Net change theorem: Worked examples — Maths deep dive

The one tool we use everywhere is the parent result: $\int_{a}^{b} F^{'} (x) d x = F (b) - F (a) .$ Here $F$ is the quantity (position, volume, cost, ...) and $F^{'}$ is its rate (velocity, flow rate, marginal cost, ...). The symbol $\int_{a}^{b}$ means "add up all the tiny pieces from $a$ to $b$ " — see Definite Integral as a Riemann Sum. The whole page builds on the parent topic.

The scenario matrix

Before working anything, let's lay out every kind of situation a net-change problem can be. Read it as a true two-axis map: the rows are the type of quantity (motion, accumulation, economics); the columns are the sign pattern of the rate — each column now holds exactly the cells that belong to it (a "changes sign" example never sits under "degenerate" or "limiting"). Each cell names the example that fills it.

Quantity type ↓ \ Sign pattern →	Rate never negative	Rate changes sign	Degenerate (zero rate / zero width)	Limiting ( $b \to \infty$ )
Motion (velocity → position)	A always forward · Ex 1	B one flip · Ex 2 ; C two flips · Ex 3 ; F symmetric round trip · Ex 5	D stationary · Ex 4a ; E zero width · Ex 4b	K decaying speed · Ex 10
Accumulation (flow → volume)	G0 pure inflow · Ex 6a	G inflow then outflow · Ex 6b	—	J decaying rate · Ex 9
Economics (marginal → total)	H rising marginal cost · Ex 7	H2 sign-changing marginal profit · Ex 8	—	—
Backwards / given data	I find missing endpoint · Ex 11	—	—	—

Reading the map: move down to pick what kind of thing is changing, move right to pick how its rate behaves. Every example below is tagged with its cell letter so you can trace it back to this grid. Notice the columns are now consistent — everything in the "changes sign" column really does change sign; everything in "degenerate" is genuinely a zero-rate or zero-width case.

Worked Examples

Ex 1 — Cell A: rate never negative

Worked example Statement

A particle moves with velocity $v (t) = 6 t$ (m/s) for $t \in [0, 4]$ ; since $6 t \geq 0$ it only ever moves forward. Find displacement and total distance.

Forecast: Since the particle never reverses, I predict displacement and distance are the same number. Guess before reading on.

Step 1 — Choose the quantity. Why? Velocity is the rate of change of position $s$ , so $v = s^{'}$ . The theorem then gives $\int v = s (4) - s (0)$ . Step 2 — Integrate. Why no split? Because $v = 6 t \geq 0$ on all of $[0, 4]$ ; the velocity never changes sign, so nothing cancels. $displacement = \int_{0}^{4} 6 t d t = [3 t^{2}]_{0}^{4} = 48 m .$ Step 3 — Distance. Why identical? Distance is $\int_{0}^{4} ∣ v ∣ d t$ , but $∣6 t ∣ = 6 t$ here (since $v \geq 0$ ), so it's the same integral $= 48$ m.

Verify: Forecast was right — both are $48$ m. Units: $(m/s) \times (s) = m$ . ✓ In the figure the whole shaded region sits above the axis, so signed area = unsigned area. See Displacement vs Distance.

Ex 2 — Cell B: one sign change

Worked example Statement

$v (t) = 2 t - 6$ (m/s) on $[0, 5]$ . Find displacement and total distance.

Forecast: The particle slows, stops, reverses, then goes forward. So displacement should be smaller in size than the distance. Guess both numbers.

Step 1 — Find where the rate is zero. Why first? The sign of $v$ can only flip where $v = 0$ ; that's the split point for distance. $2 t - 6 = 0 ⟹ t = 3.$ On $[0, 3)$ : $v < 0$ (moving back). On $(3, 5]$ : $v > 0$ (moving forward). Step 2 — Displacement (signed). Why one integral? Net change lets increases and decreases cancel — that's the point of "net." $\int_{0}^{5} (2 t - 6) d t = [t^{2} - 6 t]_{0}^{5} = (25 - 30) - 0 = - 5 m .$ Ends 5 m behind start. Step 3 — Distance (unsigned). Why split at $t = 3$ ? Because $\int ∣ v ∣$ counts backward motion as positive; we flip the sign on the negative piece. $\int_{0}^{3} (6 - 2 t) d t + \int_{3}^{5} (2 t - 6) d t .$ First $= [6 t - t^{2}]_{0}^{3} = 18 - 9 = 9$ . Second $= [t^{2} - 6 t]_{3}^{5} = (- 5) - (- 9) = 4$ . Distance $= 9 + 4 = 13$ m.

Verify: $13 > ∣ - 5 ∣$ ✓ (distance always $\geq$ |displacement|). Look at the figure: the amber area below the axis is the backward trip.

Ex 3 — Cell C: two sign changes

Worked example Statement

$v (t) = t^{2} - 4 t + 3$ (m/s) on $[0, 4]$ . Find displacement and distance.

Forecast: A parabola opening up, roots somewhere — it will dip below zero then come back. Expect two split points and distance $>$ |displacement|.

Step 1 — Zeros of the rate. Why? Same reason as Ex 2, but now factor: $t^{2} - 4 t + 3 = (t - 1) (t - 3)$ , so $v = 0$ at $t = 1$ and $t = 3$ . Signs: $[0, 1)$ positive, $(1, 3)$ negative, $(3, 4]$ positive. Step 2 — Displacement. Why one integral? We want net change, and net change lets the forward and backward parts cancel — so we integrate the signed velocity without splitting. $\int_{0}^{4} (t^{2} - 4 t + 3) d t = [\frac{t ^{3}}{3} - 2 t^{2} + 3 t]_{0}^{4} = \frac{64}{3} - 32 + 12 = \frac{64}{3} - 20 = \frac{4}{3} m .$ Step 3 — Distance: three pieces, flip the middle. Why three? Two sign changes cut $[0, 4]$ into three monotone-sign chunks. Let $G (t) = \frac{t ^{3}}{3} - 2 t^{2} + 3 t$ .

$[0, 1]$ : $G (1) - G (0) = \frac{4}{3}$ .
$[1, 3]$ : $G (3) - G (1) = 0 - \frac{4}{3} = - \frac{4}{3}$ ; take absolute value $\to \frac{4}{3}$ .
$[3, 4]$ : $G (4) - G (3) = \frac{4}{3} - 0 = \frac{4}{3}$ . Distance $= \frac{4}{3} + \frac{4}{3} + \frac{4}{3} = 4$ m.

Verify: $4 > \frac{4}{3}$ ✓. Displacement $\frac{4}{3} \approx 1.33$ ; distance $4$ . The figure shows the up-down-up pattern with the middle hump shaded amber.

Ex 4 — Cells D & E: degenerate inputs

Worked example Statement

(4a, Cell D) A particle is stationary: $v (t) = 0$ on $[0, 10]$ . (4b, Cell E) A particle has $v (t) = 3 t^{2}$ but we integrate over $[2, 2]$ . Find displacement and distance for each.

Forecast: Nothing moves in (4a). In (4b) start = end instant, so nothing accumulates. Both should give $0$ for both displacement and distance.

Step 1 — Case D, displacement. Why zero? The rate is identically zero, so every tiny piece $v Δ t = 0$ ; their sum is $0$ . $\int_{0}^{10} 0 d t = 0.$ Step 2 — Case D, distance. Why also zero? Distance uses $\int ∣ v ∣$ , and $∣0∣ = 0$ , so the unsigned integral is zero too. $\int_{0}^{10} ∣0∣ d t = \int_{0}^{10} 0 d t = 0.$ Step 3 — Case E, displacement. Why zero? By definition of the definite integral, $\int_{a}^{a} F^{'} d t = F (a) - F (a) = 0$ — zero width means zero net change, no matter how big the rate is. $\int_{2}^{2} 3 t^{2} d t = 0.$ Step 4 — Case E, distance. Why also zero? The distance integral has the same zero width $[2, 2]$ , so it collapses to zero as well — we must write it out to complete the template. $\int_{2}^{2} ∣3 t^{2} ∣ d t = \int_{2}^{2} 3 t^{2} d t = 0 (since 3 t^{2} \geq 0) .$

Verify: In D the object literally doesn't move → both $0$ ✓. In E there is no elapsed time → both $0$ ✓. These are the "trivial but must-check" corners of the matrix — the theorem still works, it just returns $0$ .

Ex 5 — Cell F: symmetric round trip

Worked example Statement

$v (t) = sin t$ (m/s) on $[0, 2 π]$ . Find displacement and distance.

Forecast: Forward for the first half, backward for the second — a perfect round trip. Displacement $0$ ; distance is the area of the two humps (one above, one below the axis) $= 4$ .

Step 1 — Sign changes. Why locate zeros first? The velocity can only flip sign where $v = 0$ , and those points are exactly where we must split for distance — so we find them before integrating anything. $sin t = 0$ at $t = 0, π, 2 π$ . On $(0, π)$ : positive. On $(π, 2 π)$ : negative. One interior split at $t = π$ . Step 2 — Displacement. Why do we expect cancellation? The two halves are mirror images, so the signed areas cancel — a single integral captures the net. $\int_{0}^{2 π} sin t d t = [- cos t]_{0}^{2 π} = (- 1) - (- 1) = 0 m .$ Step 3 — Distance. Why split at $π$ and flip the sign? Distance is $\int ∣ v ∣$ ; on $(π, 2 π)$ the velocity is negative, so $∣ v ∣ = - sin t$ there. We split at the sign-change point $t = π$ and integrate $- sin t$ on the second piece so every bit of motion counts positive. $\int_{0}^{π} sin t d t + \int_{π}^{2 π} (- sin t) d t = [- cos t]_{0}^{π} + [cos t]_{π}^{2 π} = 2 + 2 = 4 m .$

Verify: Displacement $0$ but distance $4$ — the object travelled $4$ m yet came home. ✓ Figure: two humps, equal areas, one shaded cyan (forward) one amber (back).

Ex 6 — Cells G0 & G: pure inflow vs. inflow-then-outflow

Worked example Statement

(6a, Cell G0) A tank fills at rate $r (t) = 5$ litres/min for $t \in [0, 6]$ — always positive, pure inflow. (6b, Cell G) Later the rate is $r (t) = 12 - 3 t$ L/min for $t \in [0, 6]$ , which turns negative partway. Find the net change in volume for each, and for (6b) find when it starts draining.

Forecast: In 6a water only enters, so net $=$ total added $= 30$ L. In 6b it fills then drains — net could be much smaller. Guess both.

Step 1 — Case G0, pure inflow. Why no sign worry? $r = 5 > 0$ always, so volume only rises; net change equals total added. $\int_{0}^{6} 5 d t = [5 t]_{0}^{6} = 30 L .$ Step 2 — Case G, find the sign change. Why $r = 0$ ? The rate switches from filling to draining exactly where $r (t) = 0$ ; that instant answers "when does it start draining?" $12 - 3 t = 0 ⟹ t = 4 min .$ Before $t = 4$ : filling; after: draining. Step 3 — Case G, net change of volume. Why one integral? The question asks net change, so signed accumulation is exactly right — the late outflow subtracts from the early inflow. $\int_{0}^{6} (12 - 3 t) d t = [12 t - \frac{3 t ^{2}}{2}]_{0}^{6} = (72 - 54) - 0 = 18 L .$

Verify: 6a: $30$ L, and net = total since the rate never dips negative ✓. 6b: rate flips at $t = 4$ ✓, net $+ 18$ L, smaller than 6a's pure-inflow $30$ because outflow ate into it ✓. Units $(L/min) \times min = L$ . See the figure: 6a is a flat positive band; 6b crosses the axis at $t = 4$ .

Ex 7 — Cell H: marginal cost never negative (constant cancels)

Worked example Statement

Marginal cost $C^{'} (x) = 4 x + 30$ (₹/unit), always positive. Find the added cost of going from $x = 5$ to $x = 15$ units — without knowing fixed costs.

Forecast: We don't know $C (0)$ (the fixed cost). Guess whether that matters for the change.

Step 1 — Recognise $C^{'}$ . Why? Marginal cost is literally the derivative of total cost, so $\int_{5}^{15} C^{'} d x = C (15) - C (5)$ by the theorem. Step 2 — Integrate. Why can we ignore the fixed cost here? An antiderivative is $C (x) = 2 x^{2} + 30 x + K$ for some unknown constant $K$ (the fixed cost). But the theorem evaluates $C (15) - C (5)$ , and that same $K$ appears in both and subtracts to zero — so we may pick $K = 0$ and integrate as if there were no constant. $\int_{5}^{15} (4 x + 30) d x = [2 x^{2} + 30 x]_{5}^{15} = (450 + 450) - (50 + 150) = 900 - 200 = 700.$ Step 3 — Note the cancellation. The unknown fixed cost never entered the number — confirming Step 2's reasoning.

Verify: Added cost $=$ ₹700. The fixed cost never entered — the forecast that it's irrelevant is confirmed. ✓ See Marginal Cost and Revenue.

Ex 8 — Cell H2: marginal profit that changes sign

Worked example Statement

A firm's marginal profit is $P^{'} (x) = 60 - 4 x$ (₹/unit) for $x \in [0, 20]$ . It is positive for small $x$ (each extra unit still helps) but goes negative once you overproduce. Find the net change in profit from $x = 0$ to $x = 20$ , and the production level that maximises profit.

Forecast: Profit rises then falls, so the net over $[0, 20]$ could be small or even negative. The peak is where marginal profit hits zero. Guess the peak $x$ .

Step 1 — Find the sign change. Why? Marginal profit $P^{'}$ crossing zero marks the switch from "each unit adds profit" to "each unit destroys profit" — that crossing is the profit-maximising level (this is Marginal Cost and Revenue logic). $60 - 4 x = 0 ⟹ x = 15 units .$ On $[0, 15)$ : $P^{'} > 0$ (profit growing). On $(15, 20]$ : $P^{'} < 0$ (profit shrinking). Step 2 — Net change of profit over $[0, 20]$ . Why one signed integral? We want the net change in profit, so gains and losses cancel — no splitting. $\int_{0}^{20} (60 - 4 x) d x = [60 x - 2 x^{2}]_{0}^{20} = (1200 - 800) - 0 = 400.$ Net profit change over the whole range: +₹400. Step 3 — Profit gained up to the peak. Why split here? To see how much profit was added before overproduction ate some back, integrate only up to $x = 15$ . $\int_{0}^{15} (60 - 4 x) d x = [60 x - 2 x^{2}]_{0}^{15} = (900 - 450) = 450.$ So profit climbed by ₹450 up to $x = 15$ , then the last $5$ units gave back ₹50 (since $450 - 400 = 50$ ).

Verify: Peak at $x = 15$ where $P^{'} = 0$ ✓. Net over $[0, 20]$ is $400$ ✓, which is $450 - 50$ ✓ (the overproduction loss). The figure shows $P^{'}$ crossing zero at $x = 15$ , cyan gain area then amber loss area.

Ex 9 — Cell J: limiting behaviour ( $b \to \infty$ )

Worked example Statement

A leaking battery discharges at rate $r (t) = e^{- t}$ (units/s) for $t \in [0, \infty)$ . How much charge is lost in total?

Forecast: The rate shrinks fast toward zero. Does the total stay finite even over infinite time? Guess yes/no.

Step 1 — Set up as a limit. Why a limit? " $\infty$ " is not a number we can plug in; the integral to $\infty$ means "let the upper limit $b$ grow without bound." $\int_{0}^{\infty} e^{- t} d t = lim_{b \to \infty} \int_{0}^{b} e^{- t} d t .$ Step 2 — Net change up to $b$ . Why is the antiderivative $- e^{- t}$ ? We need a function whose derivative is $e^{- t}$ . By the chain rule $\frac{d}{d t} (- e^{- t}) = - e^{- t} \cdot (- 1) = e^{- t}$ , so $- e^{- t}$ is exactly that antiderivative. $\int_{0}^{b} e^{- t} d t = [- e^{- t}]_{0}^{b} = - e^{- b} - (- 1) = 1 - e^{- b} .$ Step 3 — Take the limit. Why does it converge? As $b \to \infty$ , $e^{- b} \to 0$ , so the tail contributes nothing. $lim_{b \to \infty} (1 - e^{- b}) = 1.$ Total charge lost $= 1$ unit.

Verify: Finite answer despite infinite time — the decaying rate makes the accumulated change converge. $1 - e^{- b} \to 1$ ✓. See the figure: the shaded area under $e^{- t}$ approaches a finite value as $b$ slides right.

Ex 10 — Cell K: motion with decaying speed ( $b \to \infty$ )

Ex 11 — Cell I: exam twist (run it backwards)

Recall Which cell am I in? (self-test — every cell)

A · Rate positive throughout ::: displacement = distance (Ex 1). B · Rate changes sign once ::: split at the one zero for distance; net still one integral (Ex 2). C · Rate changes sign twice ::: three pieces for distance (Ex 3). D · Rate zero everywhere ::: displacement = distance = 0 (Ex 4a). E · Zero-width interval ( $a = b$ ) ::: both integrals are 0 regardless of the rate (Ex 4b). F · Symmetric, returns home ::: displacement 0, distance > 0 (Ex 5). G0 · Pure inflow, rate never negative ::: net = total added (Ex 6a). G · Flow that turns to outflow ::: net signed integral, smaller than pure inflow (Ex 6b). H · Rising marginal cost ::: net change subtracts the fixed cost away (Ex 7). H2 · Sign-changing marginal profit ::: peak where $P^{'} = 0$ ; net over range can be small (Ex 8). J · Accumulation with upper limit $\infty$ ::: improper integral, replace $\infty$ by $b$ , take a limit (Ex 9). K · Motion with $\infty$ limit, $v > 0$ ::: finite drift, distance = displacement (Ex 10). I · Given net + one endpoint ::: add the change to the known endpoint (Ex 11).

Connections

Parent: Net Change Theorem — this page fleshes out every case of that one equation.
Displacement vs Distance — the net-vs-total split (cells B, C, F).
Fundamental Theorem of Calculus — the machinery behind every integral here.
Definite Integral as a Riemann Sum — what $\int$ means.
Mean Value Theorem — justifies the theorem's derivation.
Telescoping Sums — the cancellation in the parent's proof.
Marginal Cost and Revenue — cells H and H2.