Derivation. Andaza lagao ki antiderivative xn+1 jaisa dikhta hai. Test karo:
dxdxn+1=(n+1)xn.
Ye (n+1) times zyada bada hai. Ye step kyun? Differentiation ne purane power+1 se multiply kar diya, toh
hum ise (n+1) se divide karke cancel karte hain:
dxd[n+1xn+1]=n+1(n+1)xn=xn.✓
∫cosxdx ki derivation. Hum chahte hain F jisme F′=cosx ho. Kyunki dxdsinx=cosx,
F=sinx lo. Ho gaya.
∫sinxdx ki derivation. Hum chahte hain F′=sinx. Hum jaante hain dxdcosx=−sinx,
jo hamari zaroorat ka negative hai. Ye step kyun?−1 se multiply karo:
dxd(−cosx)=sinx. Toh ∫sinxdx=−cosx+C.
Differentiate karna aisa hai jaise poochho "ye kitni tezi se badal raha hai?" Integrate karna detective
wali baat hai: main tumhe har pal ki speed batata hoon aur tum reconstruct karte ho ki gaadi kahan thi. Lekin ek
dikkat hai — tumhe gaadi ka starting point nahi pata, toh har answer "+ koi anjaan starting point" ke saath khatam hota hai
(ye hi +C hai). Integrate karne ke liye bas poochho "mujhe kya differentiate karna padta ye banane ke liye?"
aur likh do — phir divide karo ya sign flip karo taaki bilkul sahi aaye.
+C kya represent karta hai aur kyun aana ZAROORI hai?
Kho gaya constant — differentiation kisi bhi constant ko 0 kar deta hai, toh antiderivative use recover nahi kar sakta; +C saari possibilities ko encode karta hai.
Power rule aur uska ek exception batao.
∫xndx=n+1xn+1+C jab n=−1; n=−1 par zero se divide ho jaata.
∫x1dx=? aur ∣x∣ kyun?
ln∣x∣+C; absolute value x<0 ke liye bhi validity badhata hai kyunki dxdln∣x∣=x1 har jagah 0 ke alawa.
∫sinxdx=?
−cosx+C (minus sign dhyan se).
∫cosxdx=?
sinx+C.
∫sec2xdx=?
tanx+C.
∫exdx=? aur kyun unchanged?
ex+C; ex differentiation ka fixed point hai.
∫axdx=?
lnaax+C, dxdax se extra lna divide karke nikala.
Kya integration mein product rule hota hai?
Nahi — products ke liye substitution ya integration by parts chahiye.
Integral verify kaise karte hain hamesha?
Apna answer differentiate karo; woh integrand wapas dena chahiye.