Fundamental Theorem of Calculus — Part 1 and Part 2 — full proofs: Worked examples

4.2.4 · D3 · Maths › Calculus II — Integration › Fundamental Theorem of Calculus — Part 1 and Part 2 — full p

Shuru karne se pehle, do words jo hum baar baar use karenge:

Integrand — woh function jo integral sign ke andar hota hai (woh cheez jo add ho rahi hai). $\int_{a}^{b} f (x) d x$ mein integrand $f (x)$ hai.
Limits of integration — woh do numbers (ya expressions) $a$ (bottom) aur $b$ (top) jo batate hain kahan se adding shuru hoti hai aur kahan rukti hai.

The scenario matrix

Har FTC problem in cells mein se kisi ek mein aati hai. Neeche ke examples mein cell label kiya gaya hai jo woh hit karta hai.

Cell	Kya tricky hai	Example
A	Plain Part 2, positive area	Ex 1
B	Integrand negative ho jaata hai → signed area	Ex 2
C	Degenerate: equal limits ( $a = b$ ) → zero	Ex 3
D	Reversed limits ( $a > b$ ) → sign flip	Ex 3
E	Part 1 with variable upper limit = $x$	Ex 4
F	Variable limit ek function of $x$ hai → chain rule	Ex 5
G	Dono limits move karte hain	Ex 6
H	Integrand mein jump discontinuity hai → FTC breaks	Ex 7
I	Real-world word problem (accumulation)	Ex 8
J	Improper / limiting behaviour (infinite limit)	Ex 9
K	Exam twist: unknown limit ke liye solve karo	Ex 10

Prerequisites jinpar hum rely karte hain: Antiderivatives and Indefinite Integrals, Riemann Sums and the Definite Integral, Chain Rule, Continuity, Improper Integrals.

Cell B — signed (negative) area

Definite integral ek signed area hai: curve ke woh hisse jo $x$ -axis ke neeche hain woh negative count hote hain.

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

Worked example Ex 2 — Cell B

$\int_{0}^{π} cos x d x$ compute karo aur saath mein $cos x$ aur axis ke beech ka total geometric area bhi.

Forecast: $[0, π]$ par cosine curve $[0, \frac{π}{2}]$ mein axis ke upar hai aur $[\frac{π}{2}, π]$ mein neeche. Dono halves mirror images hain, toh signed integral zero hona chahiye, jabki geometric area positive hai.

Figure s01 mein blue curve $cos x$ hai. Green shaded lobe left par ( $0$ se $\frac{π}{2}$ tak) axis ke upar hai aur $+ 1$ count karta hai; red shaded lobe right par ( $\frac{π}{2}$ se $π$ tak) neeche hai aur $- 1$ count karta hai. Dashed vertical line sign change ko $x = \frac{π}{2}$ par mark karti hai — exactly wahan jahan hume geometric area ke liye split karna zaroori hai. Notice karo ki green aur red regions congruent reflections hain, isliye woh cancel ho jaate hain.

Step 1. $cos x$ ka antiderivative $sin x$ hai (kyunki $\frac{d}{d x} sin x = cos x$ ). Yeh step kyun? Part 2 ko ek antiderivative chahiye.

Step 2. Signed integral: $\int_{0}^{π} cos x d x = [sin x]_{0}^{π} = sin π - sin 0 = 0 - 0 = 0.$ Yeh step kyun? Positive hump ( $+ 1$ area, green) aur negative dip ( $- 1$ area, red) cancel ho jaate hain — integral signed area measure karta hai.

Step 3. Geometric area — sign change $x = \frac{π}{2}$ par split karo (dashed line) aur absolute values lo: $= 1 \int_{0}^{π /2} cos x d x + = - 1 \int_{π /2}^{π} cos x d x = 1 + 1 = 2.$ Yeh step kyun? "Area" (unsigned) ke liye har below-axis piece ko positive karna padta hai; integrand ki har root par split karna zaroori hai.

Verify: $[sin x]_{0}^{π /2} = 1 - 0 = 1$ aur $[sin x]_{π /2}^{π} = 0 - 1 = - 1$ . Magnitudes ka sum $= 2$ , signs ke saath sum $= 0$ . ✓

Cells C & D — degenerate aur reversed limits

Yeh woh "edge inputs" hain jo log exams mein bhool jaate hain.

Worked example Ex 3 — Cells C and D

(a) $\int_{5}^{5} e^{t^{2}} d t$ compute karo. (b) $\int_{3}^{1} 2 x d x$ compute karo.

Forecast: (a) zero width ki strip ka area zero hoga → $0$ . (b) "backwards" integrate karne se forward integral ka sign flip ho jaana chahiye.

Step 1 (a). Equal limits: $a = b = 5$ . $\int_{5}^{5} e^{t^{2}} d t = 0.$ Yeh step kyun? Definition se $\int_{a}^{a} f = G (a) - G (a) = 0$ — zero length par koi accumulation nahi hoti. Note karo ki humne $e^{t^{2}}$ ka antiderivative kabhi nahi dhundha (uska koi elementary antiderivative nahi hai).

Step 2 (b). Reversed limits: top $= 1$ bottom $= 3$ se neeche hai. Rule use karo $\int_{b}^{a} f = - \int_{a}^{b} f$ . $\int_{3}^{1} 2 x d x = [x^{2}]_{3}^{1} = 1 - 9 = - 8.$ Yeh step kyun? Part 2 blindly $G (top) - G (bottom) = G (1) - G (3)$ compute karta hai; ordering sign apne aap handle kar leti hai.

Verify: Forward, $\int_{1}^{3} 2 x d x = [x^{2}]_{1}^{3} = 9 - 1 = 8$ ; reverse karne par $- 8$ milta hai. ✓ Aur $\int_{5}^{5}$ kisi bhi cheez ka $0$ hota hai. ✓

Cell E — Part 1, variable upper limit equal to $x$

Worked example Ex 4 — Cell E

$\frac{d}{d x} \int_{2}^{x} 1 + t^{4} d t$ dhundho.

Forecast: Part 1 kehta hai "area accumulate hone ki rate = top edge par curve ki height." Toh answer sirf integrand hona chahiye jisme $t$ ko $x$ se replace kar diya gaya ho. Koi antiderivative required nahi.

Step 1. Part 1 ki exact form pehchano: upper limit literally $x$ hai, lower limit ek constant hai. Yeh step kyun? Part 1 tabhi apply hota hai jab top limit $x$ ho aur integrand continuous ho — $1 + t^{4}$ har jagah continuous hai.

Step 2. Answer read off karo: $\frac{d}{d x} \int_{2}^{x} 1 + t^{4} d t = 1 + x^{4} .$ Yeh step kyun? Yahi FTC Part 1 hai: $\frac{d}{d x} \int_{a}^{x} f (t) d t = f (x)$ . Lower limit $2$ ek constant hai aur differentiation mein gayab ho jaata hai.

Verify (sanity by numbers): $F (x) = \int_{2}^{x} 1 + t^{4} d t$ approximate karte hue, difference quotient $\frac{F ( 2.001 ) - F ( 2 )}{0.001}$ $\approx 1 + 2^{4} = 17 \approx 4.123$ hona chahiye. Patli si sliver ki width $0.001$ aur height $\approx 17$ hai, jo woh ratio deta hai. ✓

Cell F — variable limit jo $x$ ka function hai

Worked example Ex 5 — Cell F

$\frac{d}{d x} \int_{0}^{s i n x} ln (1 + u^{2}) d u$ dhundho.

Forecast: top limit $sin x$ hai, $x$ nahi. Part 1 rate deta hai limit ke respect mein; lekin limit khud $x$ badlne par move karti hai. Do rates ek saath chain hoti hain → ek extra factor $cos x$ expected hai.

Figure s02 do stacked rates dikhata hai. Red curve inner limit $w = sin x$ hai; ek chosen $x$ par uski steepness (red arrow) $cos x$ hai — top edge kitni tezi se slide karti hai. Blue curve outer accumulated value $H (sin x)$ hai; blue arrow mark karta hai ki stored area per unit of $w$ kitni tezi se badhti hai, jo Part 1 kehta hai $ln (1 + w^{2})$ hai. Final rate red slope aur blue slope ka product hai — exactly wohi chain rule multiply karta hai.

Step 1. Inner limit ko naam do. $w = sin x$ lo aur $H (w) = \int_{0}^{w} ln (1 + u^{2}) d u$ define karo. Yeh step kyun? $H$ (limit $w$ wala integral) aur $w = sin x$ (ek plain function) mein split karne se dono dependencies alag ho jaate hain taaki hum har rule cleanly apply kar sakein.

Step 2. Part 1 se, $H^{'} (w) = ln (1 + w^{2})$ . Yeh step kyun? Ab upper limit variable $w$ hai — exact Part 1 form.

Step 3. Chain Rule se, $\frac{d}{d x} H (sin x) = H^{'} (sin x) \cdot \frac{d}{d x} (sin x)$ . $= ln (1 + sin^{2} x) \cdot cos x .$ Yeh step kyun? Chain rule "outer rate" $H^{'}$ (blue arrow) ko "inner rate" $\frac{d w}{d x} = cos x$ (red arrow) se multiply karta hai — figure dikhata hai ki dono kyun stack hote hain.

Verify: $x = 0$ par: $sin 0 = 0$ , toh answer $ln (1) \cdot cos 0 = 0 \cdot 1 = 0$ hai. Independently, $x = 0$ ke paas top limit barely $0$ se move karti hai jahan integrand $ln (1 + u^{2}) \approx 0$ hai, toh rate $\approx 0$ hai. ✓

Cell G — dono limits moving hain

Worked example Ex 6 — Cell G

$\frac{d}{d x} \int_{x^{2}}^{x^{3}} \frac{1}{1 + t ^{2}} d t$ dhundho.

Forecast: dono ends slide karte hain. Interval ko ek fixed point par break karo, har piece ko chain rule se differentiate karo, aur combine karo — top piece positive, bottom piece negative.

Step 1. Ek constant $a$ par split karo (maano $a = 0$ ): $\int_{x^{2}}^{x^{3}} = \int_{0}^{x^{3}} - \int_{0}^{x^{2}} .$ Yeh step kyun? Part 1 aur chain rule sirf variable upper limit ke saath constant lower limit par kaam karte hain; split karne se moving lower limit ek subtracted moving upper limit ban jaati hai.

Step 2. Har ek ko Cell F ke method se differentiate karo. $g (t) = \frac{1}{1 + t ^{2}}$ ke saath: $\frac{d}{d x} \int_{0}^{x^{3}} g = g (x^{3}) \cdot 3 x^{2}, \frac{d}{d x} \int_{0}^{x^{2}} g = g (x^{2}) \cdot 2 x .$ Yeh step kyun? Har ek Part 1 + chain rule hai (Cell F).

Step 3. Subtract karo: $\frac{d}{d x} \int_{x^{2}}^{x^{3}} \frac{d t}{1 + t ^{2}} = \frac{3 x ^{2}}{1 + x ^{6}} - \frac{2 x}{1 + x ^{4}} .$ Yeh step kyun? Upper limit $+$ contribute karta hai, lower limit $-$ contribute karta hai (use subtract kiya gaya tha).

Verify: $x = 1$ par dono limits $1$ ke barabar ho jaate hain toh interval collapse ho jaata hai (Cell C) — derivative zero hone ki zaroorat nahi, lekin dono terms $\frac{3}{2} - \frac{2}{2} = \frac{1}{2}$ dete hain, ek finite number, jo smooth accumulation ke liye expected hai. ✓

Cell H — jump discontinuity: FTC ki hypothesis fail ho jaati hai

Worked example Ex 7 — Cell H

Maano $f (t) = {1, 3, t < 1 t \geq 1$ aur $F (x) = \int_{0}^{x} f (t) d t$ . Kya $F^{'} (1) = f (1)$ hai?

Forecast: Part 1 ko ek continuous integrand chahiye. Yahan $f$ $t = 1$ par jump karta hai. Toh "area accumulate hone ki rate" $x = 1$ par abruptly badlni chahiye, aur $F^{'} (1)$ exist nahi kar sakta. Guess: koi clean $F^{'} (1)$ nahi milega.

Figure s03 mein do panels hain. Left par, orange integrand $f (t)$ $t = 1$ par height $1$ se height $3$ par leap karta hai (open circle = value nahi li gayi, filled circle = value li gayi). Right par, blue accumulated area $F (x)$ $x = 1$ se pehle slope $1$ se aur baad mein slope $3$ se badhti hai — red dot us corner ko mark karta hai jahan do straight pieces milte hain. Kyunki left slope ( $1$ ) aur right slope ( $3$ ) alag hain, graph mein ek sharp kink hai, toh koi single tangent exist nahi karta.

Step 1. Har side par $F$ compute karo. $x \leq 1$ ke liye: $F (x) = \int_{0}^{x} 1 d t = x$ . $x \geq 1$ ke liye: $F (x) = \int_{0}^{1} 1 1 + \int_{1}^{x} 3 d t = 1 + 3 (x - 1) = 3 x - 2$ . Yeh step kyun? Piecewise integrands ko piece by piece integrate kiya jaata hai (interval additivity).

Step 2. $x = 1$ par one-sided slopes compare karo (figure mein corner). $F^{'} (1^{-}) = slope of x = 1, F^{'} (1^{+}) = slope of 3 x - 2 = 3.$ Yeh step kyun? Derivative tabhi exist karta hai jab left aur right slopes agree karein.

Step 3. Woh differ karte hain ( $1 \neq = 3$ ), toh $F^{'} (1)$ exist nahi karta — Part 1 ka conclusion exactly wahan fail hota hai jahan continuity fail hoti hai. Yeh step kyun? Isliye theorem continuity demand karta hai: proof mein EVT aur Squeeze dono yahan break ho gaye.

Verify: $F$ har jagah continuous hai ( $F (1) = 1$ dono formulas se: $x \to 1$ gives $1$ , $3 x - 2 \to 1$ ) lekin uska ek corner hai — continuous lekin non-differentiable, Continuity vs differentiability gap se match karta hai. ✓

Cell I — real-world accumulation

Worked example Ex 8 — Cell I

Paani ek tank mein rate $r (t) = 6 t$ litres per minute par flow karta hai, $t$ minutes mein hai. $t = 1$ aur $t = 4$ minutes ke beech kitne litres enter hote hain?

Forecast: total water = accumulated flow = rate curve ke neeche area. Kyunki rate linearly badhti hai, answer ek trapezoid ka area hona chahiye.

Step 1. Total added $= \int_{1}^{4} r (t) d t = \int_{1}^{4} 6 t d t$ . Yeh step kyun? FTC se, ek rate ko time par integrate karne se quantity ka total change milta hai — yeh Part 2 hai jo "amount function ke endpoints" ke roop mein padha jaata hai.

Step 2. $6 t$ ka antiderivative $3 t^{2}$ hai; Part 2 apply karo: $\int_{1}^{4} 6 t d t = [3 t^{2}]_{1}^{4} = 48 - 3 = 45 litres .$ Yeh step kyun? $\frac{d}{d t} (3 t^{2}) = 6 t$ , toh $3 t^{2}$ abhi-tak-ka amount function hai; endpoint values subtract karo.

Verify (units + geometry): rate L/min mein hai, time min mein → integral litres mein ✓. Trapezoid area: heights $r (1) = 6$ , $r (4) = 24$ , width $3$ : $\frac{( 6 + 24 )}{2} \cdot 3 = 45$ . ✓

Cell J — improper / limiting behaviour

Worked example Ex 9 — Cell J

Improper integral $\int_{1}^{\infty} \frac{1}{x ^{2}} d x$ evaluate karo.

Forecast: top limit infinite hai. Hum " $\infty$ plug in" nahi kar sakte. Iske bajaaye ek finite $b$ tak accumulate karo aur $b \to \infty$ hone par limit dekho. Tail $1/ x^{2}$ tezi se shrink karta hai, toh total area finite hona chahiye.

Step 1. $\infty$ ko variable $b$ se replace karo aur limit lo: $\int_{1}^{\infty} \frac{d x}{x ^{2}} = lim_{b \to \infty} \int_{1}^{b} \frac{d x}{x ^{2}} .$ Yeh step kyun? FTC (Part 2) sirf closed finite interval par apply hota hai; improper integral define hi is limit ke roop mein hota hai.

Step 2. $x^{- 2}$ ka antiderivative $- x^{- 1}$ hai; Part 2 apply karo: $\int_{1}^{b} \frac{d x}{x ^{2}} = [- \frac{1}{x}]_{1}^{b} = - \frac{1}{b} + 1 = 1 - \frac{1}{b} .$ Yeh step kyun? $\frac{d}{d x} (- x^{- 1}) = x^{- 2}$ .

Step 3. Limit lo: $lim_{b \to \infty} (1 - \frac{1}{b}) = 1.$ Yeh step kyun? $b \to \infty$ hone par, $\frac{1}{b} \to 0$ ; far tail mein bacha hua area vanish ho jaata hai.

Verify: har finite piece $1 - \frac{1}{b} < 1$ hai aur $1$ ki taraf increase karta hai — bounded aur monotone, toh value $1$ consistent hai. ✓

Cell K — exam twist: unknown limit ke liye solve karo

Active recall

Recall Kaun sa cell? Trigger ko fix se match karo.

Integrand axis ke neeche dip karta hai ::: Cell B — har zero par split karo, geometric area ke liye absolute values lo. Top aur bottom limits barabar hain ::: Cell C — integral $0$ hai chahe integrand kuch bhi ho. Bottom limit top se bada hai ::: Cell D — sign flip karo, $\int_{b}^{a} = - \int_{a}^{b}$ . Upper limit $g (x)$ hai, $x$ nahi ::: Cell F — Part 1 phir $g^{'} (x)$ se multiply karo (chain rule). Dono limits $x$ ke functions hain ::: Cell G — ek constant par split karo, do chain-rule terms subtract karo. Integrand mein jump hai ::: Cell H — Part 1 fail hota hai; $F$ ka corner ho sakta hai (wahan koi derivative nahi). Ek limit $\infty$ hai ::: Cell J — $lim_{b \to \infty}$ ke roop mein rewrite karo, phir finite piece par FTC apply karo.

Worked examples — Fundamental Theorem of Calculus — Part 1 and Part 2 — full proofs