Source: University of Texas at Austin, Calc 2 lecture notes
Difficulty: Intermediate | Prerequisites: Fundamental Theorem of Calculus (Part 2), basic antiderivative rules, and definite integral evaluation.
The Net Change Theorem is one of the most practical consequences of the Fundamental Theorem of Calculus. It tells you that if you know the rate at which something is changing, you can integrate that rate over an interval to find the total (net) change. This sits at the bridge between differentiation and integration: you learned how to take derivatives to find rates, and now you are running that process in reverse to recover accumulated quantities.
You will use this theorem constantly in applied problems: fuel consumed, population grown, distance travelled. If you are comfortable evaluating definite integrals and finding antiderivatives, you are ready for this material.
Integrating a rate-of-change function over an interval gives you the net change in the original quantity. If the rate is velocity, the integral gives displacement; if you integrate the absolute value of velocity (speed), you get total distance. The same logic applies to any rate: fuel consumption, population growth, production output.
Net Change Theorem
If f is continuous on [a, b], then ∫ₐᵇ f′(x) dx = f(b) − f(a). In plain terms, integrating the derivative (the rate of change) over an interval gives you how much the original function changed from start to finish.
Displacement
The net change in position. Calculated by integrating velocity: ∫ₐᵇ v(t) dt. Think of it as where you end up minus where you started, regardless of any backtracking in between.
Total distance
The total length of the path travelled, counting all movement regardless of direction. Calculated by integrating the absolute value of velocity: ∫ₐᵇ |v(t)| dt. In simple terms, this counts every metre you moved, even if you doubled back.
Velocity vs. speed
Velocity, v(t), is signed: it can be positive or negative depending on direction. Speed is |v(t)|, always non-negative. Integrating velocity gives displacement; integrating speed gives distance.
Rate of change
The derivative f′(x), describing how quickly a quantity changes per unit of the independent variable. When you see words like "rate," "per hour," or "per month" in a problem, you are looking at f′(x) and should integrate to find the total amount.
If f′ is continuous on [a, b], then: ∫ₐᵇ f′(x) dx = f(b) − f(a)
Read it as: "the integral of the rate of change of a function equals the net change in that function."
For total change (ignoring direction), integrate the absolute value: ∫ₐᵇ |f′(x)| dx
Recognising the right setup is half the work. When a problem gives you:
Velocity, v(t) → integrating gives displacement (net change in position)
Speed, |v(t)| → integrating gives total distance (path length)
Any rate (e.g. litres/hour, people/month) → integrating gives total production, growth, or amount
A car consumes fuel at the rate C(t) = 5 − e⁻ᵗ. How much fuel is consumed in the first 2 hours?
Set up the integral: ∫₀² (5 − e⁻ᵗ) dt
Antiderivative: 5t + e⁻ᵗ
Evaluate: [5t + e⁻ᵗ]₀² = (10 + e⁻²) − (0 + 1) = 9 + e⁻²
The rate was given; integrating it over [0, 2] yielded total fuel consumed.
Flying out of Austin at 12 pm, with distance function d(t) = 100t⁴ + t² + 1. How fast at 1 pm?
This is asking for d′(1), not an integral problem.
d′(t) = 400t³ + 2t
d′(1) = 400 + 2 = 402 mph
Notice that when the problem asks "how fast" at a point, you differentiate. When it asks "how much" over an interval, you integrate.
A town's population grows at rate p(t) = 30 + √t + (1/4)t people per month. At t = 0, population = 100,000. What is the population after 1 year (t = 12)?
Net change = ∫₀¹² (30 + √t + t/4) dt
Antiderivative: 30t + (2/3)t^(3/2) + (1/8)t²
Evaluate at 12: 30(12) + (2/3)(12)^(3/2) + (1/8)(12)²
Total population = net change + initial population (100,000)
The initial condition matters: the integral gives you how much was added, not the final total on its own.
A particle starts at rest at t = 0 with acceleration a(t) = 2 − t m/s². Find displacement and distance over [0, 6].
First, find velocity: v(t) = ∫(2 − t) dt = 2t − t²/2 + C. Since v(0) = 0, C = 0.
Displacement: ∫₀⁶ (2t − t²/2) dt = [t² − t³/6]₀⁶ = 36 − 36 = 0
The particle ends exactly where it started.
Distance: need to split at the zero of v(t). Solve 2t − t²/2 = 0 → t(2 − t/2) = 0 → t = 0 or t = 4.
v(t) > 0 on (0, 4), v(t) < 0 on (4, 6)
Distance = ∫₀⁴ (2t − t²/2) dt + ∫₄⁶ −(2t − t²/2) dt = 32/3 m
This is the classic exam distinction: displacement = 0 does not mean the particle stood still. It moved forward, then backward, covering 32/3 metres in total.
Net Change Theorem
∫ₐᵇ f′(x) dx = f(b) − f(a)
Total change (unsigned)
∫ₐᵇ |f′(x)| dx
Displacement from velocity
∫ₐᵇ v(t) dt
Total distance from velocity
∫ₐᵇ |v(t)| dt
To evaluate the absolute-value integral, find the zeros of v(t) in [a, b], split the interval there, and negate v(t) on any sub-interval where it is negative.
Engineers use the Net Change Theorem whenever they need total quantities from rate data. A flow-rate sensor on a pipe records litres per second; integrating that signal over a shift gives total volume pumped. The same logic applies to electrical charge (integrate current over time), energy consumption (integrate power over time), and revenue forecasting (integrate a sales-rate function over a quarter).
In physics, the displacement-vs-distance distinction matters for anything that reverses direction: a piston in an engine, a pendulum, or a satellite in an elliptical orbit. The total distance tells you mechanical wear; the displacement tells you net progress.
Students often assume that integrating velocity always gives distance. It does not. Integrating velocity gives displacement (net change in position). For total distance, you must integrate the absolute value of velocity.
A displacement of zero does not mean the object stayed still. It means the object returned to its starting position. The particle in the worked example above moved 32/3 metres total despite ending exactly where it began.
Students sometimes forget to add the initial condition. If a population starts at 100,000 and the integral of the growth rate gives 500, the final population is 100,500, not 500. The integral gives the change; you add it to the starting value.
Confusing "how fast" with "how much." If a problem asks how fast something is happening at a particular instant, that is a derivative evaluation, not an integration problem. Read the problem twice before setting up an integral.
⚠️ Displacement vs. distance is a near-guaranteed exam question. Know which integral to set up for each, and practise splitting the interval at velocity zeros.
⚠️ Problems that give a rate and an initial value will test whether you remember to add the initial condition after integrating.
⚠️ Watch for problems that look like integration but are really asking for an instantaneous rate (a derivative evaluation). The wording "at what rate" or "how fast at t = ..." is your cue to differentiate, not integrate.
⚠️ Be comfortable with the keyword mapping: velocity → displacement, speed → distance, rate → total amount. Exams will not always use the word "integrate" explicitly.
True or false: ∫₀⁵ v(t) dt always gives total distance travelled. (False: it gives displacement. You need ∫₀⁵ |v(t)| dt for distance.)
Fill in the blank: If a population grows at rate r(t) people per year and starts at P₀, the population at time T is ______. (P₀ + ∫₀ᵀ r(t) dt)
True or false: If displacement over an interval is zero, the object did not move. (False: the object could have moved forward and then back to its starting position.)
Fill in the blank: To find total distance when velocity changes sign, you must split the integral at the ______ of v(t). (zeros)
True or false: If a problem asks "how fast is the car going at t = 3," you should set up a definite integral. (False: you evaluate the derivative at t = 3.)
Q: A factory produces widgets at a rate of r(t) = 200 + 5t widgets per hour. How many widgets are produced during the first 8 hours?
A: ∫₀⁸ (200 + 5t) dt = [200t + (5/2)t²]₀⁸ = 1600 + 160 = 1,760 widgets.
Q: A particle has velocity v(t) = t² − 4t + 3 on [0, 5]. Set up (but do not evaluate) the integral for total distance travelled.
A: Find zeros of v(t): t² − 4t + 3 = (t − 1)(t − 3) = 0 at t = 1 and t = 3. Check signs: v > 0 on (0,1), v < 0 on (1,3), v > 0 on (3,5). Distance = ∫₀¹ v(t) dt + ∫₁³ −v(t) dt + ∫₃⁵ v(t) dt.
Q: Water flows into a tank at a rate of f(t) = 3t² + 2 litres per minute. The tank starts with 50 litres. How much water is in the tank after 4 minutes?
A: 50 + ∫₀⁴ (3t² + 2) dt = 50 + [t³ + 2t]₀⁴ = 50 + 64 + 8 = 122 litres.
Q: If ∫₂⁷ f′(x) dx = −12 and f(2) = 30, what is f(7)?
A: By the Net Change Theorem, f(7) − f(2) = −12, so f(7) = 30 − 12 = 18.
Q: A car's fuel consumption rate is C(t) = 5 − e⁻ᵗ gallons per hour. How much fuel is consumed in the first 2 hours?
A: ∫₀² (5 − e⁻ᵗ) dt = [5t + e⁻ᵗ]₀² = (10 + e⁻²) − (0 + 1) = 9 + e⁻² ≈ 9.135 gallons.
This connects directly to the Fundamental Theorem of Calculus (Part 2), which provides the theoretical foundation: the integral of f′ recovers f (up to endpoints). If you are comfortable with FTC Part 2, the Net Change Theorem is its applied interpretation.
The displacement-vs-distance distinction reappears in work and energy problems later in the course. When you study work done by a variable force (W = ∫ F dx), the same logic of signed vs. unsigned integration applies.
Initial-value problems in differential equations extend this idea further: given a rate (the DE) and a starting condition, you integrate to find the quantity over time. The population-growth example here is a preview of separable ODEs.
Net change theorem, FTC Part 2, Fundamental Theorem of Calculus applications, displacement vs distance, total distance integral, integrating velocity, integrating rate of change, absolute value integral, signed area, total accumulation, initial value problem basics, ∫ f′(x) dx = f(b) − f(a), Calc 2 integration applications, UT Austin integral calculus, speed vs velocity integral