Applications of Net Change
Textbook Reference
| Primary source | OpenStax Calculus Volume 2, Section 1.4: “Integration Formulas and the Net Change Theorem” |
| Direct link | https://openstax.org/books/calculus-volume-2/pages/1-4-integration-formulas-and-the-net-change-theorem |
| Textbook used in class | Stewart, Calculus, Section 4.4: “Indefinite Integrals and the Net Change Theorem” |
Opening Scenario
FTC Part 2 says that $\displaystyle\int_a^b F'(x)\,dx = F(b) - F(a)$. Written in words: the integral of a rate of change over an interval equals the net change in the quantity over that interval. This statement -- the Net Change Theorem -- is the same as FTC Part 2, but stated in a way that makes applied problems immediate.
Velocity is the rate of change of position. Marginal cost is the rate of change of total cost. Birth rate minus death rate is the rate of change of population. In each case, integrating the rate over a time interval gives the net change in the quantity.
Quick Reference
Net Change Theorem. If $F'(x) = f(x)$, then $$\int_a^b f(x)\,dx = F(b) - F(a) = \text{net change in } F \text{ from } a \text{ to } b.$$
Key applications:
| Rate of change $f(t)$ | Net change $\displaystyle\int_a^b f(t)\,dt$ |
|---|---|
| Velocity $v(t)$ | Displacement (net change in position) |
| Speed $|v(t)|$ | Total distance traveled |
| Acceleration $a(t)$ | Change in velocity |
| Marginal cost $C'(x)$ | Change in total cost from $x = a$ to $x = b$ units |
| Population growth rate $P'(t)$ | Change in population |
Key Concepts
1. Displacement vs. Distance Traveled
When a particle moves with velocity $v(t)$ on $[t_1, t_2]$:
- Displacement (net change in position): $\displaystyle\int_{t_1}^{t_2} v(t)\,dt$.
- Total distance traveled (always non-negative): $\displaystyle\int_{t_1}^{t_2} |v(t)|\,dt$.
These agree only when $v(t)$ does not change sign on $[t_1, t_2]$. When $v$ changes sign (the particle reverses direction), the integral of $v$ cancels some motion, but the integral of $|v|$ does not.
How to compute total distance when $v$ changes sign:
- Find the zeros of $v(t)$ on $[t_1, t_2]$. These are the reversal times.
- Split the integral at each reversal.
- Add the absolute values of each piece.
2. The Net Change Perspective Unifies Many Fields
The same formula $\int_a^b f(x)\,dx = F(b) - F(a)$ applies in many contexts:
- Charging a battery: If $I(t)$ is the current (rate of charge flow), $\displaystyle\int_0^T I(t)\,dt$ is the total charge delivered.
- Population change: If $B(t) - D(t)$ is the net birth rate, $\displaystyle\int_0^T [B(t) - D(t)]\,dt$ is the population change.
- Economics: If $C'(x)$ is the marginal cost, $\displaystyle\int_{x_1}^{x_2} C'(x)\,dx = C(x_2) - C(x_1)$ is the additional cost of producing from $x_1$ to $x_2$ units.
The specific field determines what $f$ and $F$ represent. The mathematics is identical.
3. Reading the Integral as Accumulation
An alternative reading of $\displaystyle\int_a^b f(x)\,dx$: it accumulates (adds up) infinitely many infinitesimal amounts $f(x)\,dx$ over $[a, b]$. Each $f(x)\,dx$ is the contribution to the total during the infinitesimal interval $[x, x + dx]$.
“Displacement and distance are always the same for a moving particle.” They are the same only when the particle moves in one direction throughout the interval. If the particle reverses direction, displacement can be zero even though the particle has traveled a positive total distance.
Worked Example
A particle moves along a line with velocity $v(t) = t^2 - 4t + 3$ m/s for $0 \leq t \leq 4$.
(a) Find the displacement over $[0, 4]$.
$\displaystyle\int_0^4 (t^2 - 4t + 3)\,dt = \left[\frac{t^3}{3} - 2t^2 + 3t\right]_0^4 = \left(\frac{64}{3} - 32 + 12\right) - 0 = \frac{64}{3} - 20 = \frac{4}{3}$ m.
(b) Find the total distance traveled over $[0, 4]$.
First find zeros: $t^2 - 4t + 3 = (t-1)(t-3) = 0$, so $t = 1$ and $t = 3$.
Check signs: $v(0) = 3 > 0$, $v(2) = -1 < 0$, $v(4) = 3 > 0$.
Split at the reversal times and add absolute values:
$$\text{Distance} = \left|\int_0^1 v\,dt\right| + \left|\int_1^3 v\,dt\right| + \left|\int_3^4 v\,dt\right|.$$
$\displaystyle\int_0^1(t^2-4t+3)\,dt = \left[\frac{t^3}{3}-2t^2+3t\right]_0^1 = \frac{1}{3} - 2 + 3 = \frac{4}{3}$.
$\displaystyle\int_1^3(t^2-4t+3)\,dt = \left[\frac{t^3}{3}-2t^2+3t\right]_1^3 = (9-18+9) - \frac{4}{3} = 0 - \frac{4}{3} = -\frac{4}{3}$.
$\displaystyle\int_3^4(t^2-4t+3)\,dt = \left[\frac{t^3}{3}-2t^2+3t\right]_3^4 = \frac{4}{3} - 0 = \frac{4}{3}$.
$$\text{Distance} = \frac{4}{3} + \frac{4}{3} + \frac{4}{3} = 4 \text{ m}.$$
Boxed answers: Displacement $= \dfrac{4}{3}$ m; total distance $= 4$ m.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Confusing displacement and distance | Reporting $\int_0^4 v\,dt$ as total distance when $v$ changes sign | When $v$ changes sign, compute $\int |v|\,dt$ by splitting at zeros |
| Forgetting to split the integral | Computing $\int_{t_1}^{t_2} v\,dt$ directly when $v < 0$ partway | Find zeros of $v$, split the integral, then sum absolute values |
| Misidentifying which quantity is the rate | Integrating position instead of velocity | The integrand must be a rate; position is $F$, velocity $v = F'$ is the rate |
Leveled Practice
Level 1 -- Direct Application
Problem 1. A particle’s velocity is $v(t) = 3t^2 - 6t$ for $0 \leq t \leq 3$. Find the displacement.
Show answer
$\displaystyle\int_0^3(3t^2 - 6t)\,dt = \left[t^3 - 3t^2\right]_0^3 = (27 - 27) - 0 = 0$ m.
The particle returns to its starting position.
Level 2 -- Distance with Direction Change
Problem 2. For the same $v(t) = 3t^2 - 6t$ on $[0, 3]$, find the total distance traveled.
Show answer
Zero at $t = 2$ (factor: $3t(t-2) = 0$). $v < 0$ on $(0, 2)$, $v > 0$ on $(2, 3)$.
$\displaystyle\int_0^2(3t^2-6t)\,dt = [t^3-3t^2]_0^2 = (8-12) = -4$. $|\text{\ }| = 4$.
$\displaystyle\int_2^3(3t^2-6t)\,dt = [t^3-3t^2]_2^3 = (27-27)-(8-12) = 0-(-4) = 4$.
Total distance $= 4 + 4 = 8$ m.
Mastery Checklist
Mental Model
Think of $f(x)$ as a water faucet that controls flow rate. When $f > 0$, water enters a tank; when $f < 0$, water drains. $\displaystyle\int_a^b f(x)\,dx$ is the net change in the water level -- inflow minus outflow. $\displaystyle\int_a^b |f(x)|\,dx$ is the total water that passed through the faucet in either direction. The distinction matters any time the sign of $f$ changes.
Connections
Looking back
- FTC Part 2 (Section 4.3): The Net Change Theorem is FTC Part 2 restated in terms of accumulation.
- Antiderivatives (Section 3.9): Position is the antiderivative of velocity; this connection is used directly.
Looking ahead
- Substitution (Section 4.5): More complex rate-of-change functions require substitution before the integral can be evaluated.
- Applications of integration (Chapter 5): Area between curves, volumes, arc length, and work all use the Net Change Theorem in different geometric settings.
Back to Integration Foundations | Next: The Substitution Rule