Net change theorem: Question bank — Maths deep dive

First, pin down the notation (so no symbol is a mystery)

Before any trap, let us agree on exactly what each symbol means — the traps below rely on you being crystal-clear here.

The core equation we are testing understanding of: $\int_{a}^{b} F^{'} (x) d x = F (b) - F (a)$ Read as: "integral of the rate = net (signed) change of the cumulative quantity."

Look at the figure: the orange curve is the rate $F^{'}$ . The blue shaded area under it (from $a$ to $b$ ) equals the vertical rise of the green curve $F$ from $F (a)$ to $F (b)$ . Area of the rate = rise of the quantity. That single picture is the whole theorem.

The derivation, so "Step 2 / 3 / 5" mean something

Several questions below refer to steps of the standard proof. Here they are in one place, with the picture, so you can see what each step does.

Recall The 5-step build (referenced by the Why questions)

Step 1 — Slice. Cut $[a, b]$ into $n$ thin strips. The width of each strip is $Δ x = \frac{b - a}{n}$ (that is what the symbol $Δ x$ means: one strip's width). See the vertical dashed lines in the figure below. Step 2 — Telescoping sum. Total change $= \sum_{i = 1}^{n} [F (x_{i}) - F (x_{i - 1})]$ ; interior terms cancel, leaving $F (b) - F (a)$ . Step 3 — Mean Value Theorem. In each strip there is a point $x_{i}^{*}$ with $F (x_{i}) - F (x_{i - 1}) = F^{'} (x_{i}^{*}) Δ x$ (small change $=$ slope $\times$ width). Step 4 — Riemann sum. Substitute: $F (b) - F (a) = \sum F^{'} (x_{i}^{*}) Δ x$ — a sum of strip areas. Step 5 — Limit. As $n \to \infty$ (strips get infinitely thin), that sum becomes the integral $\int_{a}^{b} F^{'} d x$ .

The figure shows Step 4 in action: each grey rectangle has height $F^{'} (x_{i}^{*})$ and width $Δ x$ ; its area is one little change. Add them all — that is the net change.

True or false — justify

Every statement below is either true or false. Decide, then reveal the reason.

The net change theorem requires you to know the value

F (a)

before you can find

F (b) - F (a)

False. The theorem gives the change directly from the rate;

F (a)

and any

+ C

cancel. You only need the initial value if you separately want the actual level

F (b)

\int_{a}^{b} v (t) d t

can be negative even though "distance" is always positive.

True. That integral is displacement (signed net change of position), which is negative when the object ends up behind its start. Distance is a different quantity,

\int_{a}^{b} ∣ v ∣ d t

For any velocity function,

\int_{a}^{b} v d t = \int_{a}^{b} ∣ v ∣ d t

False. They are equal only when

v

keeps one sign on the whole interval. When

v

changes sign, cancellation happens inside the left side but not the right, so the right side is generally larger.

The net change theorem is a brand-new result independent of the Fundamental Theorem of Calculus.

False. It is FTC Part 2, just re-read with a physical story. Same equation

\int_{a}^{b} F^{'} = F (b) - F (a)

, new interpretation.

If a marginal cost

C^{'} (x)

is always positive, then total cost

C (x)

is increasing.

True. A positive rate means the quantity only goes up, so each extra unit adds cost. The net change

\int C^{'} d x

over any interval is positive.

Total distance is always greater than or equal to the absolute value of displacement.

True. Distance counts every bit of motion positively while displacement lets backward motion cancel forward motion, so distance can only exceed (or equal)

∣ displacement ∣

. They are equal only if the object never reverses.

Doubling the interval always doubles the net change.

False. Net change is

\int_{a}^{b} F^{'}

, which depends on the shape of the rate, not just interval length. Only if

F^{'}

is constant does doubling the interval double the change.

\int_{a}^{b} v d t = 0

, the object never moved.

False. It only means the object returned to its start: forward and backward motion cancelled. It may have travelled a large distance (e.g.

v = cos t

[0, 2 π]

: displacement

0

, distance

4

Spot the error

Each line contains a real mistake. Name it and correct the reasoning.

"A particle has

v (t) = t^{2} - 4

[0, 3]

. Displacement is

\int_{0}^{3} v d t = - 3

, so distance

= ∣ - 3∣ = 3

m."

Wrong. Taking

∣ \int v ∣

ignores the sign change of

v

t = 2

. Distance needs the bars inside:

\int_{0}^{2} (4 - t^{2}) + \int_{2}^{3} (t^{2} - 4) = \frac{23}{3}

m, not

3

"To find distance I integrate

v

and take absolute value:

\int_{a}^{b} ∣ v (t) d t ∣

The notation is meaningless as written; the absolute value must wrap the integrand, giving

\int_{a}^{b} ∣ v (t) ∣ d t

. And you must split at every point where

v = 0

before integrating each piece.

"Marginal cost is

C^{'} (x) = 3 x^{2} + 50

. To get the extra cost from

x = 10

x = 20

I need

C (0)

first."

You never need

C (0)

. The net change theorem gives

C (20) - C (10) = \int_{10}^{20} C^{'} d x

directly; the unknown constant cancels in the subtraction. (See Marginal Cost and Revenue.)

\int_{0}^{3} v d t

where

v

dips below zero — I'll just integrate the part where

v > 0

since the negative bit isn't real motion."

The negative part is real motion (moving backward). Dropping it corrupts displacement (which must include the signed backward part) and distance (which counts it as positive). Neither quantity lets you discard it.

"The net change theorem needs

F^{'}

to be positive, otherwise the integral of a rate makes no sense."

False premise.

F^{'}

may be negative; the theorem handles it automatically. A negative rate simply contributes a negative amount to the net change — that is the whole meaning of net.

"I integrated and got

F (b) - F (a) = - 5

, but a change can't be negative, so the answer is

5

A net change can absolutely be negative — it means the quantity decreased. Do not strip the sign;

- 5

is the correct signed answer.

Why questions

Answer the underlying "why", not just the fact. (The steps are laid out with a picture in the collapsible box above.)

Why does the derivation use a telescoping sum in Step 2?

Because writing the total change as

\sum [F (x_{i}) - F (x_{i - 1})]

makes all interior values cancel, leaving exactly

F (b) - F (a)

. It turns "total change" into a sum of tiny changes we can approximate. (See Telescoping Sums.)

Why is the Mean Value Theorem needed in Step 3, and not just the definition of derivative?

The derivative definition is a limit (approximate for finite width); the Mean Value Theorem gives an exact point

x_{i}^{*}

with

F (x_{i}) - F (x_{i - 1}) = F^{'} (x_{i}^{*}) Δ x

, so the sum is exact before we take the limit.

Why does taking

n \to \infty

in Step 5 turn the sum into an integral?

Because

\sum F^{'} (x_{i}^{*}) Δ x

(each term a strip of width

Δ x

) is precisely a Riemann sum, whose limit as

Δ x \to 0

defines the definite integral.

Why must the absolute value go inside the integral for total distance?

Cancellation between forward and backward motion happens moment-by-moment inside the integral. Putting bars outside lets that cancellation occur first; putting them inside forbids it, so every bit is counted positive. (See Displacement vs Distance.)

Why doesn't the constant of integration ever matter in a net change problem?

Because we compute

F (b) - F (a)

; the

+ C

appears in both terms and subtracts away. Net change asks "how much did it change", which is independent of the baseline.

Why do we integrate the rate rather than the quantity itself?

Because we are usually given the rate (velocity, marginal cost, flow rate) and want the quantity's change. The theorem says integrating a rate reconstructs exactly that change — the integral undoes the differentiation.

Edge cases

Boundary and degenerate scenarios the theorem must still handle.

What is the net change over an interval of zero width,

\int_{a}^{a} F^{'} d x

Zero. No interval means no accumulated change:

F (a) - F (a) = 0

. The theorem gives this automatically.

What if the rate

F^{'} (x) = 0

for the whole interval?

Net change is

0

— the quantity is constant. Both displacement and distance are zero because nothing moved.

A velocity is negative on the entire interval. Do displacement and distance differ?

Their magnitudes are equal but signs differ: displacement is negative (moved backward the whole time), distance is its positive magnitude,

\int ∣ v ∣ = - \int v

. Equality of magnitudes holds because

v

never changed sign.

What happens at the exact instant

v = 0

(a turning point) when splitting for distance?

That single point contributes zero area, so it doesn't matter which piece you assign it to. Its role is only to mark where to split so each piece keeps one sign.

Can total distance ever equal

∣

displacement

∣

even though the object clearly moved?

Yes — whenever the velocity never changes sign. The object moves in one direction only, so nothing cancels and distance

= ∣

displacement

∣

F^{'}

has a jump discontinuity (piecewise rate), does the theorem still apply?

Yes, as long as

F^{'}

is integrable; the integral just sums the pieces. A finite jump doesn't affect area, so

\int_{a}^{b} F^{'} = F (b) - F (a)

still holds with

F

continuous.

A rate is given in litres per minute but you integrate over an interval measured in seconds. What breaks?

Units break: the net change comes out

60 \times

too large or small. The theorem is unit-agnostic algebraically, but the rate's time unit must match the integration variable's unit.

What if the rate

F^{'}

blows up to infinity inside the interval (a vertical asymptote, e.g.

F^{'} (x) = 1/ x

near

0

You now have an improper integral. Net change is still

F (b) - F (a)

provided that limit exists — take

\int

as a limit approaching the bad point. If the improper integral diverges, the net change is genuinely infinite (or undefined).

Can the net change be finite even though the rate is unbounded on the interval?

Yes. An unbounded rate can still enclose a finite signed area (e.g.

\int_{0}^{1} x^{- 1/2} d x = 2

is finite despite

F^{'} \to \infty

0

). Unbounded rate does not automatically mean infinite net change.

Connections

Net change theorem — the parent topic these traps stress-test; every equation here lives there.
Displacement vs Distance — the net-vs-total distinction behind half these items (see the signed-area figure).
Fundamental Theorem of Calculus — why net change is not a new theorem (same $\int F^{'} = F (b) - F (a)$ ).
Mean Value Theorem — the exactness in Step 3 of the boxed derivation.
Definite Integral as a Riemann Sum — why the Step 5 limit is an integral (the grey rectangles).
Telescoping Sums — the cancellation in Step 2.
Marginal Cost and Revenue — the economics no-constant-needed trap.