Fundamental Theorem of Calculus — Part 1 and Part 2 — full proofs: Question bank

4.2.4 · D5 · Maths › Calculus II — Integration › Fundamental Theorem of Calculus — Part 1 and Part 2 — full p

parent note se dono statements ke reminders jinke baare mein tum reason kar rahe ho:

Part 1: $f$ continuous, $F (x) = \int_{a}^{x} f (t) d t$ $\Rightarrow$ $F^{'} (x) = f (x)$ .
Part 2: $G^{'} = f$ (with $f$ continuous) $\Rightarrow$ $\int_{a}^{b} f = G (b) - G (a)$ .

True or false — justify

Har item ek claim hai; answer batata hai ki dono theorems / hypotheses mein se kaun sa actually kaam kar raha hai.

FTC Part 1 ko

f

differentiable chahiye, kyunki hum

F

ka derivative lete hain.

False — hum derivative lete hain $F$ ka,

f

ka nahi. Humein sirf

f

continuous chahiye; tab EVT aur Squeeze

F^{'} = f

force karte hain. Actually

F

f

se zyada smooth ho jaata hai.

Agar

f

sirf integrable ho lekin jump ho, toh

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

phir bhi continuous rahega.

True — ek shrinking interval par area accumulate karna

\to 0

hota hai, isliye

F

har jagah continuous hai. Lekin

F

jump par differentiable nahi bhi ho sakta, isliye Part 1 ka conclusion

F^{'} = f

wahan fail ho sakta hai.

Part 2 ke liye tumhe specific antiderivative

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

use karna zaroori hai; koi doosra antiderivative alag answer dega.

False — koi bhi antiderivative

G

kaam karta hai. Do antiderivatives ek constant se differ karte hain (Mean Value Theorem se prove hota hai), aur woh constant

G (b) - G (a)

mein cancel ho jaata hai.

Equation

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

kehti hai ki differentiation aur integration inverse operations hain.

"Integrate-then-differentiate" direction mein True: Part 1 dikhata hai ki accumulation function ko differentiate karne par original height

f

wapas milti hai. Reverse direction (

\int_{a}^{x} F^{'} d t = F (x) - F (a)

) Part 2 hai.

Kyunki

\int_{a}^{a} f (t) d t = 0

hai, lower limit

a

F^{'} (x)

ki value ke liye irrelevant hai.

Derivative ke liye True:

a

badalne se

F

ek constant se shift ho jaata hai (dono lower limits ke beech ki area), aur constants differentiation mein vanish ho jaate hain. Isliye

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

a

ki parwah kiye bina.

Agar do functions ka ek interval par same derivative ho, toh woh equal hain.

False — woh ek constant se differ karte hain, necessarily zero se nahi. Yahi "

+ C

" hai jo indefinite antiderivatives ko ek family banata hai, aur yahi Part 2 mein cancel hota hai.

FTC humein antiderivative dhundh kar har definite integral compute karne deta hai.

Practice mein False — Part 2 ko ek aisa antiderivative chahiye jo tum likh sako.

\int_{0}^{x} e^{- t^{2}} d t

jaise functions ka koi elementary antiderivative nahi hai, isliye Part 2 useless hai; sirf Part 1 (ya Riemann sums) apply hota hai.

Spot the error

Har statement mein ek flawed step hai. Use name karo.

\frac{d}{d x} \int_{0}^{x^{2}} f (t) d t = f (x^{2})

Chain Rule factor missing hai. Upper limit

x^{2}

hai,

x

nahi, isliye answer

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

hai. Part 1 rate deta hai limit ki unit per; limit khud

2 x

speed se move karti hai.

(F - G)^{'} = 0

, therefore

F - G = 0

, so

F = G

Zero derivative constant deta hai, zero nahi:

F - G = C

C = 0

conclude karne ke liye ek specific data point chahiye. Isliye Part 2

F = G

claim karne ki jagah endpoints subtract karta hai.

"Kyunki

f

[x, x + h]

par min

m_{h}

aur max

M_{h}

attain karta hai, hume milta hai

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

— yeh sab

h

ke liye hold karta hai."

Inequality direction assume karta hai ki

h > 0

h < 0

ke liye interval

[x + h, x]

hai aur width factor sign flip karta hai, isliye inequalities reverse ho jaati hain. Conclusion phir bhi follow karta hai, lekin

h < 0

ko alag handle karna padta hai.

\frac{d}{d x} \int_{x}^{b} f (t) d t = f (x)

by Part 1."

Wrong sign. Yahan

x

lower limit hai. Ise

- \int_{b}^{x} f

likhein, tab Part 1

- f (x)

deta hai. Lower limit badhane se area remove hoti hai, isliye derivative height ka negative hai.

f

continuous hai, isliye EVT se

(x, x + h)

par max hai."

EVT ko ek closed, bounded interval

[x, x + h]

chahiye; open interval par max attain nahi bhi ho sakta. Proof deliberately closed sliver use karta hai.

"Part 2 prove karne ke liye hum sirf cite kar sakte hain ki antiderivatives ek constant se differ karte hain — koi theorem nahi chahiye, yeh obvious hai."

"Derivative zero

\Rightarrow

constant" ek theorem hai, axiom nahi, aur ek interval par iska honest proof Mean Value Theorem use karta hai. Ise skip karna real logical hinge chhupa deta hai.

Why questions

f

sirf integrable kyun nahi, balki continuous kyun hona chahiye, Part 1 ke conclusion

F^{'} = f

ke liye?

Proof mein continuity do baar use hoti hai: EVT ko sliver par max/min guarantee karne ke liye chahiye, aur final Squeeze ko

f (u), f (v) \to f (x)

chahiye jab

u, v \to x

, jo exactly Continuity at

x

hai.

Antiderivative ka

+ C

definite integral ke answer mein kabhi kyun nahi aata?

Part 2

G (b) - G (a)

evaluate karta hai; agar

G

G + C

se replace karein, toh dono endpoints ko

C

milta hai aur woh subtract ho jaata hai. Definite integral ek difference hai, vertical shifts se blind.

Part 1 guarantee kyun karta hai ki ek continuous

f

ka antiderivative exist karta hai, chahe uska koi formula na ho?

Woh explicitly ek construct karta hai:

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

ek genuine function hai jiska derivative

f

hai. Existence integral se aati hai, closed form likhne ki ability se nahi.

Proof mein Part 2 se pehle Part 1 kyun chahiye?

Part 2 ke Step 1 mein integral-defined

F

use hota hai aur

F^{'} = f

jaanna zaroori hai — woh fact precisely Part 1 hai. Part 1 existence supply karta hai jise Part 2 phir ek arbitrary

G

se compare karta hai.

[x, x + h]

par

f

ki average height

m_{h}

aur

M_{h}

ke beech kyun trapped hai?

Width

h

se divide kiya hua integral ek average value hai; ek continuous function ka ek interval par average kभी us interval par apne minimum se neeche ya maximum se upar nahi ja sakta, isliye woh

m_{h}

aur

M_{h}

ke beech rehta hai.

"Water tank" picture (level curve ki height ki rate se badhta hai) Part 1 ko exactly kyun capture karta hai?

Height

f (x)

incoming flow rate hai; water level accumulated volume

F (x)

hai. Rate-of-level-change equals inflow ka matlab hai

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

— physical language mein Part 1.

Part 1

\int_{0}^{x} sin (t^{2}) d t

ko differentiate kyun kar sakta hai, jabki integral ka koi elementary formula nahi?

Part 1 derivative directly integrand ko upper limit par (

sin (x^{2})

) se read off karta hai, bina integral evaluate kiye. Yahi iska superpower hai: yeh antidifferentiation ko bypass karta hai.

Edge cases

x = a

par

\frac{d}{d x} \int_{a}^{x} f

kya hai? Kya Part 1 wahan bhi apply hota hai?

Haan — left endpoint par tum one-sided derivative use karte ho (sirf

h > 0

makes sense), aur wohi Squeeze argument right-hand derivative

f (a)

deta hai. Proof ka

h < 0

branch yahan simply omit ho jaata hai.

Agar

f (x) = c

(ek constant) ho, toh Part 1 kya predict karta hai, aur kya yeh direct computation se match karta hai?

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

, isliye

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

. Match karta hai — ek flat height ka accumulation linearly badhta hai, aur growth rate wahi height hai.

Unbounded interval par Part 2 ka kya hota hai, jaise

\int_{1}^{\infty} x^{- 2} d x

Ordinary Part 2 finite

[a, b]

assume karta hai jahan

f

continuous ho. Infinite limits ke liye Improper Integrals chahiye:

[G (x)]_{1}^{b}

compute karo phir

lim_{b \to \infty}

lo. Yahan

lim_{b \to \infty} (1 - 1/ b) = 1

Agar

f

mein

x_{0} \in (a, b)

par ek single jump discontinuity ho, toh kya tum phir bhi

\int_{a}^{b} f

find kar sakte ho?

Haan, split karke:

\int_{a}^{x_{0}} + \int_{x_{0}}^{b}

, har piece par Part 2 apply karo jahan

f

continuous hai. Lekin

F

x_{0}

par differentiable nahi hai, isliye tum ek antiderivative ko jump ke across use nahi kar sakte.

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

kya equal hai aur structurally important kyun hai?

Yeh

0

equal hai — zero width, zero area. Isse

F (a) = 0

milta hai,

F

ko correct base value milti hai taaki

F (b) - F (a) = \int_{a}^{b} f

ko koi correction term na chahiye.

b < a

ke liye, kya

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

abhi bhi valid hai?

Haan, orientation convention

\int_{a}^{b} = - \int_{b}^{a}

se. Part 2 ka algebra

G (b) - G (a)

automatically sign flip produce karta hai, isliye formula order ki parwah kiye bina hold karta hai.

Agar

G^{'} = f

ho lekin

G

[a, b]

mein kahin discontinuous ho, toh kya Part 2 fail ho sakta hai?

Ek antiderivative jo

[a, b]

par har jagah

G^{'} = f

satisfy karta hai woh automatically differentiable, hence continuous, wahan hai — isliye aisa ho nahi sakta. Danger ek "piecewise antiderivative" mein hai jo jump ke saath glue ki gayi ho, jiska

G^{'} = f

gluing point par nahi hota.