4.2.1
Intuition Badi picture (WHY yeh exist karta hai)Differentiation ek function F rate of change F ′ Antiderivative ulta sawaal poochhta hai: "Mujhe har point par slope pata hai — toh original function kya tha?"
Twist yeh hai: sirf slope jaankar aap kabhi bhi height nahi pata kar sakte. Ek curve ko seedha upar ya neeche slide karne se uski slope kahin bhi nahi badlti. Toh jawab ek function nahi balki parallel curves ka poora family hai — isliye == + C ==
Definition AntiderivativeEk function F f antiderivative hai interval I F ′ ( x ) = f ( x ) for all x ∈ I . ∫ f ( x ) d x = F ( x ) + C , == C == arbitrary constant of integration hai.
HAR piece ka MATLAB
∫ ⋯ d x F ( x ) ek particular antiderivative.+ C infinitely many antiderivatives ko package karta hai.
Hum + C assume nahi karte; hum prove karte hain ki yeh hona chahiye aur yahi ek-maatra freedom hai.
Step 1 — Kam se kam do antiderivatives jo ek constant se differ karti hain, dono kaam karti hain.
Maano F ′ ( x ) = f ( x ) C G ( x ) = F ( x ) + C G ′ ( x ) = F ′ ( x ) + d x d ( C ) = f ( x ) + 0 = f ( x ) . Yeh step kyun? Ek constant ka derivative 0 F + C
Step 2 — Koi AUR nahi hai (family complete hai).
Maano F G f antiderivatives hain. H ( x ) = G ( x ) − F ( x ) H ′ ( x ) = G ′ ( x ) − F ′ ( x ) = f ( x ) − f ( x ) = 0. Yeh step kyun? Jis function ka derivative poore interval par zero ho, uski slope har jagah zero hai → woh upar ya neeche ja hi nahi sakta → woh constant hai (yeh Mean Value Theorem ka consequence hai).
⇒ H ( x ) = C ⇒ G ( x ) = F ( x ) + C .
Intuition Geometric meaning (Dual Coding)Saare members ek curve ke vertical translates hain. Kisi bhi fixed x tangent slope f ( x ) C
Worked example Example 1 — power function∫ x 2 d x
Pehle andaaza lagao: x 3 3 x 2 d x d ( 3 x 3 ) = 3 3 x 2 = x 2 . ✓ Yeh step kyun? Main differentiation forward run karta hoon aur constant factor ko adjust karta hoon taaki result x 2 ∫ x 2 d x = 3 x 3 + C . Verify: d x d ( 3 x 3 + C ) = x 2
Worked example Example 2 — ek particular solution (condition use karke)f ( x ) = cos x F F ( 0 ) = 5
General family: ∫ cos x d x = sin x + C d x d sin x = cos x F ( 0 ) = sin 0 + C = 0 + C = 5 ⇒ C = 5 Yeh step kyun? Point ( 0 , 5 ) ek curve select karta hai.
F ( x ) = sin x + 5.
Worked example Example 3 — domain matter karta hai (sneaky case)∫ x 1 d x = ln ∣ x ∣ + C
∣ x ∣ d x d ln ∣ x ∣ = x 1 x = 0 Subtlety: domain { x = 0 } do alag intervals hai — ( − ∞ , 0 ) ( 0 , ∞ ) Poora general answer hai
{ ln ( − x ) + C 1 , ln ( x ) + C 2 , x < 0 x > 0. Yeh step kyun? "Koi aur antiderivative nahi" wala theorem ek connected interval maangta hai; gap hone par har piece ko apna alag constant milta hai.
Common mistake "Maine differentiate kiya aur sahi answer aaya, toh constant matter nahi karta —+ C Kyun sahi lagta hai: C Fix: + C kaunsa curve mean kar rahe ho. Indefinite integral mein isse chhod dena mathematically galat hai (aap claim kar rahe ho ki ek unique answer hai jabki infinitely many hain). Yeh sirf definite integrals ∫ a b C − C = 0
∫ x 2 d x = 3 x 3 ∫ x − 1 d x = 0 x 0 Kyun sahi lagta hai: power rule n + 1 x n + 1 Fix: yeh rule n = − 1 kaam nahi karta (zero se division). Wohi exceptional case hai jahan logarithm aata hai: ∫ x − 1 d x = ln ∣ x ∣ + C
Common mistake "Zero derivative ka matlab constant hai, chahe domain mein hole kyun na ho."Kyun sahi lagta hai: "slope 0 Fix: yeh sirf ek single interval par sach hai. Disconnected domain par har piece alag height par baith sakta hai (Example 3).
Recall Feynman: 12-saal ke bacche ko samjhaoSocho tumhe har kadam par pahaad ki steepness pata hai, lekin height nahi pata ki tumne kahan se shuru kiya. Tum pahaad ki shape bilkul sahi draw kar sakte ho — lekin shayad tumne 1 metre upar se shuru kiya ho, ya 100 metre upar se, aur shape bilkul ek jaisi dikhegi. Woh unknown starting height hi + C ek jagah ki height batani padegi.
+ C "Slopes Shy hoti hain Starting heights ke baare mein." Differentiation start chhupa deta hai; integration isse + C Constant → derivative mar jaata hai → toh kuch bhi ho sakta tha → + C
f Ek function F F ′ ( x ) = f ( x )
Indefinite integral mein + C Kyunki d x d ( C ) = 0 F ( x ) + C
Prove karo ki ek hi f H = G − F H ′ = G ′ − F ′ = f − f = 0 H =
∫ x 2 d x = ? 3 x 3 + C
∫ cos x d x = ? sin x + C
Kis n ∫ x n d x = n + 1 x n + 1 + C n = − 1 ∫ x − 1 d x = ln ∣ x ∣ + C
+ C Definite integral ∫ a b [ F + C ] a b = F ( b ) − F ( a )
∫ 1/ x d x ln ∣ x ∣ Kyunki d x d ln ∣ x ∣ = 1/ x x = 0
Family mein se ek member kaunsi extra information select karti hai? Ek initial/boundary condition jaise F ( x 0 ) = y 0
Geometrically, family ke members aapas mein kaise related hain? Vertical translates jinka har x f ( x )
Mean Value Theorem — "zero derivative ⇒ constant" justify karta hai, jo + C Fundamental Theorem of Calculus — antiderivatives ko definite integrals se jodta hai.Power Rule (Integration) — n = − 1 Differential Equations — Initial Value Problems — C Logarithmic & Exponential Integrals — ln ∣ x ∣ Definite Integral — jahan + C
Differentiation gives slope F prime
F prime x equals f x on I
Two antiderivatives differ H equals G minus F
Parallel vertical translates
Initial condition e.g. F 0 equals 5
