U-substitution — technique, change of limits for definite integrals: Question bank

4.2.6 · D5 · Maths › Calculus II — Integration › U-substitution — technique, change of limits for definite in

Reminder of the players (taaki is page ka koi bhi symbol behjha na lage):

$u = g (x)$ — woh nickname jo hum ek inner function ko dete hain (woh cheez jo ek power/root/exp/log/trig ke andar hoti hai).
$d u = g^{'} (x) d x$ — woh differential, woh rule jo "stuff ki miqdar" ko honest rakhta hai jab hum rename karte hain. Ye $\frac{d u}{d x} = g^{'} (x)$ se aata hai, jo $u$ ki slope hai, $d x$ se multiply ki gayi.
$F$ — $f$ ka ek antiderivative: ek aisi function jiska derivative hi $f$ hai, yaani $F^{'} = f$ . " $F^{'} = f$ " ko padho as "har point par $F$ ki slope wahan $f$ ke barabar hai." (Dekho Antiderivatives & Indefinite Integrals.)
$\int_{a}^{b}$ — ek definite integral: " $x$ , $a$ se $b$ tak jaata hai." Bina numbers ke sada $\int$ indefinite hota hai (jawab ek function $+ C$ hota hai).

Why questions — the geometric heart

Upar diye area-mapping figure ko dekhte hue in sawaalon ke jawaab do.

U-substitution area ko same kyun rakhta hai jabki interval ki width change ho jaati hai?

Kyunki

d u = g^{'} (x) d x

height ko usi factor se rescale karta hai jitna interval stretch hota hai:

u

mein ek wider strip proportionally shorter integrand carry karta hai, isliye width

\times

height (yaani area) preserve rehta hai. Ye poore theorem ka visual meaning hai.

Inner function ka derivative "pehle se wahan hona" u-sub ke kaam karne ke liye kyun zaroori hai?

Kyunki u-sub Chain Rule ko ulta karta hai, jiska output hamesha inner-function-times-inner-derivative hota hai; agar woh derivative absent hai, toh undo karne ke liye koi chain-rule pattern nahi hai — geometrically, height rescale karne ke liye koi factor hi nahi hoga.

Constants ko integral sign ke bahar pull kyun kiya ja sakta hai lekin variables ko nahi?

Integration linear hai:

\int c f = c \int f

sirf tab hold karta hai jab

c

fixed ho. Ek variable interval ke across change hota hai, isliye woh

d x

ke saath interact karta hai aur use fixed multiplier ki tarah treat nahi kiya ja sakta.

Definite integral ke liye limits kyun change hoti hain lekin indefinite ke liye "

+ C

" kyun aata hai?

Ek definite integral ek fixed range par ek number maangta hai, isliye us range ko nayi

u

-axis par re-express karna hoga; ek indefinite integral ek function maangta hai, isliye hum original variable restore karte hain aur lost constant ke liye

C

add karte hain.

\frac{d u}{d x}

ko ek fraction ki tarah treat karna (taaki

d u = g^{'} (x) d x

mile) actually legitimate kyun hai?

Ye shorthand hai jo u-sub theorem ke khud se justified hai — theorem prove karta hai ki swap

f (g (x)) g^{'} (x) d x \to f (u) d u

valid hai, isliye fraction-manipulation ek sahi bookkeeping notation hai, koi hand-wave nahi.

Do students alag-alag

u

choose karke dono succeed kyun ho sakte hain?

Kayi inner functions ek chain-rule pattern expose kar sakte hain; jab tak chosen

u

ka derivative available ho (ek constant tak), koi bhi valid choice same antiderivative tak pahunch jaati hai.

Trigonometric Substitution ko u-substitution ka special case kyun maana jaata hai, koi nayi idea nahi?

Ye fir bhi variable ko rename karna hi hai (

x = a sin θ

, wagera) matching differential

d x

ke saath; bas ye substitution inner-function-derivative pairing ki jagah trig identities ko exploit karne ke liye choose karta hai.

True or false — justify karo

U-substitution sirf chain rule ko, ulta run karna hai.

True — antiderivative

F (g (x))

ko differentiate karne par

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

milta hai (

F^{'} = f

use karke); us pattern ko integrate karne par

F (g (x))

milta hai, jo exactly u-sub recover karta hai. Dekho Chain Rule.

Har integral kisi na kisi clever

u

choice se solve ho sakti hai.

False — u-sub sirf tab kaam karta hai jab integrand mein ek inner function saath mein (apne derivative ke constant multiple ke saath) chupi ho. Kayi integrals ke liye Integration by Parts ya doosre tools chahiye.

Ek definite integral mein ya toh limits change karo ya back-substitute karo, lekin kabhi dono nahi.

True — dono poore finishing methods hain; dono karna endpoints ko do baar translate karta hai aur jawab kharaab kar deta hai.

Ek baar jab sab kuch

u

mein likh do,

d x

simply drop kiya ja sakta hai kyunki ye "sirf notation hai."

False —

d x

integrand ka hissa hai aur ise

d u = g^{'} (x) d x

ke zariye convert karna hoga; ek

u

-integral jo

d u

par khatam nahi hoti woh simplified nahi, incomplete hai.

Agar

u = g (x)

ek constant function hai, toh bhi u-substitution kaam karta hai.

False — tab

g^{'} (x) = 0

hoga, isliye

d u = 0

aur inner ka koi derivative present nahi; method ke paas reverse karne ke liye kuch nahi.

\int \frac{1}{u} d u = ln u + C

hamesha correct hai.

False — ye

ln ∣ u ∣ + C

hai.

\frac{1}{u}

ki domain

u = 0

par split hoti hai positive aur negative branches mein; absolute value hi negative branch ko cover karti hai, jahan

ln u

defined bhi nahi hota.

Limits change karne se ek "

0

2

" integral aise ban sakti hai jahan lower limit upper limit se zyaada ho.

True — agar

g

us interval par decreasing hai, toh

g (a) > g (b)

; ye theek hai, FTC reversed limits ko sign flip karke handle karta hai (neeche ka worked figure aur Definite Integral & Fundamental Theorem of Calculus dekho).

d u

match karwaane ke liye

\frac{1}{3}

ka ek stray factor integral se bahar pull kiya ja sakta hai.

True —

\frac{1}{3}

ek constant hai, aur integration constants ke liye linear hai, isliye constant factors integral ke andar ya bahar freely move ho sakte hain.

d u

match karwaane ke liye

\frac{1}{x}

ka ek stray factor integral se bahar pull kiya ja sakta hai.

False —

\frac{1}{x}

ek variable hai, constant nahi; ye integral nahi chhor sakta. Tumhe

u = g (x)

x

ke liye solve karke substitute karna hoga.

Sahi substitution ke baad nayi integrand mein fir bhi

x

aa sakta hai.

False (agar poori tarah se kiya ho) — ek completed substitution har

x

-piece ko eliminate kar deta hai; ek aatka hua

x

ya toh incomplete conversion ya galat

u

choice signal karta hai.

Ek definite integral ke liye, limits change karne ke baad double-check ke liye back-substitute karna chahiye.

False — limits already change karne ke baad back-substituting endpoints ko double-translate karta hai; changed-limit answer pehle se ek pure number hai aur use back-sub ki zaroorat nahi.

\int \frac{1}{g ( x )} g^{'} (x) d x

hamesha theek hai jab tak

g^{'}

present ho.

False — tumhe

g (x) > 0

bhi chahiye interval mein throughout, warna

g (x)

(aur

1/ g (x)

) reals se bahar chali jaati hai; ek valid substitution ko har

u

-value ko nayi integrand ke domain ke andar rakhna chahiye.

Spot the error — misconception ke hisaab se group kiya gaya

Group A — $d x \to d u$ swap (height-rescale) skip kar diya gaya.

\int 2 x (x^{2} + 1)^{5} d x

: let

u = x^{2} + 1

, so

\int u^{5} d x = \frac{u ^{6}}{6} + C

." Slip kahan hai?

Integral mein abhi bhi

d x

likha hai,

d u

nahi —

2 x d x

ko kabhi convert nahi kiya gaya, isliye height-rescale missing hai. Sahi tarike se

d u = 2 x d x

, jo

\int u^{5} d u

deta hai.

\int x e^{x^{2}} d x

: let

u = x^{2}

d u = 2 x d x

, so this is

\int e^{u} d u = e^{x^{2}} + C

." Missing piece dhundho.

Hamare paas sirf

x d x = \frac{1}{2} d u

hai,

2 x d x = d u

nahi, isliye ek constant factor

\frac{1}{2}

drop ho gaya; sahi jawab

\frac{1}{2} e^{x^{2}} + C

hai.

Group B — inner derivative actually present nahi hai.

\int sin (x^{2}) d x

: let

u = x^{2}

d u = 2 x d x

, so

\int sin u d u

." Ye galat kyun hai?

Integrand mein

d u

supply karne ke liye koi

2 x d x

nahi hai; inner function ka derivative absent hai, isliye u-sub apply nahi hota (is integral ka koi elementary answer nahi hai).

\int cos x d x

: let

u = cos x

d u = - sin x d x

." Ye poor choice kyun hai?

Integrand mein

d u

ke saath pair karne ke liye koi

sin x

nahi hai, aur integral pehle se elementary hai; koi present derivative se match na karne wala nickname sirf cheezein mushkil banata hai.

Group C — nayi $u$ -axis par limits aur orientation.

\int_{0}^{2} x x^{2} + 1 d x

with

u = x^{2} + 1

gives

\frac{1}{3} u^{3/2}_{0}^{2}

." Kya galat hua?

Limits change nahi ki gayi — purana interval

[0, 2]

nayi

u

-axis par rakha gaya.

u = x^{2} + 1

ke saath:

x = 0 \Rightarrow u = 1

x = 2 \Rightarrow u = 5

, isliye bounds

1

5

hone chahiye.

\int_{1}^{- 1} \frac{d u}{1 - u}

ek substitution se — lower limit (

1

) upper (

- 1

) se zyaada hai, isliye main unhe freely swap kar lunga." Kya ye safe hai?

Tum sirf tabhi swap kar sakte ho agar negate bhi karo:

\int_{1}^{- 1} = - \int_{- 1}^{1}

. Sign flip ke bina silently swap karna value badal deta hai (neeche ka orientation figure dekho).

Group D — domain/absolute-value oversights.

\int_{- 2}^{- 1} \frac{1}{x} d x = ln x_{- 2}^{- 1}

." Kya break hota hai?

ln x

negative

x

ke liye undefined hai; is interval par antiderivative

ln ∣ x ∣

hai, isliye value

ln ∣ - 1 ∣ - ln ∣ - 2 ∣ = - ln 2

hai.

Orientation: decreasing $u$ limits kyun flip karta hai

Worked flip:

\int_{0}^{π} sin x cos x d x

with

u = sin x

. Nayi limits kya hain, aur value kya hai?

x = 0 \Rightarrow u = 0

aur

x = π \Rightarrow u = sin π = 0

; limits coincide karti hain, isliye integral

\int_{0}^{0} u d u = 0

hai — sweep

u = 1

tak jaati hai aur

0

par wapas aati hai, aur dono halves exactly cancel ho jaate hain.

Edge cases

Agar

u = g (x)

dono endpoints par same value leta hai (jaise

g (a) = g (b)

), toh definite integral kya hai?

Zero — nayi limits coincide karti hain, aur

\int_{c}^{c} f (u) d u = 0

hota hai

f

chahe kuch bhi ho, kyunki range ek point par collapse ho gayi hai.

Kya hoga agar substitution

u = g (x)

[a, b]

par one-to-one nahi hai (woh andar se turn around karta hai)?

Tum blindly

g (a)

aur

g (b)

ko limits ki tarah use nahi kar sakte; turning point par interval split karo taaki

g

har piece par monotonic ho, phir har piece par alag-alag substitute karo. Dekho Definite Integral & Fundamental Theorem of Calculus.

Kya hoga agar

g

[a, b]

ke kuch hisse ko us domain ke bahar map kare jahan integrand sense nahi banaata (jaise jahan

g (x)

1/ u

undefined ho)?

Wahaan substitution invalid hai — u-sub ke liye zaroori hai ki

g

continuous ho aur poore interval mein har

u

-value ko nayi integrand ke domain ke andar rakhe; warna tum interval ko ek aisi region tak restrict ya split karo jahan ye kaam kare.

U-substitution ke liye

g

continuous hona zaroori hai, ya sirf monotonic?

Ise

[a, b]

par

g

(aur

g^{'}

) continuous chahiye taaki differential

d u = g^{'} (x) d x

aur FTC apply ho sakein; monotonicity ek extra comfort hai jo tumhe directly

g (a), g (b)

use karne deti hai, lekin continuity non-negotiable requirement hai.

Agar chosen

u

d u

galat sign ke saath aaye (jaise

d u = - sin x d x

lekin tumhare paas

+ sin x d x

hai), toh kya karo?

Sign ko ek constant

- 1

ke roop mein absorb karo:

sin x d x = - d u

likho; negative ek constant hai aur integral ke bahar safely chali jaati hai.

\int \frac{g ^{'} ( x )}{g ( x )} d x

integrate karte waqt, jawab

ln ∣ g (x) ∣ + C

kyun hona chahiye na ki

ln g (x) + C

u = g (x)

ke saath ye

\int \frac{d u}{u} = ln ∣ u ∣ + C

ban jaata hai; absolute value un intervals ko cover karta hai jahan

g (x) < 0

ho, jahan

ln g (x)

undefined hoga.

Definite integral mein limits change karne par constant of integration

C

ka kya hota hai?

Ye bilkul gayab ho jaata hai —

F (u)

ko dono limits par evaluate karke subtract karna kisi bhi

+ C

ko cancel kar deta hai, yahi wajah hai ki definite answers bare numbers hote hain.

Agar

g^{'} (x) = 0

ek interior point par ho lekin

g

overall abhi bhi monotonic ho, toh kya u-sub valid hai?

Haan — ek single zero-slope point substitution nahi todta jab tak

g

interval par still one-to-one ho; change-of-limits formula ko sirf endpoint values

g (a), g (b)

chahiye.

Kya u-substitution kabhi ek integral ko zyaada mushkil bana sakti hai?

Haan —

u

ka ek poor choice ek aatka hua

x

chhor sakta hai jise messily re-express karna padega, jo original se kuch zyaada mushkil produce karta hai;

u

choose karna ek judgement hai jo inner-derivative pair spot karne se guide hota hai.

Recall Ek-line survival checklist

Kya maine $d x$ ko $d u$ mein convert kiya (height-rescale)? Kya maine $u$ choose kiya jiska derivative present ho (ek constant tak)? Definite integrals ke liye, kya maine limits change kiye aur back-substitute nahi kiya? $1/ u$ ya $\cdot$ ke liye, kya maine absolute values / domain ka dhyan rakha? Agar koi variable $d u$ se match nahi karta, kya maine $x$ ke liye $u = g (x)$ solve kiya instead of bahar pull karne ke?

Connections

U-substitution — technique, change of limits for definite integrals
Chain Rule — woh rule jise u-sub invert karta hai.
Antiderivatives & Indefinite Integrals — jahan $+ C$ aur $ln ∣ u ∣$ rehte hain.
Definite Integral & Fundamental Theorem of Calculus — limit-changing aur sign flips justify karta hai.
Integration by Parts — contrast: product rule ko invert karta hai.
Trigonometric Substitution — ek specialized u-sub.

Why questions — the geometric heart

True or false — justify karo

Spot the error — misconception ke hisaab se group kiya gaya

Orientation: decreasing u limits kyun flip karta hai

Edge cases

Connections

Orientation: decreasing $u$ limits kyun flip karta hai