Visual walkthrough — Riemann sums — left, right, midpoint; formal definition of definite integral

4.2.3 · D2 · Maths › Calculus II — Integration › Riemann sums — left, right, midpoint; formal definition of d

Is note ke end tak tum dekh chuke hoge ki ${\int }_0^1{x}^{2}dx$ kaise $\frac{1}{3}$ ban jaata hai.

Step 1 — Ek curve, ek region, uske liye koi formula nahi

KYA. Hamare paas ek curve hai. Horizontal position $x$ par height ko naam dete hain $f(x)$ — padho ise "$f$ of $x$", ek machine jo ek number $x$ khaati hai aur wahan ki height wapas deti hai. Hum woh area chahte hain jo is curve aur flat ground (horizontal axis) ke beech trapped hai, left wall $x=a$ se lekar right wall $x=b$ tak.

KYUN. Ek rectangle ka area width $\times$ height hota hai — bilkul aasaan. Curvy top ka aisa koi formula nahi hota. Yahi ek obstacle integration ke existence ki wajah hai. Neeche sab kuch usi se bachne ki ek lambi trick hai.

PICTURE. Neeche shaded region wahi hai jiska area hum chahte hain. Dhyan do ki uska top edge curved hai — yahi poora problem hai.

Step 2 — Region ko vertical strips mein kato

KYA. Region ko vertical cuts se slice karo. Hum ground ke saath cut-points rakhte hain aur unhe left se right naam dete hain: $a=x_0<x_1<x_2<\cdots <x_n=b.$ Yahan $x_0$ sirf left wall $a$ ka doosra naam hai, $x_n$ right wall $b$ ka doosra naam hai, aur beech waale interior cuts hain. Cut-points ki yeh list partition kehlati hai. Har slice do neighbours ke beech hoti hai: strip number $i$ $[x_{i-1},x_i]$ par rehti hai.

KYUN. Ek patli strip almost ek rectangle hoti hai — uski top barely curve karti hai. Toh agar hum har patli strip ko ek rectangle se approximate kar sakein, hum ek impossible area ko bahut saare aasaan areas se trade kar lete hain.

PICTURE. Neeche, $n=4$ strips hain. Figure par seedha term-by-term meaning padho: strip $i$ ki width hai $\Delta x_i=x_i-x_{i-1}$ (symbol "$\Delta$" ka matlab sirf "mein change" hota hai, toh $\Delta x_i$ = strip $i$ ke across $x$ kitna move karta hai).

Step 3 — Har strip ko ek rectangle mein badlo (height chuno)

KYA. Ek strip ka top curved hota hai, isliye uski koi single "height" nahi hoti. Hum choose karte hain: strip $i$ ke andar kahin ek sample point $x_i^{\ast }$ lete hain, wahan curve measure karte hain, aur $f(x_i^{\ast })$ ko poori strip ke liye flat height ki tarah use karte hain.

KYUN. Jis moment hum ek flat height decide kar lete hain, strip ek genuine rectangle ban jaati hai: area $=f(x_i^{\ast })\cdot \Delta x_i$. Yeh pehla area hai jise hum actually compute kar sakte hain.

PICTURE. Red dot dekho: hum sample point par curve ki height padhte hain, phir us height ko strip ke across flat extend karte hain, ek rectangle create karte hain. Top-right corner curve ke upar ek taraf pooch jaata hai aur doosri taraf neeche gap rehta hai — yeh mismatch hamaari error hai, aur yeh strips ke patli hone par shrink hoti hai.

Step 4 — Height choose karne ke teen honest tarike

KYA. Strip ke andar sample point $x_i^{\ast }$ kahaan hona chahiye? Teen natural answers, har ek alag-named sum deta hai:

KYUN. Koi single "correct" spot nahi hai — yahi freedom ka point hai. Hum prove karna chahte hain ki final answer is arbitrary choice par depend nahi karta. Isliye pehle humein choices ko alag dekhna hoga, phir unhe limit mein agree karte dekhna hoga.

PICTURE. Wohi badhta hua curve, teen tarike se sampled. Ek aise curve ke liye jo increase karta hai, left rule ek too-low height padhta hai (underestimate, upar red gap), right rule ek too-high height padhta hai (overestimate, red overshoot), aur midpoint beech mein baith jaata hai — centre ke ek taraf overshoot roughly doosri taraf ke gap ko cancel kar deta hai.

Step 5 — Saare rectangles ko add karo (the Riemann sum)

KYA. Hamare paas ab $n$ rectangles hain. Unke areas add karo. Woh total Riemann sum hai: $S_n={\sum }_{i=1}^{n}f(x_i^{\ast })\Delta x_i.$

KYUN. Poore region ka area (approximately) strips ke areas ka sum hai — yeh sirf "pieces ka area poore ka area add karta hai" wali baat hai. Sum symbol $\sum$ "inhe add karo jab $i$ $1,2,\dots ,n$ run kare" ka bookkeeping hai.

PICTURE. Symbols ko wahan decode karo jahan woh rahte hain: $\underset{\text{running total}}{\underbrace{S_n}}=\underset{\begin{array}{c}\text{add over}\\ \text{all strips}\end{array}}{\underbrace{\sum _{i=1}^n}}\underset{\text{height of strip }i}{\underbrace{f(x_i^{\ast })}}\underset{\text{width of strip }i}{\underbrace{\Delta x_i}}.$ Neeche, saare chaar rectangles draw kiye gaye hain aur unke areas literally ek bar mein stack kiye gaye hain — us bar ki height $S_4$ hai, true area ka hamaara pehla estimate.

Step 6 — Aur patla, aur patla, aur patla…

KYA. $n$ badhao (zyada strips), jisse har width $\Delta x$ shrink ho. Dekho rectangle-tops ka staircase curve ko tighter aur tighter hug karta jaata hai.

KYUN. Har rectangle ki error woh choti sliver hai uski flat top aur curve ke beech. Patli strips $\Rightarrow$ chhoti slivers $\Rightarrow$ chhoti total error. Hum approximation ko truth ki taraf squeeze kar rahe hain.

PICTURE. Wohi region aur curve, $n=2$, phir $n=4$, phir $n=8$ right sums ke saath draw ki gayi. Red "error slivers" visibly shrink hoti hain. Staircase curve se indistinguishable ho jaati hai.

Step 7 — Limit lo: definite integral paida hota hai

KYA. Jaise-jaise $\Vert P\Vert \to 0$, sums ek number $I$ par close in karte hain. Agar wahi number $I$ chahe hum sample points kaisi bhi chune (left, right, midpoint, kuch bhi), reach hota hai, toh hum $f$ ko integrable kehte hain aur us number ko definite integral naam dete hain: ${\int }_a^{b}f(x)dx={\lim }_{\Vert P\Vert \to 0}{\sum }_{i=1}^{n}f(x_i^{\ast })\Delta x_i.$

KYUN "every choice"? Agar left, right, aur midpoint teeno same limit ki taraf march karein, toh answer hamare arbitrary sampling ka artifact nahi ho sakta — yeh $f$ ki ek real property hai. (Ek well-behaved curve ke liye — dekho Continuity and Integrability — yeh hamesha kaam karta hai; pathological ones ke liye fail ho sakta hai.)

PICTURE. Reborn notation ko, figure par symbol by symbol decode karo: $\underset{\text{a long "S" = Sum}}{\underbrace{\int }}\underset{\text{height}}{\underbrace{f(x)}}\underset{\begin{array}{c}\text{infinitely thin}\\ \text{width, }\lim \Delta x\end{array}}{\underbrace{dx}}$ Red staircase aur black curve merge ho gayi hain — limit hi area hai.

Step 8 — Degenerate aur edge cases jo tum skip nahi kar sakte

Har woh scenario jo reader ko mil sakta hai — dikhaya gaya, assume nahi kiya gaya.

Case A · Empty interval, $a=b$. Fill karne ki koi width nahi: $\Delta x=\frac{b-a}{n}=0$, har rectangle ek line hai, total area $=0$. Toh ${\int }_a^{a}f(x)dx=0$. Samajh aata hai — koi region hi nahi.

Case B · Curve ground ke neeche, $f(x)<0$. "Height" negative hai, toh $f(x_i^{\ast })\Delta x$ negative hai — rectangle negative area count karta hai. Integral Signed area return karta hai: above-ground positive, below-ground negative, jahan overlap hote hain wahan cancel karte hain.

Case C · Ek straight top, $f$ linear. Yahan midpoint rule kisi bhi $n$ ke liye exact hota hai — koi limit ki zaroorat nahi. Woh choti triangle jo sample height ke ek taraf curve ke upar pooch jaati hai, doosri taraf ke gap ke congruent hoti hai, toh woh perfectly cancel ho jaati hain. (Left aur right abhi bhi sirf approximate hain.)

Case D · Ek "wild" function. Agar alag-alag sample choices alag-alag limits dein, toh koi single $I$ exist nahi karta aur $f$ not Riemann integrable hai. Classic villain Dirichlet function hai (rationals par 1, irrationals par 0): rational samples lo aur har sum $b-a$ hai; irrational samples lo aur har sum $0$ hai. Do limits $\Rightarrow$ koi integral nahi.

PICTURE. Teen panels: (A) collapsed zero-width case, (B) ek curve ground cross karti hai jisme ek red rectangle neeche negative counted, (C) ek straight line jisme midpoint rectangle par do cancelling red triangles hain.

Ek picture mein poora summary

Left se right, poori derivation ek single strip mein: curve → strips mein kato → har ek ko ek rectangle mein flatten karo → sum karo → widths shrink karo → limit = exact area = integral.

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