Net change theorem: Worked examples — Hinglish deep dive

4.2.5 · D3 · Maths › Calculus II — Integration › Net change theorem

Ek hi tool hai jo hum har jagah use karte hain, woh hai parent result: $\int_{a}^{b} F^{'} (x) d x = F (b) - F (a) .$ Yahan $F$ woh quantity hai (position, volume, cost, ...) aur $F^{'}$ uski rate hai (velocity, flow rate, marginal cost, ...). Symbol $\int_{a}^{b}$ ka matlab hai "saare chote pieces ko $a$ se $b$ tak jodo" — dekho Definite Integral as a Riemann Sum. Poora page parent topic par build karta hai.

Scenario matrix

Kuch bhi work karne se pehle, chalte hain har tarah ki situation lay out karte hain jo ek net-change problem mein aa sakti hai. Isse ek true two-axis map ki tarah padho: rows hain quantity ka type (motion, accumulation, economics); columns hain rate ka sign pattern — har column mein exactly wahi cells hain jo us pattern ki hain (ek "changes sign" example kabhi "degenerate" ya "limiting" ke neeche nahi baithega). Har cell us example ka naam deta hai jo use fill karta hai.

Quantity type ↓ \ Sign pattern →	Rate kabhi negative nahi	Rate sign badalta hai	Degenerate (zero rate / zero width)	Limiting ( $b \to \infty$ )
Motion (velocity → position)	A hamesha forward · Ex 1	B ek flip · Ex 2 ; C do 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 phir 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 missing endpoint dhundo · Ex 11	—	—	—

Map padhna: neeche jao kya cheez change ho rahi hai choose karne ke liye, *daayein jao uski rate kaise behave karti hai choose karne ke liye. Har example neeche uske cell letter se tagged hai taaki tum ise grid par trace kar sako. Dhyaan do ki columns ab consistent hain — "changes sign" column mein sab kuch sach mein sign change karta hai; "degenerate" mein sab kuch genuinely zero-rate ya zero-width case hai.

Worked Examples

Ex 1 — Cell A: rate kabhi negative nahi

Worked example Statement

Ek particle velocity $v (t) = 6 t$ (m/s) se move karta hai $t \in [0, 4]$ ke liye; kyunki $6 t \geq 0$ hai, woh sirf aage hi move karta hai. Displacement aur total distance nikalo.

Forecast: Kyunki particle kabhi reverse nahi karta, mujhe predict karna hai ki displacement aur distance ek hi number honge. Padhne se pehle guess karo.

Step 1 — Quantity choose karo. Kyun? Velocity position $s$ ki rate of change hai, toh $v = s^{'}$ . Theorem phir deta hai $\int v = s (4) - s (0)$ . Step 2 — Integrate karo. Split kyun nahi? Kyunki $v = 6 t \geq 0$ poore $[0, 4]$ par; velocity kabhi sign nahi badlati, toh kuch cancel nahi hota. $displacement = \int_{0}^{4} 6 t d t = [3 t^{2}]_{0}^{4} = 48 m .$ Step 3 — Distance. Identical kyun? Distance hai $\int_{0}^{4} ∣ v ∣ d t$ , lekin $∣6 t ∣ = 6 t$ yahan (kyunki $v \geq 0$ ), toh yeh wahi integral hai $= 48$ m.

Verify: Forecast sahi nikla — dono $48$ m hain. Units: $(m/s) \times (s) = m$ . ✓ Figure mein poora shaded region axis ke upar hai, toh signed area = unsigned area. Dekho Displacement vs Distance.

Ex 2 — Cell B: ek sign change

Worked example Statement

$v (t) = 2 t - 6$ (m/s) on $[0, 5]$ . Displacement aur total distance nikalo.

Forecast: Particle slow hoga, rukega, reverse karega, phir aage jaayega. Toh displacement size mein distance se chota hona chahiye. Dono numbers guess karo.

Step 1 — Rate ka zero dhundo. Pehle kyun? $v$ ka sign sirf wahan flip ho sakta hai jahan $v = 0$ ; wahi distance ka split point hai. $2 t - 6 = 0 ⟹ t = 3.$ $[0, 3)$ par: $v < 0$ (peeche ja raha hai). $(3, 5]$ par: $v > 0$ (aage ja raha hai). Step 2 — Displacement (signed). Ek integral kyun? Net change forward aur backward parts ko cancel hone deta hai — yahi "net" ka matlab hai. $\int_{0}^{5} (2 t - 6) d t = [t^{2} - 6 t]_{0}^{5} = (25 - 30) - 0 = - 5 m .$ Start se 5 m peeche hai. Step 3 — Distance (unsigned). $t = 3$ par split kyun? Kyunki $\int ∣ v ∣$ backward motion ko positive count karta hai; negative piece par sign flip karte hain. $\int_{0}^{3} (6 - 2 t) d t + \int_{3}^{5} (2 t - 6) d t .$ Pehla $= [6 t - t^{2}]_{0}^{3} = 18 - 9 = 9$ . Doosra $= [t^{2} - 6 t]_{3}^{5} = (- 5) - (- 9) = 4$ . Distance $= 9 + 4 = 13$ m.

Verify: $13 > ∣ - 5 ∣$ ✓ (distance hamesha $\geq$ |displacement|). Figure dekho: axis ke neeche amber area backward trip hai.

Ex 3 — Cell C: do sign changes

Worked example Statement

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

Forecast: Ek parabola upar ki taraf opening, kahin roots hain — woh zero ke neeche jaayega phir wapas aayega. Do split points aur distance $>$ |displacement| expect karo.

Step 1 — Rate ke zeros. Kyun? Ex 2 jaisi wajah, lekin ab factor karo: $t^{2} - 4 t + 3 = (t - 1) (t - 3)$ , toh $v = 0$ at $t = 1$ aur $t = 3$ . Signs: $[0, 1)$ positive, $(1, 3)$ negative, $(3, 4]$ positive. Step 2 — Displacement. Ek integral kyun? Hum net change chahte hain, aur net change forward aur backward parts ko cancel hone deta hai — toh hum signed velocity ko bina split kiye integrate karte hain. $\int_{0}^{4} (t^{2} - 4 t + 3) d t = [\frac{t ^{3}}{3} - 2 t^{2} + 3 t]_{0}^{4} = \frac{64}{3} - 32 + 12 = \frac{64}{3} - 20 = \frac{4}{3} m .$ Step 3 — Distance: teen pieces, beech wala flip karo. Teen kyun? Do sign changes $[0, 4]$ ko teen monotone-sign chunks mein kaatate hain. Maano $G (t) = \frac{t ^{3}}{3} - 2 t^{2} + 3 t$ .

$[0, 1]$ : $G (1) - G (0) = \frac{4}{3}$ .
$[1, 3]$ : $G (3) - G (1) = 0 - \frac{4}{3} = - \frac{4}{3}$ ; absolute value lo $\to \frac{4}{3}$ .
$[3, 4]$ : $G (4) - G (3) = \frac{4}{3} - 0 = \frac{4}{3}$ . 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 $4$ . Figure mein up-down-up pattern dikha hai beech wala hump amber shaded hai.

Ex 4 — Cells D & E: degenerate inputs

Worked example Statement

(4a, Cell D) Ek particle stationary hai: $v (t) = 0$ on $[0, 10]$ . (4b, Cell E) Ek particle ka $v (t) = 3 t^{2}$ hai lekin hum $[2, 2]$ par integrate karte hain. Dono ke liye displacement aur distance nikalo.

Forecast: (4a) mein kuch move nahi karta. (4b) mein start = end instant hai, toh kuch accumulate nahi hota. Dono ko displacement aur distance dono ke liye $0$ dena chahiye.

Step 1 — Case D, displacement. Zero kyun? Rate identically zero hai, toh har tiny piece $v Δ t = 0$ ; unka sum $0$ hai. $\int_{0}^{10} 0 d t = 0.$ Step 2 — Case D, distance. Woh bhi zero kyun? Distance $\int ∣ v ∣$ use karta hai, aur $∣0∣ = 0$ , toh unsigned integral bhi zero hai. $\int_{0}^{10} ∣0∣ d t = \int_{0}^{10} 0 d t = 0.$ Step 3 — Case E, displacement. Zero kyun? Definite integral ki definition se, $\int_{a}^{a} F^{'} d t = F (a) - F (a) = 0$ — zero width matlab zero net change, chahe rate kitni bhi badi ho. $\int_{2}^{2} 3 t^{2} d t = 0.$ Step 4 — Case E, distance. Woh bhi zero kyun? Distance integral ki wahi zero width $[2, 2]$ hai, toh woh bhi zero par collapse ho jaata hai — template complete karne ke liye ise likhhna zaroori hai. $\int_{2}^{2} ∣3 t^{2} ∣ d t = \int_{2}^{2} 3 t^{2} d t = 0 (since 3 t^{2} \geq 0) .$

Verify: D mein object literally nahi hila → dono $0$ ✓. E mein koi elapsed time nahi → dono $0$ ✓. Ye matrix ke "trivial but must-check" corners hain — theorem phir bhi kaam karta hai, bas $0$ return karta hai.

Ex 5 — Cell F: symmetric round trip

Worked example Statement

$v (t) = sin t$ (m/s) on $[0, 2 π]$ . Displacement aur distance nikalo.

Forecast: Pehle half mein forward, doosre mein backward — ek perfect round trip. Displacement $0$ ; distance do humps ka area hai (ek upar, ek neeche axis ke) $= 4$ .

Step 1 — Sign changes. Pehle zeros kyun locate karo? Velocity sirf wahan sign flip kar sakti hai jahan $v = 0$ , aur wahi points exactly hain jahan distance ke liye split karna zaroori hai — toh kuch bhi integrate karne se pehle inhe dhundho. $sin t = 0$ at $t = 0, π, 2 π$ . $(0, π)$ par: positive. $(π, 2 π)$ par: negative. Ek interior split $t = π$ par. Step 2 — Displacement. Cancellation kyun expect karte hain? Dono halves mirror images hain, toh signed areas cancel ho jaate hain — ek single integral net capture karta hai. $\int_{0}^{2 π} sin t d t = [- cos t]_{0}^{2 π} = (- 1) - (- 1) = 0 m .$ Step 3 — Distance. $π$ par kyun split karo aur sign flip karo? Distance hai $\int ∣ v ∣$ ; $(π, 2 π)$ par velocity negative hai, toh $∣ v ∣ = - sin t$ wahan. Hum sign-change point $t = π$ par split karte hain aur doosre piece par $- sin t$ integrate karte hain taaki motion ka har bit positive count ho. $\int_{0}^{π} sin t d t + \int_{π}^{2 π} (- sin t) d t = [- cos t]_{0}^{π} + [cos t]_{π}^{2 π} = 2 + 2 = 4 m .$

Verify: Displacement $0$ lekin distance $4$ — object $4$ m travel kiya phir ghar aa gaya. ✓ Figure: do humps, equal areas, ek cyan shaded (forward) ek amber (back).

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

Worked example Statement

(6a, Cell G0) Ek tank rate $r (t) = 5$ litres/min se $t \in [0, 6]$ ke liye bharta hai — hamesha positive, pure inflow. (6b, Cell G) Baad mein rate $r (t) = 12 - 3 t$ L/min hai $t \in [0, 6]$ ke liye, jo beech mein negative ho jaata hai. Dono ke liye volume mein net change nikalo, aur (6b) ke liye woh batao jab drain hona shuru hota hai.

Forecast: 6a mein sirf paani enter karta hai, toh net $=$ total added $= 30$ L. 6b mein pehle bharta hai phir drain hota hai — net bahut chhota ho sakta hai. Dono guess karo.

Step 1 — Case G0, pure inflow. Sign ki chinta kyun nahi? $r = 5 > 0$ hamesha, toh volume sirf badhta hai; net change total added ke barabar hai. $\int_{0}^{6} 5 d t = [5 t]_{0}^{6} = 30 L .$ Step 2 — Case G, sign change dhundo. $r = 0$ kyun? Rate exactly wahan filling se draining mein switch hota hai jahan $r (t) = 0$ ; woh instant jawaab deta hai "kab drain hona shuru hota hai?" $12 - 3 t = 0 ⟹ t = 4 min .$ $t = 4$ se pehle: filling; baad mein: draining. Step 3 — Case G, volume mein net change. Ek integral kyun? Question net change poochh raha hai, toh signed accumulation exactly sahi hai — baad ka outflow early inflow se subtract hota hai. $\int_{0}^{6} (12 - 3 t) d t = [12 t - \frac{3 t ^{2}}{2}]_{0}^{6} = (72 - 54) - 0 = 18 L .$

Verify: 6a: $30$ L, aur net = total kyunki rate kabhi negative nahi gayi ✓. 6b: rate $t = 4$ par flip hoti hai ✓, net $+ 18$ L, 6a ke pure-inflow $30$ se chhota kyunki outflow ne ise khaaya ✓. Units $(L/min) \times min = L$ . Figure dekho: 6a ek flat positive band hai; 6b $t = 4$ par axis cross karta hai.

Ex 7 — Cell H: marginal cost kabhi negative nahi (constant cancel ho jaata hai)

Worked example Statement

Marginal cost $C^{'} (x) = 4 x + 30$ (₹/unit), hamesha positive. $x = 5$ se $x = 15$ units jaane ka added cost nikalo — bina fixed costs jaane.

Forecast: Hum $C (0)$ (fixed cost) nahi jaante. Guess karo ki kya woh change ke liye matter karta hai.

Step 1 — $C^{'}$ ko pehchano. Kyun? Marginal cost literally total cost ka derivative hai, toh $\int_{5}^{15} C^{'} d x = C (15) - C (5)$ theorem se. Step 2 — Integrate karo. Fixed cost yahan kyun ignore kar sakte hain? Ek antiderivative hai $C (x) = 2 x^{2} + 30 x + K$ kisi unknown constant $K$ ke liye (fixed cost). Lekin theorem $C (15) - C (5)$ evaluate karta hai, aur wahi $K$ dono mein aata hai aur subtract hokar zero ho jaata hai — toh hum $K = 0$ choose kar sakte hain aur aise integrate kar sakte hain jaise koi constant ho hi nahi. $\int_{5}^{15} (4 x + 30) d x = [2 x^{2} + 30 x]_{5}^{15} = (450 + 450) - (50 + 150) = 900 - 200 = 700.$ Step 3 — Cancellation note karo. Unknown fixed cost number mein kabhi enter hi nahi hua — Step 2 ki reasoning confirm ho gayi.

Verify: Added cost $=$ ₹700. Fixed cost kabhi enter nahi hua — forecast ki woh irrelevant hai, confirmed. ✓ Dekho Marginal Cost and Revenue.

Ex 8 — Cell H2: marginal profit jo sign badalta hai

Worked example Statement

Ek firm ka marginal profit $P^{'} (x) = 60 - 4 x$ (₹/unit) hai $x \in [0, 20]$ ke liye. Chhote $x$ ke liye positive hai (har extra unit abhi bhi help karta hai) lekin negative ho jaata hai jab overproduce karte ho. $x = 0$ se $x = 20$ tak profit mein net change nikalo, aur woh production level nikalo jo profit maximise karta hai.

Forecast: Profit pehle badhega phir girega, toh $[0, 20]$ par net chhota ya negative bhi ho sakta hai. Peak wahan hai jahan marginal profit zero hogi. Peak $x$ guess karo.

Step 1 — Sign change dhundo. Kyun? Marginal profit $P^{'}$ ka zero cross karna "har unit profit add karta hai" se "har unit profit destroy karta hai" ki switch ko mark karta hai — wahi crossing profit-maximising level hai (yeh Marginal Cost and Revenue logic hai). $60 - 4 x = 0 ⟹ x = 15 units .$ $[0, 15)$ par: $P^{'} > 0$ (profit badh raha hai). $(15, 20]$ par: $P^{'} < 0$ (profit ghatt raha hai). Step 2 — $[0, 20]$ par profit mein net change. Ek signed integral kyun? Hum profit mein net change chahte hain, toh gains aur losses cancel hote hain — koi splitting nahi. $\int_{0}^{20} (60 - 4 x) d x = [60 x - 2 x^{2}]_{0}^{20} = (1200 - 800) - 0 = 400.$ Poore range par net profit change: +₹400. Step 3 — Peak tak gained profit. Yahan kyun split karo? Ye dekhne ke liye ki peak se pehle kitna profit add hua, sirf $x = 15$ tak integrate karo. $\int_{0}^{15} (60 - 4 x) d x = [60 x - 2 x^{2}]_{0}^{15} = (900 - 450) = 450.$ Toh profit ₹450 badha $x = 15$ tak, phir aakhri $5$ units ne ₹50 wapas liya (kyunki $450 - 400 = 50$ ).

Verify: Peak $x = 15$ par jahan $P^{'} = 0$ ✓. $[0, 20]$ par net $400$ hai ✓, jo $450 - 50$ hai ✓ (overproduction loss). Figure mein $P^{'}$ zero cross karta hai $x = 15$ par, cyan gain area phir amber loss area.

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

Worked example Statement

Ek leaking battery rate $r (t) = e^{- t}$ (units/s) se discharge hoti hai $t \in [0, \infty)$ ke liye. Total kitna charge lost hota hai?

Forecast: Rate tezi se zero ki taraf shrink karti hai. Kya total infinite time mein bhi finite rehta hai? Guess karo yes/no.

Step 1 — Limit ki tarah set up karo. Limit kyun? " $\infty$ " koi number nahi hai jise hum plug in kar sakein; $\infty$ tak integral ka matlab hai "upper limit $b$ ko grow karne do bina bound ke." $\int_{0}^{\infty} e^{- t} d t = lim_{b \to \infty} \int_{0}^{b} e^{- t} d t .$ Step 2 — $b$ tak net change. Antiderivative $- e^{- t}$ kyun hai? Hume ek function chahiye jiski derivative $e^{- t}$ ho. Chain rule se $\frac{d}{d t} (- e^{- t}) = - e^{- t} \cdot (- 1) = e^{- t}$ , toh $- e^{- t}$ exactly wahi antiderivative hai. $\int_{0}^{b} e^{- t} d t = [- e^{- t}]_{0}^{b} = - e^{- b} - (- 1) = 1 - e^{- b} .$ Step 3 — Limit lo. Converge kyun karta hai? Jab $b \to \infty$ , $e^{- b} \to 0$ , toh tail kuch contribute nahi karta. $lim_{b \to \infty} (1 - e^{- b}) = 1.$ Total charge lost $= 1$ unit.

Verify: Finite answer infinite time ke baavajood — decaying rate accumulated change ko converge karati hai. $1 - e^{- b} \to 1$ ✓. Figure dekho: $e^{- t}$ ke neeche shaded area ek finite value ki taraf jaata hai jab $b$ right slide karta hai.

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

Ex 11 — Cell I: exam twist (ulta chalao)

Recall Main kis cell mein hoon? (self-test — har cell)

A · Rate poori tarah positive ::: displacement = distance (Ex 1). B · Rate ek baar sign change karta hai ::: distance ke liye ek zero par split karo; net phir bhi ek integral (Ex 2). C · Rate do baar sign change karta hai ::: distance ke liye teen pieces (Ex 3). D · Rate har jagah zero ::: displacement = distance = 0 (Ex 4a). E · Zero-width interval ( $a = b$ ) ::: dono integrals 0 hain chahe rate kuch bhi ho (Ex 4b). F · Symmetric, ghar wapas aata hai ::: displacement 0, distance > 0 (Ex 5). G0 · Pure inflow, rate kabhi negative nahi ::: net = total added (Ex 6a). G · Flow jo outflow mein badal jaata hai ::: net signed integral, pure inflow se chhota (Ex 6b). H · Rising marginal cost ::: net change fixed cost ko subtract kar deta hai (Ex 7). H2 · Sign-changing marginal profit ::: peak jahan $P^{'} = 0$ ; range par net chhota ho sakta hai (Ex 8). J · Accumulation with upper limit $\infty$ ::: improper integral, $\infty$ ko $b$ se replace karo, limit lo (Ex 9). K · Motion with $\infty$ limit, $v > 0$ ::: finite drift, distance = displacement (Ex 10). I · Given net + ek endpoint ::: known endpoint mein change add karo (Ex 11).

Connections

Parent: Net Change Theorem — ye page us ek equation ke har case ko explain karta hai.
Displacement vs Distance — net-vs-total split (cells B, C, F).
Fundamental Theorem of Calculus — yahan har integral ke peeche ki machinery.
Definite Integral as a Riemann Sum — $\int$ ka matlab kya hai.
Mean Value Theorem — theorem ki derivation ko justify karta hai.
Telescoping Sums — parent ke proof mein cancellation.
Marginal Cost and Revenue — cells H aur H2.