Net change theorem

WHAT is it?

This is literally the Fundamental Theorem of Calculus, Part 2, just re-read with a physical story. Same equation, new clothes.

WHY is it true? (Derive from scratch)

We build it from the definition of the integral — no memorising.

Step 1 — Slice the interval. Why? Because change over a long interval is hard, but change over a tiny step is easy. Partition $[a,b]$ into $n$ pieces: $a=x_0 < x_1 < \dots < x_n = b, \qquad \Delta x = \frac{b-a}{n}.$

Step 2 — Net change is a telescoping sum. The total change from $a$ to $b$ is the sum of all the little changes. Why? Because the middle values cancel: $F(b)-F(a) = \sum_{i=1}^{n}\Big[F(x_i)-F(x_{i-1})\Big].$ (Write it out: $(F(x_1)-F(x_0))+(F(x_2)-F(x_1))+\dots$ — everything cancels except $F(b)-F(a)$ .)

Step 3 — Approximate each little change by a rate × time. Why? The Mean Value Theorem guarantees a point $x_i^*$ in each slice where $F(x_i)-F(x_{i-1}) = F'(x_i^*)\,\Delta x.$ This says "small change = (slope there) × (width)". Pure definition of derivative made exact.

Step 4 — Recognise a Riemann sum. Substitute Step 3 into Step 2: $F(b)-F(a) = \sum_{i=1}^{n} F'(x_i^*)\,\Delta x.$

Step 5 — Take the limit. As $n\to\infty$ , the right side is the definition of the integral: $F(b)-F(a) = \lim_{n\to\infty}\sum_{i=1}^{n} F'(x_i^*)\,\Delta x = \int_a^b F'(x)\,dx. \qquad\blacksquare$

HOW to use it (the recipe)

Identify the quantity $F$ and its rate $F'$ (which one are you given?).
Integrate the rate over the interval.
The answer = $F(\text{end}) - F(\text{start})$ = net change.
If you want total distance / total amount moved, integrate $|F'(x)|$ instead.

Worked Examples

Worked example 1 — Velocity → displacement vs distance

A particle moves with velocity $v(t)=t^2-4$ (m/s) for $t\in[0,3]$ . Find displacement and total distance.

Displacement $=\int_0^3 (t^2-4)\,dt$ . Why integrate v? Because $v=s'$ , so by the theorem $\int v = s(3)-s(0)$ = net change in position. $=\Big[\tfrac{t^3}{3}-4t\Big]_0^3 = (9-12)-0 = -3\text{ m}.$ Net: it ended 3 m behind the start.

Total distance. Why split? Because $v<0$ on $[0,2)$ and $v>0$ on $(2,3]$ ; distance counts both as positive. $v=0$ at $t=2$ . $\int_0^2 (4-t^2)\,dt + \int_2^3 (t^2-4)\,dt.$ First $=\big[4t-\tfrac{t^3}{3}\big]_0^2 = 8-\tfrac83=\tfrac{16}{3}$ . Second $=\big[\tfrac{t^3}{3}-4t\big]_2^3 = (-3)-(\tfrac83-8)=(-3)-(-\tfrac{16}{3})=\tfrac{7}{3}$ . Total $=\tfrac{16}{3}+\tfrac{7}{3}=\tfrac{23}{3}\approx 7.67$ m. (Bigger than $|{-3}|$ — sanity check passes.)

Forecast-then-Verify

Recall Before computing: a particle has

$v(t)=\cos t$ on $[0,2\pi]$ . Forecast displacement and distance, then check. Forecast: It goes forward then back, returning home → displacement $\approx 0$ ; distance should be the "area of all the humps" $=4$ . Verify: Displacement $=\int_0^{2\pi}\cos t\,dt=[\sin t]_0^{2\pi}=0$ . ✓ Distance $=\int_0^{2\pi}|\cos t|\,dt = 4\int_0^{\pi/2}\cos t\,dt = 4[\sin t]_0^{\pi/2}=4$ . ✓

Common Mistakes (Steel-manned)

Recall Feynman: explain to a 12-year-old

Imagine a speedometer in a car. It never tells you where you are — only how fast you're going right now. But if you write down your speed every single second and add up "speed × one second" for the whole trip, you find out exactly how far the car moved. The net change theorem says: add up all the little "how fast × tiny time" pieces and you get the total change in position. And if you sometimes drive backwards, those count as negative — unless you only care about how much road you covered, in which case you treat backward driving as positive too.

Connections

Fundamental Theorem of Calculus — net change theorem is FTC Part 2 in physical disguise.
Definite Integral as a Riemann Sum — the limit we used in Step 5.
Mean Value Theorem — justified Step 3 (small change = slope × width).
Displacement vs Distance — the net-vs-total distinction.
Telescoping Sums — the cancellation trick in Step 2.
Marginal Cost and Revenue — economics application.

State the net change theorem.

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

: the integral of a rate equals the net change of the quantity.

Which theorem is the net change theorem secretly identical to?

The Fundamental Theorem of Calculus, Part 2 — reinterpreted physically.

In Step 2 of the derivation, what trick collapses the sum of little changes?

A telescoping sum:

\sum [F(x_i)-F(x_{i-1})] = F(b)-F(a)

What does the Mean Value Theorem provide in the derivation?

A point

x_i^*

with

F(x_i)-F(x_{i-1}) = F'(x_i^*)\Delta x

, turning each little change into rate×width.

How do you get displacement from velocity?

\int_a^b v(t)\,dt

(signed, can be negative).

How do you get total distance from velocity?

\int_a^b |v(t)|\,dt

— split at sign changes and add positive pieces.

Why don't you need the constant of integration in a net change problem?

Because

F(b)-F(a)

subtracts it away;

+C

cancels.

Why is

|\int v|

wrong for total distance?

Because cancellation happens inside the integral; the bars must go inside:

\int|v|

v(t)=t²−4 on [0,3]: displacement?

\int_0^3(t^2-4)dt = -3

v(t)=cos t on [0,2π]: displacement and distance?

Displacement 0; distance 4.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, net change theorem ka core idea bahut simple hai: agar tumhe pata hai koi cheez har instant pe kitni tezi se badal rahi hai (yaani rate, jo derivative $F'$ hai), to us rate ka integral le lo — wahi tumhe total net change de dega, $\int_a^b F'(x)\,dx = F(b)-F(a)$ . Ye actually Fundamental Theorem of Calculus Part 2 hi hai, bas physics ki bhasha mein. Speedometer wali example yaad rakho: speedometer batata hai sirf speed, location nahi — par speed ko time ke upar add karte jao to total distance/displacement mil jaata hai.

Sabse important baat jahan students marks gawate hain: displacement vs total distance. Agar velocity $v(t)$ kabhi negative ho jaati hai (object peeche jaa raha hai), to $\int v\,dt$ se aage-peeche cancel ho jaata hai aur tumhe net displacement milta hai. Lekin agar total distance chahiye — jitna road actually cover hua — to absolute value andar daalo: $\int |v|\,dt$ , aur jahan $v=0$ hota hai wahan integral ko todo. Yaad rakho: "bars before integral" — $|\int v|$ galat hai, $\int|v|$ sahi.

Ek aur cool baat: net change nikalte waqt tumhe initial value ya $+C$ ki zaroorat hi nahi padti, kyunki $F(b)-F(a)$ mein constant khud cancel ho jaata hai. Isiliye marginal cost wale problem mein $C(0)$ jaanna zaroori nahi — sirf change chahiye. Jab bhi koi "rate" diya ho (water flow rate, marginal cost, velocity), socho "ye kis quantity ka derivative hai?" — phir integrate karke seedha net change paa lo. Bas, "rate in, change out".

WHAT is it?

WHY is it true? (Derive from scratch)

HOW to use it (the recipe)

Worked Examples

Forecast-then-Verify

Common Mistakes (Steel-manned)

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Calculus II — Integration

Connections