4.2.5
Intuition Ek hi breath mein badi baatAgar tumhe pata ho ki koi cheez har instant par kitni fast change ho rahi hai, toh us poore interval mein un saari choti changes ko jodne se tumhe us cheez mein hua kul (net) change milta hai. Kisi rate ka integral hi net change hota hai.
Definition Net Change TheoremAgar F ′ ( x ) F ( x ) x = a x = b F net change uski rate ke integral ke barabar hota hai:
∫ a b F ′ ( x ) d x = F ( b ) − F ( a ) net word matter karta hai: increases aur decreases cancel ho jaate hain. Yeh total signed change hai, na ki total distance travelled.
Yeh literally Fundamental Theorem of Calculus, Part 2 hi hai, bas ise ek physical story ke saath re-read kiya gaya hai. Same equation, naye kapde.
Hum ise integral ki definition se banate hain — kuch yaad karne ki zaroorat nahi.
Step 1 — Interval ko slice karo. Kyun? Kyunki ek lambe interval pe change mushkil lagta hai, lekin ek tiny step pe change aasaan hota hai. [ a , b ] n a = x 0 < x 1 < ⋯ < x n = b , Δ x = n b − a .
Step 2 — Net change ek telescoping sum hai. a b F ( b ) − F ( a ) = ∑ i = 1 n [ F ( x i ) − F ( x i − 1 ) ] . ( F ( x 1 ) − F ( x 0 )) + ( F ( x 2 ) − F ( x 1 )) + … F ( b ) − F ( a )
Step 3 — Har choti change ko rate × time se approximate karo. Kyun? Mean Value Theorem guarantee karta hai ki har slice mein ek point x i ∗ F ( x i ) − F ( x i − 1 ) = F ′ ( x i ∗ ) Δ x .
Step 4 — Ek Riemann sum pehchano. Step 3 ko Step 2 mein substitute karo:
F ( b ) − F ( a ) = ∑ i = 1 n F ′ ( x i ∗ ) Δ x .
Step 5 — Limit lo. Jaise-jaise n → ∞ hi integral ki definition ban jaati hai:
F ( b ) − F ( a ) = lim n → ∞ ∑ i = 1 n F ′ ( x i ∗ ) Δ x = ∫ a b F ′ ( x ) d x . ■
Quantity F rate F ′ Rate ko interval pe integrate karo.
Answer = F ( end ) − F ( start )
Agar tumhe total distance / total amount moved chahiye, toh ∣ F ′ ( x ) ∣
Intuition Net vs. Total (woh ek distinction jisme log marks kho dete hain)
∫ a b v ( t ) d t = displacement (position ka net change — negative bhi ho sakta hai).∫ a b ∣ v ( t ) ∣ d t = total distance (motion ka har bit positive count hota hai).
Worked example 1 — Velocity → displacement vs distanceEk particle velocity v ( t ) = t 2 − 4 t ∈ [ 0 , 3 ]
Displacement = ∫ 0 3 ( t 2 − 4 ) d t v integrate kyun karein? Kyunki v = s ′ ∫ v = s ( 3 ) − s ( 0 ) = [ 3 t 3 − 4 t ] 0 3 = ( 9 − 12 ) − 0 = − 3 m . 3 m peeche start se khatam hua.
Total distance. Split kyun karein? Kyunki [ 0 , 2 ) v < 0 ( 2 , 3 ] v > 0 v = 0 t = 2 ∫ 0 2 ( 4 − t 2 ) d t + ∫ 2 3 ( t 2 − 4 ) d t . = [ 4 t − 3 t 3 ] 0 2 = 8 − 3 8 = 3 16 = [ 3 t 3 − 4 t ] 2 3 = ( − 3 ) − ( 3 8 − 8 ) = ( − 3 ) − ( − 3 16 ) = 3 7 = 3 16 + 3 7 = 3 23 ≈ 7.67 (− 3 ∣
Worked example 2 — Tank mein paani beh raha haiPaani r ( t ) = 20 − 2 t t ∈ [ 0 , 5 ]
r integrate kyun karein? r V ∫ 0 5 r d t = V ( 5 ) − V ( 0 ) ∫ 0 5 ( 20 − 2 t ) d t = [ 20 t − t 2 ] 0 5 = ( 100 − 25 ) − 0 = 75 L . Yeh step (limits 0 aur 5) kyun? Yeh question ke start/end times hain — theorem bilkul unhi do instants ke beech change fill karta hai.
Worked example 3 — Marginal cost → cost increaseMarginal cost C ′ ( x ) = 3 x 2 + 50 x = 10 x = 20
Kyun? Marginal cost literally C ′ C ( 20 ) − C ( 10 ) ∫ 10 20 ( 3 x 2 + 50 ) d x = [ x 3 + 50 x ] 10 20 = ( 8000 + 1000 ) − ( 1000 + 500 ) = 9000 − 1500 = 7500. = ₹7500 . C ( 0 )
Recall Compute karne se pehle: ek particle kav ( t ) = cos t [ 0 , 2 π ] Forecast: Yeh aage jaata hai phir wapas aata hai, ghar wapas return karta hai → displacement ≈ 0 = 4 Verify: Displacement = ∫ 0 2 π cos t d t = [ sin t ] 0 2 π = 0 = ∫ 0 2 π ∣ cos t ∣ d t = 4 ∫ 0 π /2 cos t d t = 4 [ sin t ] 0 π /2 = 4
Common mistake "Displacement aur distance same hote hain."Kyun sahi lagta hai: jab koi object sirf aage move karta hai, toh woh hote hain equal, toh easy problems mein tumhe kabhi fark nahi dikhta.
Fix: Jab bhi rate sign change kare tab yeh differ karte hain. Net change ∫ v ∫ ∣ v ∣ v = 0
Common mistake "Mujhe constant of integration / initial value chahiye."Kyun sahi lagta hai: antiderivative nikalte waqt aam taur pe + C C Fix: Net change mein F ( b ) F ( a ) + C Change nikalte waqt tumhe initial amount ki kabhi zaroorat nahi.
v Kyun sahi lagta hai: distance positive hota hai, toh answer pe ∣ ⋅ ∣ Fix: ∫ v = ∫ ∣ v ∣ andar daalni hoti hai, sign changes pe split karke, integrate karne se pehle .
Recall Feynman: ek 12-year-old ko samjhaoEk car mein speedometer imagine karo. Yeh tumhe kabhi nahi batata ki tum kahan ho — sirf yeh batata hai ki abhi kitni fast ja rahe ho. Lekin agar tum apni speed har second likhte jao aur poore trip ke liye "speed × ek second" jodto jao, toh pata lagta hai car ne exactly kitna move kiya. Net change theorem kehta hai: "kitna fast × thoda time" ke saare chote pieces jodo aur tumhe position mein total change milega. Aur agar tum kabhi peeche drive karo, toh woh negative count hote hain — jab tak tumhe sirf yeh nahi chahiye ki kitni road cover ki, tab backward driving bhi positive count hogi.
"Rate in, change out." Aur: N et ko N o constant chahiye (+ C "bars before integral" — ∫ ∣ f ∣ ∣ ∫ f ∣
Fundamental Theorem of Calculus — net change theorem hi FTC Part 2 hai physical disguise mein.Definite Integral as a Riemann Sum — woh limit jo humne Step 5 mein use ki.Mean Value Theorem — Step 3 justify kiya (small change = slope × width).Displacement vs Distance — net-vs-total distinction.Telescoping Sums — Step 2 mein cancellation trick.Marginal Cost and Revenue — economics application.
Net change theorem state karo. ∫ a b F ′ ( x ) d x = F ( b ) − F ( a )
Net change theorem secretly kis theorem ke identical hai? Fundamental Theorem of Calculus, Part 2 — physically reinterpret kiya gaya.
Derivation ke Step 2 mein kaunsi trick choti changes ke sum ko collapse karti hai? Telescoping sum: ∑ [ F ( x i ) − F ( x i − 1 )] = F ( b ) − F ( a )
Derivation mein Mean Value Theorem kya provide karta hai? Ek point x i ∗ F ( x i ) − F ( x i − 1 ) = F ′ ( x i ∗ ) Δ x
Velocity se displacement kaise nikalte hain? ∫ a b v ( t ) d t
Velocity se total distance kaise nikalte hain? ∫ a b ∣ v ( t ) ∣ d t
Net change problem mein constant of integration ki zaroorat kyun nahi? Kyunki F ( b ) − F ( a ) + C
Total distance ke liye ∣ ∫ v ∣ Kyunki integral ke andar cancellation hoti hai; bars andar jaane chahiye: ∫ ∣ v ∣
v(t)=t²−4 on [0,3]: displacement? ∫ 0 3 ( t 2 − 4 ) d t = − 3
v(t)=cos t on [0,2π]: displacement aur distance? Displacement 0; distance 4.
