quantity is changing and what its rate of change is.

Step 2: Write the integral: $\int_a^b (\text{rate})\, d(\text{variable})$

Step 3: Add interpretation: “This represents the net change in [quantity] from $a$ to $b$.”

Units Analysis

The units of $\int_a^b F'(x)\, dx$ are:

$$(\text{units of } F') \times (\text{units of } x) = \text{units of } F$$

Example: If velocity is in m/s and time is in seconds: $$\int_0^{10} v(t)\, dt \quad \text{has units} \quad \frac{\text{m}}{\text{s}} \times \text{s} = \text{m}$$

The result is a distance (displacement), as expected!

Practice Problems

Level 1 Basic Interpretation

If $w'(t)$ represents the rate of growth of a child’s weight in pounds per year, what does $\int_5^{10} w'(t)\, dt$ represent?

Thought Process

Identify the rate: $w'(t)$ = rate of weight change (lbs/year)
Apply Net Change Theorem: $\int_5^{10} w'(t)\, dt = w(10) - w(5)$
Interpret: This is the change in weight from age 5 to age 10

Show Answer

By the Net Change Theorem: $$\int_5^{10} w'(t)\, dt = w(10) - w(5)$$

This represents the net change in the child’s weight (in pounds) between ages 5 and 10.

Or equivalently: how much weight the child gained from age 5 to age 10.

Level 2 Marginal Cost Application

A company’s marginal cost function is $C'(x) = 20 - 0.02x$ dollars per unit, where $x$ is the number of units produced.

What does $\int_{100}^{200} C'(x)\, dx$ represent, and compute its value.

Thought Process

Identify: $C'(x)$ = marginal cost = cost to produce one more unit
Net Change Theorem: $\int_{100}^{200} C'(x)\, dx = C(200) - C(100)$
Meaning: Total additional cost to increase production from 100 to 200 units
Compute: Find antiderivative and evaluate

Show Answer

Interpretation: By the Net Change Theorem, $\int_{100}^{200} C'(x)\, dx = C(200) - C(100)$

This represents the increase in total cost when increasing production from 100 units to 200 units.

Computation: $$\int_{100}^{200} (20 - 0.02x)\, dx = \left[20x - 0.01x^2\right]_{100}^{200}$$

$$= (20(200) - 0.01(200)^2) - (20(100) - 0.01(100)^2)$$

$$= (4000 - 400) - (2000 - 100)$$

$$= 3600 - 1900 = \boxed{\$1700}$$

It costs an additional \$1700 to produce units 101 through 200.

Level 3 Water Tank Problem

Water flows from a tank at a rate of $r(t) = 200 - 4t$ liters per minute, where $t$ is measured in minutes with $0 \leq t \leq 50$.

(a) What does the integral $\int_0^{10} r(t)\, dt$ represent?

(b) Find the amount of water that flows out during the first 10 minutes.

Thought Process

(a) Since $r(t)$ is the outflow rate, the integral gives total water that flowed out.

(b) Integrate: $\int_0^{10} (200 - 4t)\, dt$

Note: This is not the net change in water level (which would be negative of this, since water is leaving).

Show Answer

(a) Interpretation:

The integral $\int_0^{10} r(t)\, dt$ represents the total volume of water (in liters) that flows out of the tank during the first 10 minutes.

(b) Computation:

$$\int_0^{10} (200 - 4t)\, dt = \left[200t - 2t^2\right]_0^{10}$$

$$= (200(10) - 2(10)^2) - (0 - 0)$$

$$= 2000 - 200 = \boxed{1800 \text{ liters}}$$

Level 4 Population Growth with Initial Condition

A honeybee population starts with 100 bees and increases at a rate of $n'(t) = 50e^{0.1t}$ bees per week.

(a) Write an expression for the population after 15 weeks.

(b) Find the population after 15 weeks.

Thought Process

(a) Initial population + net change = final population $$n(15) = n(0) + \int_0^{15} n'(t)\, dt = 100 + \int_0^{15} 50e^{0.1t}\, dt$$

(b) Evaluate the integral:

Antiderivative of $50e^{0.1t}$ is $\frac{50}{0.1}e^{0.1t} = 500e^{0.1t}$

Show Answer

(a) Expression using Net Change Theorem:

$$n(15) = n(0) + \int_0^{15} n'(t)\, dt$$

$$= 100 + \int_0^{15} 50e^{0.1t}\, dt$$

(b) Evaluation:

$$\int_0^{15} 50e^{0.1t}\, dt = \left[\frac{50}{0.1}e^{0.1t}\right]_0^{15} = \left[500e^{0.1t}\right]_0^{15}$$

$$= 500e^{1.5} - 500e^0 = 500(e^{1.5} - 1)$$

$$= 500(4.4817 - 1) \approx 500(3.4817) \approx 1741$$

Total population: $$n(15) = 100 + 1741 = \boxed{1841 \text{ bees}}$$

Level 5 Units and Dimensional Analysis

The power consumption of San Francisco on a certain day is given by $P(t)$ megawatts, where $t$ is measured in hours starting at midnight.

(a) What are the units of $\int_0^{24} P(t)\, dt$?

(b) If $P(t) = E'(t)$ where $E(t)$ is energy consumed, explain what $\int_0^{24} P(t)\, dt$ represents physically.

(c) Using the Midpoint Rule with data points $P(1) = 440, P(3) = 400, P(5) = 420, P(7) = 620, P(9) = 790, P(11) = 840, P(13) = 850, P(15) = 840, P(17) = 810, P(19) = 690, P(21) = 670, P(23) = 550$ (all in MW), estimate the energy used that day.

Thought Process

(a) Units: (megawatts) × (hours) = megawatt-hours (MWh), a unit of energy

(b) By Net Change Theorem: $\int_0^{24} E'(t)\, dt = E(24) - E(0)$ = total energy consumed

Show Answer

(a) Units:

$$\text{Units of } \int_0^{24} P(t)\, dt = (\text{megawatts}) \times (\text{hours}) = \boxed{\text{megawatt-hours (MWh)}}$$

Megawatt-hours is a standard unit of energy (1 MWh = 1000 kWh).

(b) Physical interpretation:

Since power is the rate of change of energy ($P = \frac{dE}{dt}$), by the Net Change Theorem:

$$\int_0^{24} P(t)\, dt = E(24) - E(0)$$

This represents the total energy consumed by San Francisco during that 24-hour period.

(c) Midpoint Rule approximation:

Using 12 subintervals with $\Delta t = 2$ hours:

$$\int_0^{24} P(t)\, dt \approx [P(1) + P(3) + P(5) + \cdots + P(23)] \cdot \Delta t$$

Sum of midpoint values: $$440 + 400 + 420 + 620 + 790 + 840 + 850 + 840 + 810 + 690 + 670 + 550 = 7920$$

$$\approx 7920 \times 2 = \boxed{15{,}840 \text{ MWh}}$$

The city consumed approximately 15,840 megawatt-hours of energy that day.

Common Misconceptions

Common misconception

$\int_a^b F'(x)\,dx$ gives the value of $F$ at some representative point, not the total change.

This is the height-vs-slope error. The integral of the rate of change $F'$ accumulates all the small changes over $[a, b]$; the result is the net change $F(b) - F(a)$, not a function value. For example, if $v(t)$ is velocity, $\int_0^{10} v(t)\,dt$ is the total displacement over $10$ seconds, not the speed at any moment. The function value $F(t_0)$ and the integral of $F'$ over an interval are unrelated in general.

Mastery Checklist

I can state the Net Change Theorem: $\int_a^b F'(x)\, dx = F(b) - F(a)$
I understand why it’s called “net” change (accounts for both increases and decreases)
I can interpret integrals in context (cost, population, volume, etc.)
I can set up the correct integral from a word problem
I can verify units using dimensional analysis

Mental Model

The Odometer Analogy

Think of $F'(x)$ as your speedometer reading and $F(x)$ as your odometer.

The speedometer tells you how fast you’re going at each moment
The integral of speedometer readings over time = net change in odometer
If you drive forward 50 miles, then back 30 miles, the odometer shows 80 miles more, but your displacement is only 20 miles

The Net Change Theorem captures displacement (where you end up relative to where you started), not total distance traveled.

Connections

Looking back:

FTC Part 2: the theorem this is based on
Computing Indefinite Integrals: needed to evaluate

Looking ahead:

Displacement vs Total Distance: when you need total movement, not net
Work and Energy: major application of net change

Real-world connections:

Economics: Integrating marginal cost/revenue gives total cost/revenue change
Biology: Integrating growth rate gives population change
Physics: Integrating power gives energy; integrating force gives impulse

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Computing Indefinite Integrals	Skills Index	Displacement vs Total Distance

Last updated: 2026-01-22