Net change theorem: Question bank — Hinglish deep dive

4.2.5 · D5 · Maths › Calculus II — Integration › Net change theorem

Pehle, notation clearly samajh lo (taaki koi bhi symbol mystery na rahe)

Kisi bhi trap se pehle, hum exactly agree kar lete hain ki har symbol ka matlab kya hai — neeche ke traps tab hi kaam karenge jab ye symbols tumhare dimag mein crystal-clear honge.

Wo core equation jis ki humein understanding test karni hai: $\int_{a}^{b} F^{'} (x) d x = F (b) - F (a)$ Padhein: "rate ka integral = cumulative quantity ka net (signed) change."

Figure dekho: orange curve rate $F^{'}$ hai. Uske neeche blue shaded area (from $a$ to $b$ ) green curve $F$ ke vertical rise ke barabar hai, $F (a)$ se $F (b)$ tak. Rate ka area = quantity ka rise. Woh ek picture hi poora theorem hai.

Derivation, taaki "Step 2 / 3 / 5" ka kuch matlab ho

Neeche ke kai questions standard proof ke steps refer karte hain. Ye sab ek jagah hain, picture ke saath, taaki tum dekh sako ki har step kya karta hai.

Recall 5-step build (Why questions mein refer hota hai)

Step 1 — Slice. $[a, b]$ ko $n$ patli strips mein kaato. Har strip ki width $Δ x = \frac{b - a}{n}$ hai (yahi symbol $Δ x$ ka matlab hai: ek strip ki width). Neeche figure mein vertical dashed lines dekho. Step 2 — Telescoping sum. Total change $= \sum_{i = 1}^{n} [F (x_{i}) - F (x_{i - 1})]$ ; interior terms cancel ho jaate hain, sirf $F (b) - F (a)$ bachta hai. Step 3 — Mean Value Theorem. Har strip mein ek point $x_{i}^{*}$ hota hai jahan $F (x_{i}) - F (x_{i - 1}) = F^{'} (x_{i}^{*}) Δ x$ (small change $=$ slope $\times$ width). Step 4 — Riemann sum. Substitute karo: $F (b) - F (a) = \sum F^{'} (x_{i}^{*}) Δ x$ — strip areas ka sum. Step 5 — Limit. Jab $n \to \infty$ (strips infinitely patli hoti jaati hain), woh sum $\int_{a}^{b} F^{'} d x$ integral ban jaata hai.

Figure Step 4 dikhata hai: har grey rectangle ki height $F^{'} (x_{i}^{*})$ hai aur width $Δ x$ ; uska area ek choti si change hai. Sab add karo — wahi net change hai.

True or false — justify karo

Neeche har statement ya to true hai ya false. Decide karo, phir reason reveal karo.

Net change theorem ke liye zaruri hai ki

F (b) - F (a)

nikalne se pehle

F (a)

ki value pata ho.

False. Theorem seedha rate se change deta hai;

F (a)

aur koi bhi

+ C

cancel ho jaate hain. Initial value tabhi chahiye jab alag se actual level

F (b)

jaanna ho.

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

negative ho sakta hai chahe "distance" hamesha positive ho.

True. Woh integral displacement hai (position ka signed net change), jo negative hota hai jab object apne start se peeche end ho. Distance alag quantity hai,

\int_{a}^{b} ∣ v ∣ d t

Kisi bhi velocity function ke liye,

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

False. Ye tabhi equal hote hain jab

v

poore interval mein ek hi sign rakhe. Jab

v

sign change karta hai, cancellation left side ke andar hoti hai lekin right side mein nahi, isliye right side generally bada hota hai.

Net change theorem FTC se bilkul alag ek brand-new result hai.

False. Ye hai hi FTC Part 2, bas physical story ke saath re-read kiya gaya. Same equation

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

, naya interpretation.

Agar marginal cost

C^{'} (x)

hamesha positive hai, to total cost

C (x)

increasing hai.

True. Positive rate ka matlab quantity sirf upar jaati hai, isliye har extra unit cost add karta hai. Kisi bhi interval par net change

\int C^{'} d x

positive hoga.

Total distance hamesha displacement ki absolute value se greater than or equal to hoti hai.

True. Distance har motion ko positively count karta hai jabki displacement mein backward motion forward motion ko cancel kar sakta hai, isliye distance sirf exceed kar sakta hai (ya equal ho sakta hai)

∣ displacement ∣

ke. Equal tabhi hote hain jab object kabhi reverse na kare.

Interval ko double karne se net change hamesha double ho jaata hai.

False. Net change

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

hai, jo rate ki shape par depend karta hai, sirf interval length par nahi. Sirf tab jab

F^{'}

constant ho, interval double karne se change double hoga.

Agar

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

hai, to object kabhi hila hi nahi.

False. Iska sirf yahi matlab hai ki object apne start par wapas aa gaya: forward aur backward motion cancel ho gayi. Ho sakta hai usne bahut bada distance cover kiya ho (e.g.

v = cos t

[0, 2 π]

: displacement

0

, distance

4

Error pakdo

Har line mein ek real mistake hai. Use naam do aur reasoning correct karo.

"Ek particle ka

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

[0, 3]

hai. Displacement

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

hai, isliye distance

= ∣ - 3∣ = 3

m."

Galat.

∣ \int v ∣

lena

v

ke sign change ko

t = 2

par ignore karta hai. Distance ke liye bars andar chahiye:

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

m, na ki

3

"Distance nikalne ke liye main

v

integrate karta hoon aur absolute value leta hoon:

\int_{a}^{b} ∣ v (t) d t ∣

Ye notation jaisa likha gaya meaningless hai; absolute value integrand ko wrap karni chahiye, jo deta hai

\int_{a}^{b} ∣ v (t) ∣ d t

. Aur integrate karne se pehle har us point par split karna hoga jahan

v = 0

ho.

"Marginal cost

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

hai.

x = 10

x = 20

tak extra cost nikalne ke liye mujhe pehle

C (0)

chahiye."

C (0)

kabhi nahi chahiye. Net change theorem seedha deta hai

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

; unknown constant subtraction mein cancel ho jaata hai. (Dekho Marginal Cost and Revenue.)

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

jahan

v

zero se neeche jaata hai — main sirf woh part integrate karunga jahan

v > 0

ho kyunki negative part real motion nahi hai."

Negative part real motion hai (peeche ki taraf jaana). Use drop karna displacement ko corrupt karta hai (jisme signed backward part hona chahiye) aur distance ko bhi (jo use positively count karta hai). Koi bhi quantity use discard karne nahi deti.

"Net change theorem ke liye

F^{'}

positive honi chahiye, warna rate ka integral koi sense nahi banata."

Galat premise.

F^{'}

negative ho sakti hai; theorem ise automatically handle karta hai. Negative rate simply net change mein negative amount contribute karta hai — yahi net ka poora matlab hai.

"Maine integrate kiya aur

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

mila, lekin change negative nahi ho sakta, isliye answer

5

hai."

Net change bilkul negative ho sakta hai — iska matlab quantity decrease hui. Sign mat haatao;

- 5

sahi signed answer hai.

Why questions

Sirf fact nahi, underlying "why" answer karo. (Steps upar collapsible box mein picture ke saath diye gaye hain.)

Derivation mein Step 2 mein telescoping sum kyun use kiya jaata hai?

Kyunki total change ko

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

likhne se saare interior values cancel ho jaate hain, exactly

F (b) - F (a)

bachta hai. Ye "total change" ko tiny changes ke sum mein convert karta hai jinhein hum approximate kar sakte hain. (Dekho Telescoping Sums.)

Step 3 mein Mean Value Theorem kyun chahiye, derivative ki definition kyun nahi chalti?

Derivative definition ek limit hai (finite width ke liye approximate); Mean Value Theorem ek exact point

x_{i}^{*}

deta hai jahan

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

, isliye sum exact hota hai limit lene se pehle.

Step 5 mein

n \to \infty

lene se sum integral kyun ban jaata hai?

Kyunki

\sum F^{'} (x_{i}^{*}) Δ x

(har term width

Δ x

ki strip) exactly ek Riemann sum hai, jiska limit jab

Δ x \to 0

hota hai woh definite integral ko define karta hai.

Total distance ke liye absolute value integral ke andar kyun honi chahiye?

Forward aur backward motion ke beech cancellation integral ke andar moment-by-moment hoti hai. Bars bahar rakhne se cancellation pehle ho jaati hai; andar rakhne se woh forbidden ho jaati hai, isliye har bit positive count hota hai. (Dekho Displacement vs Distance.)

Net change problem mein constant of integration kabhi matter kyun nahi karta?

Kyunki hum

F (b) - F (a)

compute karte hain;

+ C

dono terms mein aata hai aur subtract ho jaata hai. Net change poochta hai "kitna change hua", jo baseline se independent hai.

Hum quantity ki jagah rate ko integrate kyun karte hain?

Kyunki hume usually rate diya hota hai (velocity, marginal cost, flow rate) aur quantity ka change chahiye hota hai. Theorem kehta hai rate integrate karna exactly woh change reconstruct karta hai — integral differentiation ko undo karta hai.

Edge cases

Boundary aur degenerate scenarios jinhein theorem ko phir bhi handle karna hoga.

Zero width ke interval par net change kya hoga,

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

Zero. Koi interval nahi matlab koi accumulated change nahi:

F (a) - F (a) = 0

. Theorem yeh automatically deta hai.

Agar poore interval par rate

F^{'} (x) = 0

ho to?

Net change

0

hai — quantity constant hai. Displacement aur distance dono zero hain kyunki kuch hila hi nahi.

Velocity poore interval par negative hai. Kya displacement aur distance alag honge?

Unke magnitudes equal honge lekin signs alag: displacement negative hai (poora time peeche gaya), distance uska positive magnitude hai,

\int ∣ v ∣ = - \int v

. Magnitudes ki equality isliye hoti hai kyunki

v

ne kabhi sign nahi badla.

Distance ke liye split karte waqt exactly us instant par kya hota hai jab

v = 0

ho (turning point)?

Woh single point zero area contribute karta hai, isliye koi fark nahi padta ki use kis piece mein assign karo. Uska role sirf kahan split karna hai yeh mark karna hai taaki har piece ek sign rakhe.

Kya total distance kabhi

∣

displacement

∣

ke equal ho sakti hai chahe object clearly hila ho?

Haan — jab bhi velocity kabhi sign nahi change karti. Object sirf ek direction mein move karta hai, isliye kuch cancel nahi hota aur distance

= ∣

displacement

∣

Agar

F^{'}

mein jump discontinuity ho (piecewise rate), kya theorem tab bhi apply hota hai?

Haan, jab tak

F^{'}

integrable ho; integral bas pieces ko sum kar leta hai. Finite jump area ko affect nahi karta, isliye

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

tab bhi hold karta hai jab

F

continuous ho.

Rate litres per minute mein diya gaya hai lekin seconds mein measure kiye gaye interval par integrate kar rahe ho. Kya toot jaata hai?

Units toot jaate hain: net change

60 \times

bahut bada ya chota aa jaata hai. Theorem algebraically unit-agnostic hai, lekin rate ka time unit integration variable ke unit se match karna chahiye.

Agar rate

F^{'}

interval ke andar infinity tak blow up ho jaaye (vertical asymptote, e.g.

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

near

0

Ab tumhare paas improper integral hai. Net change tab bhi

F (b) - F (a)

hai bas agar woh limit exist kare —

\int

ko bad point ke paas limit ke roop mein lo. Agar improper integral diverge kare, net change genuinely infinite (ya undefined) hai.

Kya net change finite ho sakta hai chahe rate interval par unbounded ho?

Haan. Ek unbounded rate tab bhi finite signed area enclose kar sakta hai (e.g.

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

finite hai chahe

F^{'} \to \infty

0

). Unbounded rate ka matlab automatically infinite net change nahi hota.

Connections

Net change theorem — parent topic jise ye traps stress-test karti hain; yahan ki har equation wahan rehti hai.
Displacement vs Distance — in aadhe items ke peeche net-vs-total distinction (signed-area figure dekho).
Fundamental Theorem of Calculus — isliye net change naya theorem nahi hai (same $\int F^{'} = F (b) - F (a)$ ).
Mean Value Theorem — boxed derivation ke Step 3 mein exactness.
Definite Integral as a Riemann Sum — isliye Step 5 ka limit ek integral hai (grey rectangles).
Telescoping Sums — Step 2 mein cancellation.
Marginal Cost and Revenue — economics no-constant-needed trap.