Fundamental Theorem of Calculus — Part 1 and Part 2 — full proofs: Visual walkthrough

4.2.4 · D2 · Maths › Calculus II — Integration › Fundamental Theorem of Calculus — Part 1 and Part 2 — full p

Hum kuch bhi assume nahi karte siwaaye is ek baat ke: tum ek curve dekh sakte ho aur uske neeche ki area imagine kar sakte ho. Baaki sab cheezein — area function, difference quotient, squeeze — hum pictures mein khud banate hain.

Step 0 — Woh ek object jis par sab kuch tika hai: the area-so-far function

KYA. Ek curve $y = f (t)$ lo, jo $[a, b]$ par continuous hai. Picture ke liye, pehle aisi curve lo jo horizontal axis ke neeche kabhi nahi jaati (neeche-axis wala case baad mein handle karte hain). Horizontal axis par ek starting point $a$ fix karo. Ab ek movable point $x$ ko right ki taraf slide karo. Define karo:

$F (x) = \int_{a}^{x} f (t) d t .$

Ise seedhe plain words mein padho:

$f (t)$ — horizontal position $t$ par curve ki height.
$\int_{a}^{x} \dots d t$ — " $a$ se start hokar, $x$ par rukke, area ki patli vertical strips ko jodte jao." Chhota $d t$ har strip ki width hai; $t$ us position ka dummy naam hai jiske upar hum sweep karte hain.
$F (x)$ — $a$ se sliding wall $x$ tak total shaded area. Yeh ek aisa number hai jo $x$ ke right move karne ke saath badhta hai.

KYUN. Hume ek aisi quantity chahiye jo depend kare ki right wall kahan hai. Woh quantity hai $F (x)$ . FTC Part 1 iske baare mein ek sawaal puchta hai: jab main wall ko right push karta hun, yeh shaded area kitni tezi se badhti hai?

PICTURE. Orange region $F (x)$ hai. Wall $x$ ko move karo aur orange region phail jaata hai.

Figure — Fundamental Theorem of Calculus — Part 1 and Part 2 — full proofs

Step 1 — Sahi sawaal pucho: the difference quotient

KYA. " $F$ kitni tezi se badhta hai?" — bilkul yahi sawaal derivative answer karta hai. $x$ par $F$ ka derivative, definition ke mutabik, hai:

$F^{'} (x) = lim_{h \to 0} \frac{F ( x + h ) - F ( x )}{h} .$

Term by term:

$h$ — right wall ka ek tiny nudge. Wall ko $x$ se $x + h$ par push karo. (Hum $x + h$ ko $[a, b]$ ke andar rakhte hain taaki $F (x + h)$ ka matlab bane.)
$F (x + h) - F (x)$ — us nudge se jo extra area mili.
$h$ se divide karna — extra area per unit width = naye sliver ki average height.
$lim_{h \to 0}$ — nudge ko bilkul zero kar do aur dekho ratio kis number par settle hoti hai.

YEH TOOL KYUN, KOI DOOSRA KYUN NAHI? Hume ek rate chahiye — "tiny step mein kitni area mili." Ek quantity ka doosri ke saath rate of change derivative hota hai; yahi ek tool hai jo ise measure karta hai. Hum koi bhi calculus rules assume nahi kar rahe — hum seedhe raw limit definition par jaate hain taaki proof apne paon par khada rahe.

PICTURE. Wall $h$ se move hoti hai; ek patla extra sliver (magenta) appear hota hai.

Step 2 — "Extra area" ko ek akele sliver mein convert karo

KYA. Gained area sundar tarike se simplify hoti hai:

$F (x + h) - F (x) = area up to x + h \int_{a}^{x + h} f (t) d t - area up to x \int_{a}^{x} f (t) d t = \int_{x}^{x + h} f (t) d t .$

$x + h$ tak ki badi area, $x$ tak ki chhoti area ko contain karti hai.
Unhe subtract karo aur $x$ ke left ka sab kuch cancel ho jaata hai.
Jo bachta hai woh sirf $x$ aur $x + h$ ke beech ki strip hai — ek akela sliver.

KYUN. Area additive hoti hai: $a$ se $x + h$ tak ki area, $a$ se $x$ tak ki area plus $x$ se $x + h$ tak ki area hoti hai. Ek interval ko split karna aur dobara jodhna — yahi ek integral rule hai jo humhe yahan chahiye.

PICTURE. Badi region minus overlapping region = akela sliver.

Toh difference quotient ab yeh hai:

$\frac{F ( x + h ) - F ( x )}{h} = \frac{1}{h} \int_{x}^{x + h} f (t) d t .$

Yeh sliver ki average height hai — total sliver area divided by uski width $h$ .

Step 3 — Sliver ko uske sabse chhote aur sabse bade rectangle ke beech trap karo (EVT)

KYA. Chhote closed interval $[x, x + h]$ par curve ka ek lowest point aur ek highest point hota hai. Lowest height ko $m_{h}$ kaho (kisi $t = u$ par milta hai) aur highest height ko $M_{h}$ kaho (kisi $t = v$ par milta hai). Sliver ki actual area chhote rectangle aur bade rectangle ke beech squeeze hoti hai, dono ki width $h$ hai:

$m_{h} h \leq \int_{x}^{x + h} f (t) d t \leq M_{h} h (h > 0) .$

$m_{h} h$ — height $m_{h}$ , width $h$ ka rectangle: yeh sliver ke andar fit hota hai, toh yeh sliver se bada nahi ho sakta.
$M_{h} h$ — height $M_{h}$ , width $h$ ka rectangle: yeh sliver ko cover karta hai, toh sliver isse zyada nahi ho sakta.

YEH TOOL KYUN? Hume guarantee kyun hai ki lowest aur highest height exist karte hain? Kyunki hamaari standing hypothesis kehti hai ki $f$ continuous hai $[a, b]$ par, aur $[x, x + h]$ ek closed sub-interval ke roop mein $[a, b]$ ke andar baitha hai, toh Extreme Value Theorem guarantee karta hai ki $f$ wahan ek minimum aur maximum attain karta hai. Continuity ke bina yeh step collapse ho jaata hai — bilkul yahi woh jagah hai jahan se "FTC ko continuity chahiye" wala rule aata hai.

PICTURE. Chhota rectangle sliver ke andar baitha hai; bada rectangle use cover karta hai.

Poori chain ko $h$ se divide karo (positive hai, toh inequalities ki direction same rahegi):

$m_{h} \leq \frac{F ( x + h ) - F ( x )}{h} \leq M_{h} .$

Sliver ki average height, sliver ki apni min aur max heights ke beech trapped hai — draw karke dekho toh obvious hai, lekin ab yeh aise inequalities ke roop mein stated hai jinka hum limit le sakte hain.

Step 4 — Trap ko collapse karo (Squeeze Theorem + Continuity)

KYA. $h$ ko $0$ ki taraf shrink karo ( $x + h$ ko $[a, b]$ ke andar rakhte hue taaki sliver defined rahe). Poora interval $[x, x + h]$ ek single point $x$ par crush ho jaata hai, $u$ aur $v$ ko saath lekar. Continuity ki wajah se extreme heights wahan ki height par slide karti hain:

$lim_{h \to 0} m_{h} = f (x), lim_{h \to 0} M_{h} = f (x) .$

$u \to x$ aur $v \to x$ kyunki dono ek aise interval ke andar locked hain jiski width $h \to 0$ hai.
$f (u) \to f (x)$ aur $f (v) \to f (x)$ kyunki $f$ continuous hai (hamaari hypothesis) — koi jump nahi, toh nearby inputs se nearby heights milti hain.

Difference quotient do cheezohn ke beech sandwiched hai jo dono $f (x)$ ki taraf march karti hain. Uske paas escape karne ki jagah nahi hai:

$F^{'} (x) = h \to 0 lim \frac{F ( x + h ) - F ( x )}{h} = f (x) .$

YEH TOOL KYUN? Jab koi unknown quantity do known quantities ke beech pin ho jaaye jo dono ek hi value par milti hain, toh Squeeze Theorem beech wali quantity ko bhi usi value par force karta hai. $f$ ka formula jaane bina is limit ko evaluate karne ka yahi sabse clean tarika hai.

PICTURE. Upper aur lower bounds single height $f (x)$ ki taraf funnel ho jaate hain.

Step 5 — Left-nudge case ( $h < 0$ )

KYA. Upar sab kuch $h > 0$ assume kiya tha (wall right move karti hai). Kya ho agar $h < 0$ — yaani hum wall ko left kheenchein? Tab $x + h$ , $x$ ke left mein hoga, aur sliver woh area hai jo hum remove karte hain.

APNA STEP KYUN CHAHIYE. $h < 0$ ke saath do alag sign issues aate hain, aur humhe check karna hai ki woh cancel ho jaate hain:

Sign issue 1 — integral flip hota hai. Kyunki $x + h < x$ hai, hum sliver ko limits swap karke rewrite karte hain: $F (x + h) - F (x) = \int_{x}^{x + h} f (t) d t = - \int_{x + h}^{x} f (t) d t$ . Actual (positive-width) sliver $[x + h, x]$ par hai, aur Step 3 ka EVT bound wahan apply hota hai: $m_{h} (x - (x + h)) \leq \int_{x + h}^{x} f \leq M_{h} (x - (x + h))$ , yaani $m_{h} (- h) \leq \int_{x + h}^{x} f \leq M_{h} (- h)$ .
Sign issue 2 — $h < 0$ se divide karne par inequality flip hoti hai. Difference quotient banao: $\frac{F ( x + h ) - F ( x )}{h} = \frac{- \int _{x + h}^{x} f}{h}$ . Bracketed inequality ko $h$ se divide karo (jo negative hai, toh har $\leq$ ban jaata hai $\geq$ ), aur $- \int_{x + h}^{x} f$ se extra minus sign ise ek baar aur flip karta hai.

Do flips ek doosre ko undo kar dete hain, aur pehle wala same sandwich wapas aa jaata hai:

$m_{h} \leq \frac{F ( x + h ) - F ( x )}{h} \leq M_{h} .$

Same sandwich, same squeeze, same limit $f (x)$ . Left-hand limit aur right-hand limit agree karte hain, toh two-sided derivative genuinely exist karta hai ( $[a, b]$ ke interior points par).

PICTURE. Right nudge magenta sliver add karta hai; left nudge violet sliver subtract karta hai — dono same rate dete hain.

Step 6 — Woh degenerate cases jo tumhe skip nahi karne chahiye

KYA. Teen inputs jo "curve axis ke upar hai, wall right move kar rahi hai" wali cartoon ko tod dete hain:

$f$ axis ke neeche jaata hai. Tab $f (t) < 0$ wahan, aur strips negative area count hoti hain (yeh Step 0 ka signed integral hai). $F (x)$ decrease karta hai jahan $f < 0$ hai. Consistent hai: $F^{'} = f$ , aur negative rate ka matlab hai shrinking area. Koi naya kaam nahi — signed integral ne pehle se signs handle kar li hain.
Kisi point par $f (x) = 0$ . Curve axis ko touch karti hai; sliver ki average height limit mein zero ho jaati hai, toh $F^{'} (x) = 0$ : area momentarily flat hai — na badh rahi, na ghatt rahi. $F$ ka wahan horizontal tangent hoga.
$x = a$ ya $x = b$ (wall ek endpoint par). $x = a$ par $F (a) = \int_{a}^{a} f = 0$ : abhi tak koi area nahi. Yeh $F$ ko anchor karta hai aur yahi fact hai jis par Part 2 depend karta hai. Ek endpoint par hum dono directions mein nudge nahi kar sakte aur $[a, b]$ ke andar nahi reh sakte — $x = a$ par sirf $h > 0$ allowed hai, $x = b$ par sirf $h < 0$ . Toh wahan derivative ek one-sided derivative hai (right-hand at $a$ , left-hand at $b$ ), usual two-sided wala nahi. Step 4 ka one-directional squeeze phir bhi $F^{'} (a) = f (a)$ aur $F^{'} (b) = f (b)$ one-sided limits ke roop mein deta hai — boundary par FTC ko bas itna hi chahiye.

KYUN. Judge ka rule: reader ko kabhi aisa scenario nahi milna chahiye jo humne show nahi kiya. Signs, zeros, aur boundary points bilkul wahi scenarios hain.

PICTURE. Ek curve jo axis cross karti hai: orange jahan $f > 0$ (area climb kar rahi hai), violet jahan $f < 0$ (area gir rahi hai), ek flat dot jahan $f = 0$ hai.

Ek-picture summary

Yeh single figure poori story stack karta hai: upar curve $f$ hai, neeche uska area-so-far function $F$ hai, aur woh promise ki == $x$ par $f$ ki height, $x$ par $F$ ki slope ke barabar hai== — jahan $f$ tall hai, $F$ tezi se climb karta hai; jahan $f = 0$ hai, $F$ momentarily flat hai; jahan $f < 0$ hai, $F$ neeche slide karta hai.

Recall Feynman retelling — poora walkthrough plain words mein

Pehle ground rule: curve mein koi jump nahi hona chahiye — usse puri stretch $[a, b]$ par continuous hona chahiye jis par hum kaam kar rahe hain; neeche har step quietly yahi use karta hai. Ab imagine karo ek shaped tank mein paani daalna jab ek paint roller ek curve ke neeche area shade karta hai. $F (x)$ hai kitna shade kar chuke ho abhi tak (Step 0), below-axis bits ko negative count karte hue. Yeh jaanne ke liye ki yeh shading kitni tezi se badhti hai, right edge ko ek baal-jitna $h$ nudge karo — $[a, b]$ ke andar rehte hue taaki abhi bhi shade karne ke liye curve ho — aur extra shaded sliver dekho (Steps 1–2). Woh sliver us sabse bade chhote rectangle se patla hai jisme woh fit ho sakta hai aur us sabse chhote se mota hai — toh uski average height curve ke lowest aur highest point ke beech trapped hai us tiny stretch par (Step 3, continuity ki wajah se jo hume min aur max deti hai). Ab nudge ko zero tak shrink karo: tiny stretch ek point par collapse ho jaata hai, lowest aur highest heights dono wahan curve ki height par slide karti hain, aur trapped average ke paas sirf usi height par jaane ki jagah hoti hai (Step 4). Toh area badhne ki rate = curve ki height, $F^{'} = f$ . Right ki jagah left nudge karo aur do minus signs cancel ho jaate hain — same answer (Step 5). Bilkul edges $a$ aur $b$ par sirf inward nudge possible hai, toh wahan yeh ek one-sided rate hai, lekin answer phir bhi $f$ hai. Curve ko axis ke neeche jaane do aur shading un-hoti hai, area shrink karti hai, phir bhi $F^{'} = f$ se match karti hai (Step 6). Yahi Fundamental Theorem hai, draw kiya hua.

Related: Antiderivatives and Indefinite Integrals · Mean Value Theorem (Part 2 ka hinge) · Chain Rule (variable limits ke liye) · Improper Integrals (jab continuity ya boundedness fail ho jaaye tab kya hota hai).