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

4.2.4 · Maths › Calculus II — Integration

HUM KYA PROVE KAR RAHE HAIN?

DO parts kyun? Part 1 guarantee karta hai ki ek antiderivative exist karta hai (use integral ke roop mein construct karta hai). Part 2 usi antiderivative ko use karke definite integral compute karta hai. Part 1 existence deta hai; Part 2 shortcut deta hai.

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

Part 1 KO PROVE KAISE KAREN — pehle principles se

Hum sirf definition of the derivative use karte hain, kuch nahi.

Step 1 — Difference quotient likho. $F^{'} (x) = lim_{h \to 0} \frac{F ( x + h ) - F ( x )}{h} .$ Ye step kyun? Derivative IS yahi limit hai definition se — hum derive kar rahe hain, assume nahi kar rahe.

Step 2 — Integrals ki additivity use karke numerator simplify karo. $F (x + h) - F (x) = \int_{a}^{x + h} f (t) d t - \int_{a}^{x} f (t) d t = \int_{x}^{x + h} f (t) d t .$ Ye step kyun? $\int_{a}^{x + h} = \int_{a}^{x} + \int_{x}^{x + h}$ (interval splitting). " $x$ tak ka area" cancel ho jaata hai, sirf $x$ se $x + h$ tak ka patla sliver bachta hai.

Step 3 — Extreme Value Theorem se sliver ko squeeze karo. Kyunki $f$ closed interval $[x, x + h]$ par continuous hai, woh wahan minimum $m_{h} = f (u)$ aur maximum $M_{h} = f (v)$ attain karta hai. Sliver ka area sabse chote aur sabse bade rectangle ke beech hota hai: $m_{h} h \leq \int_{x}^{x + h} f (t) d t \leq M_{h} h (h > 0) .$ $h$ se divide karo: $m_{h} \leq \frac{F ( x + h ) - F ( x )}{h} \leq M_{h} .$ Ye step kyun? Ek continuous function apne min se neeche nahi ja sakta ya max se oopar nahi ja sakta, isliye sliver ki average height unke beech trapped hai.

Step 4 — Limit lo (Squeeze Theorem). Jab $h \to 0$ , points $u, v \in [x, x + h] \to x$ , aur continuity se $f (u) \to f (x)$ , $f (v) \to f (x)$ . Toh dono bounds collapse ho jaate hain: $lim_{h \to 0} m_{h} = f (x) = lim_{h \to 0} M_{h} .$ Squeeze Theorem se beech wala term $f (x)$ pe force ho jaata hai: $F^{'} (x) = f (x) .$ (Jab $h < 0$ hota hai tab inequalities flip ho jaati hain — same conclusion milta hai.) $■$

Part 2 KO PROVE KAISE KAREN — Part 1 + MVT par build karke

Step 1 — Part 1 use karke ek antiderivative lo. $F (x) = \int_{a}^{x} f (t) d t$ define karo. Part 1 se, $F^{'} = f$ .

Step 2 — Kisi arbitrary antiderivative $G$ se compare karo. Maano $G^{'} = f$ bhi. Toh $(F - G)^{'} = f - f = 0$ on $[a, b]$ . Ye step kyun? Hume integral-defined specific $F$ ko kisi bhi arbitrary $G$ se connect karna hai jo student find kare.

Step 3 — Zero derivative ⇒ constant (iske liye Mean Value Theorem chahiye). Agar $(F - G)^{'} = 0$ har jagah ho, toh kisi bhi $x$ ke liye, MVT deta hai $(F - G) (x) - (F - G) (a) = (F - G)^{'} (c) (x - a) = 0$ . Isliye $F - G$ constant hai: $G (x) = F (x) + C$ . Ye step kyun? "Derivative zero ⇒ constant" free nahi milta — yeh MVT ka consequence hai. Yeh proof ka honest hinge hai.

Step 4 — Endpoints par evaluate karo. $G (b) - G (a) = (F (b) + C) - (F (a) + C) = F (b) - F (a) .$ Ab $F (b) = \int_{a}^{b} f d t$ aur $F (a) = \int_{a}^{a} f d t = 0$ . Isliye $\int_{a}^{b} f (x) d x = G (b) - G (a) .$ $■$

Worked examples

Common mistakes (steel-manned)

Active recall

Recall Feynman: ek 12-saal ke bacche ko samjhao

Socho tum ek tank mein paani daal rahe ho aur water level dekh rahe ho. Is instant par water curve ki height level kitni tez badhti hai — yeh Part 1 hai (bharne ki rate = curve ki height). Ab, start se end tak total paani sirf (end par level) minus (start par level) hai — tumhe har drop nahi dekhna, sirf dono readings check karo — yeh Part 2 hai. Bharna (integrating) aur badhne ki speed measure karna (differentiating) ek hi sikke ke do pehlu hain.

What does FTC Part 1 state?

Agar

f

continuous hai aur

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

, toh

F^{'} (x) = f (x)

What does FTC Part 2 state?

Agar

G^{'} = f

aur

f

continuous ho, toh

\int_{a}^{b} f d x = G (b) - G (a)

In the Part 1 proof, what simplifies

F (x + h) - F (x)

\int_{x}^{x + h} f

Additivity of integrals (interval splitting).

Which theorem provides

m_{h} h \leq \int_{x}^{x + h} f \leq M_{h} h

Extreme Value Theorem (closed interval par max/min), phir bounded rectangles.

Which theorem forces the difference quotient to

f (x)

h \to 0

Squeeze Theorem,

f

ki continuity use karke.

In Part 2, why is

G (x) = F (x) + C

(F - G)^{'} = f - f = 0

, aur zero derivative ⇒ constant by Mean Value Theorem.

Why does the constant

C

not appear in the final answer of Part 2?

Woh

G (b) - G (a)

mein cancel ho jaata hai.

Compute

\frac{d}{d x} \int_{0}^{x^{2}} sin t d t

sin (x^{2}) \cdot 2 x

(Part 1 + chain rule).

What single hypothesis on

f

powers both proofs?

Continuity on

[a, b]

What is

\int_{a}^{a} f (t) d t

and why does it matter?

0

; yeh area function

F

ki base value set karta hai.

Connections

Riemann Sums and the Definite Integral — $\int_{a}^{b}$ ka matlab FTC se pehle.
Antiderivatives and Indefinite Integrals — Part 2 ka $G$ .
Mean Value Theorem — Part 2, Step 3 ka hinge.
Extreme Value Theorem / Squeeze Theorem — Part 1 ke engines.
Continuity — indispensable hypothesis.
Chain Rule — Part 1 ko variable limits tak extend karta hai (Leibniz rule).
Improper Integrals — jahan continuity fail hoti hai aur FTC ko care chahiye.