U-substitution — technique, change of limits for definite integrals: Worked examples

4.2.6 · D3 · Maths › Calculus II — Integration › U-substitution — technique, change of limits for definite in

Humara ek tool ka reminder (parent se):

Yahan $g (x)$ woh inner function hai jise hum $u$ kehte hain, aur $g^{'} (x)$ uska derivative hai (uski "slope"). Dekho Chain Rule — kyun andar ka derivative zaroor present hona chahiye.

Do antiderivative rules jo hum baar baar use karte hain (from Antiderivatives & Indefinite Integrals):

Cell	Kya mushkil banaata hai	Covered by
A. Perfect match	andar ka derivative wahan hi baitha hai	Ex 1
B1. Positive constant mismatch	tumhare paas $x d x$ hai par chahiye $c x d x$ ( $c > 0$ )	Ex 2
B2. Negative constant mismatch	$g^{'}$ mein minus sign hai — tumhe " $-$ " drag karna padega	Ex 3
C. Leftover variable	ek stray $x$ cancel nahi hoga — $x$ ko $u$ ke terms mein solve karo	Ex 4
*D. Definite, limits cross* (decreasing inside)**	andar ka function decrease karta hai → $u (b) < u (a)$ , reversed limits	Ex 5
E. Degenerate limits ( $u (a) = u (b)$ )	naye limits equal hain → answer $0$ hai bina kisi kaam ke	Ex 6
F1. Trig inside	inner function ek trig function hai jiska derivative present hai	Ex 7
F2. Logarithm $g^{'} / g$ form	numerator exactly denominator ka derivative hai	Ex 8
G. Real-world word problem	tumhe pehle integral banana padega	Ex 9
H. Exam twist (rewrite before you can see $u$ )	koi obvious inner function nahi jab tak split/manipulate na karo	Ex 10

Ex 2 — Cell B1: positive constant fix karo

Worked example Statement

Evaluate $\int cos (5 x) d x$ .

Forecast: Inside hai $5 x$ ; uska derivative hai $5$ , ek constant. Humare paas saamne $5$ nahi hai. Kyunki $5$ ek constant hai, hum ise conjure kar sakte hain. Guess karo ki answer mein $5$ hoga ya $\frac{1}{5}$ .

Choose $u = 5 x$ . Kyun? Yeh $cos$ ka inner function hai.
Differentiate: $d u = 5 d x \Rightarrow d x = \frac{1}{5} d u$ . Kyun? Humein $d x$ replace karna hai, lekin sirf $d x$ (na ki $5 d x$ ) appear karta hai, toh uske liye solve karo.
Replace: $\int cos (u) \cdot \frac{1}{5} d u = \frac{1}{5} \int cos u d u$ . Kyun? $\frac{1}{5}$ ek constant hai, toh yeh integral se bahar ja sakta hai (linearity — dekho Antiderivatives & Indefinite Integrals). Koi variable kabhi nahi ja sakta.
Integrate: $\frac{1}{5} sin u + C$ . Kyun? $\int cos u d u = sin u$ (basic antiderivative — $\frac{d}{d u} sin u = cos u$ ka reverse).
Back-substitute: $\frac{1}{5} sin (5 x) + C$ .

Verify: $\frac{d}{d x} \frac{1}{5} sin (5 x) = \frac{1}{5} cos (5 x) \cdot 5 = cos (5 x)$ . ✓

Ex 3 — Cell B2: NEGATIVE constant mismatch (minus carry karo)

Worked example Statement

Evaluate $\int x e^{- x^{2}} d x$ .

Forecast: Exponential ke andar hai $- x^{2}$ ; uska derivative hai $- 2 x$ — ek negative constant times $x$ . Humare paas sirf $+ x d x$ hai. Dekho kaise ek " $-$ " sign drag karna padta hai. Guess karo ki answer $+ \frac{1}{2}$ se shuru hoga ya $- \frac{1}{2}$ se.

Choose $u = - x^{2}$ . Kyun? Yeh $e^{(\cdot)}$ ka inner function hai.
Differentiate: $d u = - 2 x d x \Rightarrow x d x = - \frac{1}{2} d u$ . Kyun? Us chunk ke liye solve karo jo actually hamare paas hai, $x d x$ . $- 2$ se divide karne par minus sign bana rehta hai — yahi Cell B2 ka poora point hai.
Replace: $\int e^{u} \cdot (- \frac{1}{2}) d u = - \frac{1}{2} \int e^{u} d u$ . Kyun? $- \frac{1}{2}$ ek constant hai, toh yeh integral se bahar jaata hai, apna sign saath lekar.
Integrate: $- \frac{1}{2} e^{u} + C$ . Kyun? $\int e^{u} d u = e^{u}$ (exponential apna khud ka antiderivative hai).
Back-substitute: $- \frac{1}{2} e^{- x^{2}} + C$ .

Verify: $\frac{d}{d x} (- \frac{1}{2} e^{- x^{2}}) = - \frac{1}{2} e^{- x^{2}} \cdot (- 2 x) = x e^{- x^{2}}$ . ✓ Dono minus signs multiply hokar plus bante hain — exactly integrand recover ho jaata hai.

Ex 4 — Cell C: ek stray variable jo cancel nahi hoga

Worked example Statement

Evaluate $\int x x + 1 d x$ .

Forecast: Root ke andar hai $x + 1$ ; uska derivative hai $1$ . Lekin ek extra akela $x$ multiply kar raha hai — aur $x$ ek constant nahi hai, toh hum ise bahar nahi nikal sakte. Iske saath kya karein? Predict: hum $x$ ko $u$ mein rewrite karenge.

Choose $u = x + 1$ . Kyun? Yeh root ke andar hai; uska derivative $1$ makes $d u = d x$ clean.
Differentiate: $d u = d x$ .
Stray $x$ handle karo: $u = x + 1$ se hum paate hain $x = u - 1$ . Yeh step kyun? Bache hue $x$ ko constant ki tarah bahar nahi nikala ja sakta (yeh classic mistake hai). Bajaay, har $x$ ko $u$ ke terms mein express karo, is wale ko bhi.
Replace: $\int (u - 1) u d u = \int (u^{3/2} - u^{1/2}) d u$ . Kyun? $u = u^{1/2}$ distribute karo taaki hum $u$ ki powers integrate kar sakein.
Integrate: $\frac{u ^{5/2}}{5/2} - \frac{u ^{3/2}}{3/2} = \frac{2}{5} u^{5/2} - \frac{2}{3} u^{3/2}$ . Kyun? Power rule term-by-term apply kiya $n = \frac{3}{2}$ aur $n = \frac{1}{2}$ ke saath.
Back-substitute: $\frac{2}{5} (x + 1)^{5/2} - \frac{2}{3} (x + 1)^{3/2} + C$ .

Verify: Differentiate karo: $\frac{2}{5} \cdot \frac{5}{2} (x + 1)^{3/2} - \frac{2}{3} \cdot \frac{3}{2} (x + 1)^{1/2} = (x + 1)^{3/2} - (x + 1)^{1/2} = (x + 1)^{1/2} [(x + 1) - 1] = x x + 1$ . ✓

Ex 5 — Cell D: definite integral jahan limits CROSS karti hain (decreasing inside)

Worked example Statement

Evaluate $\int_{0}^{1} \frac{x}{( 4 - x ^{2} ) ^{2}} d x$ .

Forecast: Inside hai $u = 4 - x^{2}$ , jo $x$ badhne ke saath decrease karta hai. Toh jab $x$ , $0 \to 1$ chalta hai, $u$ , $4 \to 3$ chalta hai — naya upper limit ( $3$ ) naye lower limit ( $4$ ) se neeche hai. Limits literally cross karti hain. Dekho hum reversed range kaise handle karte hain. Ek chota positive answer guess karo.

Neeche wala figure inner function $u = 4 - x^{2}$ (pale yellow) ko $x \in [0, 1]$ ke across plot karta hai. Blue dot left endpoint $x = 0$ mark karta hai, jo $u = 4$ par high hai; pink dot right endpoint $x = 1$ mark karta hai, jo $u = 3$ tak neeche aa gaya hai. Curve par white arrow $x$ badhne ke saath downhill sweep dikhata hai — exactly yahi reason hai ki naye $u$ -limits reversed aate hain ( $4 \to 3$ chal rahe hain). Ise kyun include kiya? "Limits cross" ko kuch aisa banana jise tum dekh sako: endpoints substitution ke under apna up/down order swap kar lete hain.

Choose $u = 4 - x^{2}$ , $d u = - 2 x d x \Rightarrow x d x = - \frac{1}{2} d u$ . Kyun? Squared bracket ke andar; uska derivative $- 2 x$ saamne wale $x$ ke proportional hai, ek minus ke saath (Cell B2 flavour, definite integral ke andar reuse hua).
Limits translate karo (variable ab $u$ hai; purane numbers range describe nahi karte — dekho Definite Integral & Fundamental Theorem of Calculus):
- $x = 0 \Rightarrow u = 4 - 0 = 4$
- $x = 1 \Rightarrow u = 4 - 1 = 3$ Toh $u$ $4$ se $3$ tak neeche jaata hai — picture yeh reversed sweep dikhati hai.
Replace (limits jis order mein aaye usi mein rakho): $\int_{4}^{3} \frac{1}{u ^{2}} \cdot (- \frac{1}{2}) d u = - \frac{1}{2} \int_{4}^{3} u^{- 2} d u$ . Kyun? Limits ko silently flip mat karo; $4 \to 3$ exactly waise likho jaise substitution ne produce kiya.
Swap rule se flip karo: $- \frac{1}{2} \int_{4}^{3} = + \frac{1}{2} \int_{3}^{4}$ . Kyun? $\int_{a}^{b} = - \int_{b}^{a}$ : limits swap karne se sign flip hota hai, jo leading minus ko cancel kar deta hai. Yeh crossed limits handle karne ka clean tarika hai.
Integrate: $\frac{1}{2} \cdot \frac{u ^{- 1}}{- 1} = - \frac{1}{2 u}$ , $3$ se $4$ tak evaluate kiya. Kyun? Power rule with $n = - 2$ : $\int u^{- 2} d u = \frac{u ^{- 1}}{- 1}$ .
Evaluate: $- \frac{1}{2 \cdot 4} - (- \frac{1}{2 \cdot 3}) = - \frac{1}{8} + \frac{1}{6} = \frac{- 3 + 4}{24} = \frac{1}{24}$ .

Verify: $\frac{1}{24} \approx 0.0417$ — chota aur positive jaise forecast kiya tha; original ka $0$ se $1$ tak numeric integration $0.041 \overline{6}$ deta hai. ✓

Ex 6 — Cell E: degenerate limits ( $u (a) = u (b)$ ), aur iska geometry kyun hai

Worked example Statement

Evaluate $\int_{- 1}^{1} 2 x (x^{2} + 3)^{3} d x$ .

Forecast: Inner function $u = x^{2} + 3$ . $x = - 1$ par, $u = 4$ ; $x = 1$ par, $u = 4$ phir. Naye limits equal hain. Koi bhi algebra karne se pehle answer predict karo.

Choose $u = x^{2} + 3$ , toh $d u = 2 x d x$ . Kyun? $2 x$ exactly inside ka derivative hai — $u$ -piece par perfect match.
Limits translate karo:
- $x = - 1 \Rightarrow u = (- 1)^{2} + 3 = 4$
- $x = 1 \Rightarrow u = (1)^{2} + 3 = 4$
Replace: $\int_{4}^{4} u^{3} d u$ . Kyun? Upar aur neeche ke limits identical hain.
Integrate & evaluate: zero-width interval par integral $0$ hai. Kyun? $\int_{c}^{c} (kuch bhi) = 0$ — koi width nahi, koi area nahi.

Verify (do tarike se): (a) $\int_{4}^{4} = 0$ by definition. (b) Integrand $2 x (x^{2} + 3)^{3}$ ek odd function hai ( $x \to - x$ flip karo aur yeh sign badal leta hai), symmetric interval $[- 1, 1]$ par integrate kiya, toh positive aur negative areas exactly cancel ho jaate hain. ✓

Ex 7 — Cell F1: trig inside

Ex 8 — Cell F2: $g^{'} / g$ logarithm pattern

Worked example Statement

Evaluate $\int \frac{2 x + 1}{x ^{2} + x + 7} d x$ .

Forecast: Bottom $x^{2} + x + 7$ dekho. Uska derivative $2 x + 1$ hai — exactly top! Jab bhi numerator denominator ka derivative ho, answer ek natural log hota hai. $ln ∣ denominator ∣$ predict karo.

Choose $u = x^{2} + x + 7$ . Kyun? Uska derivative numerator mein baitha hai — special " $\int \frac{g ^{'}}{g} = ln ∣ g ∣$ " signature.
Differentiate: $d u = (2 x + 1) d x$ . Kyun? Poora numerator times $d x$ , $d u$ ban jaata hai.
Replace: $\int \frac{1}{u} d u$ . Kyun? Denominator $x^{2} + x + 7$ , $u$ ban jaata hai, aur poora numerator $(2 x + 1) d x$ , $d u$ ban jaata hai — $\frac{d u}{u}$ bacha rehta hai, ek pure $u$ -integral bina kisi $x$ ke.
Integrate: $ln ∣ u ∣ + C$ . Kyun? Log rule — yeh $n = - 1$ case hai jo power rule nahi pakad sakta, toh $\int \frac{1}{u} d u = ln ∣ u ∣$ .
Back-substitute: $ln ∣ x^{2} + x + 7∣ + C$ . (Absolute value optional hai kyunki $x^{2} + x + 7 > 0$ always, lekin habit ke liye rakho.)

Verify: $\frac{d}{d x} ln (x^{2} + x + 7) = \frac{2 x + 1}{x ^{2} + x + 7}$ . ✓

Ex 9 — Cell G: ek real-world word problem (build then substitute)

Worked example Statement

Paani ek tank mein $r (t) = \frac{6 t}{( t ^{2} + 1 ) ^{2}}$ litres per minute ki rate se flow karta hai. Pehle $2$ minutes mein kitna paani enter karta hai?

Forecast: "Rate se total amount" ka matlab hai rate ko time interval par integrate karo — yeh hai Definite Integral & Fundamental Theorem of Calculus. Humein chahiye $\int_{0}^{2} r (t) d t$ . Inner function $t^{2} + 1$ ; derivative $2 t$ ; hamare paas $6 t$ hai — ek constant fix. Thoda $3$ litre se kam answer guess karo.

Set up: $V = \int_{0}^{2} \frac{6 t}{( t ^{2} + 1 ) ^{2}} d t$ . Kyun? Accumulated volume = time window par inflow rate ka integral (units: $\frac{L}{min} \times min = L$ ).
Choose $u = t^{2} + 1$ , $d u = 2 t d t \Rightarrow 6 t d t = 3 d u$ . Kyun? Squared bracket ke andar; constant fix karo ( $6 t = 3 \cdot 2 t$ ).
Limits translate karo:
- $t = 0 \Rightarrow u = 0 + 1 = 1$
- $t = 2 \Rightarrow u = 4 + 1 = 5$
Replace: $\int_{1}^{5} \frac{3}{u ^{2}} d u = 3 \int_{1}^{5} u^{- 2} d u$ . Kyun? Denominator $(t^{2} + 1)^{2}$ , $u^{2}$ ban jaata hai, aur poora numerator $6 t d t$ , step 2 se $3 d u$ ban jaata hai. $3$ ek constant hai, toh yeh integral se bahar ja sakta hai (koi variable nahi ja sakta) — ek clean power of $u$ bachti hai.
Integrate: $3 \cdot \frac{u ^{- 1}}{- 1} = - \frac{3}{u}$ , $1$ se $5$ tak. Kyun? Power rule with $n = - 2$ .
Evaluate: $- \frac{3}{5} - (- \frac{3}{1}) = - \frac{3}{5} + 3 = \frac{- 3 + 15}{5} = \frac{12}{5} = 2.4 litres$ . Kyun? Fundamental Theorem: upper $u = 5$ plug karo phir $- \frac{3}{u}$ mein se lower $u = 1$ subtract karo. Lower term par dono minus signs mila kar $+ 3$ banaate hain, aur $3 - \frac{3}{5}$ , $5$ ke common denominator par $\frac{12}{5}$ deta hai.

Verify: $2.4$ L positive hai, $3$ se kam jaise forecast kiya tha, aur units check karte hain ( $L/min \cdot min$ ). $[0, 2]$ par $r (t)$ ka numeric integration $2.4$ deta hai. ✓

Ex 10 — Cell H: exam twist (rewrite karo $u$ dikhne se pehle)

Worked example Statement

Evaluate $\int \frac{1}{x ln x} d x$ ( $x > 1$ ke liye).

Forecast: Koi obvious "inner function times uska derivative" nahi hai... jab tak yeh notice na karo ki $\frac{1}{x}$ , $ln x$ ka derivative hai. Toh awkward-looking $ln x$ actually real inner function hai. Ek log-of-a-log predict karo.

Mentally rewrite karo: $\frac{1}{x ln x} = \frac{1}{ln x} \cdot \frac{1}{x}$ . Yeh step kyun? Split karne se "function $ln x$ aur uska derivative $\frac{1}{x}$ " wali pair reveal hoti hai — chain-rule fingerprint jo hidden tha.
Choose $u = ln x$ , $d u = \frac{1}{x} d x$ . Kyun? $ln x$ inner function hai; uska derivative $\frac{1}{x} d x$ present hai.
Replace: $\int \frac{1}{u} d u$ . Kyun? Factor $\frac{1}{ln x}$ , $\frac{1}{u}$ ban jaata hai, aur trailing $\frac{1}{x} d x$ , $d u$ ban jaata hai — toh saare $x$ -pieces vanish ho jaate hain, $\frac{d u}{u}$ bacha rehta hai.
Integrate: $ln ∣ u ∣ + C$ . Kyun? Log rule — phir se $n = - 1$ case.
Back-substitute: $ln ∣ ln x ∣ + C$ .

Verify: $\frac{d}{d x} ln (ln x) = \frac{1}{l n x} \cdot \frac{1}{x} = \frac{1}{x l n x}$ . ✓

Matrix fill karna — ek quick recap

Recall Kaun se example ne kaun si cell ko cover kiya? (one-to-one)

A perfect match ::: Ex 1 B1 positive constant mismatch ::: Ex 2 B2 negative constant mismatch (minus carry karo) ::: Ex 3 C stray variable → $x = u - 1$ solve karo ::: Ex 4 D definite, limits cross ( $u (b) < u (a)$ ) → $\int_{a}^{b} = - \int_{b}^{a}$ se flip karo ::: Ex 5 E degenerate equal limits 0 dete hain ::: Ex 6 F1 trig inside ::: Ex 7 F2 logarithm $g^{'} / g$ ::: Ex 8 G real-world (integral banao) ::: Ex 9 H exam twist (pehle rewrite karo) ::: Ex 10

Connections

यही cases Hinglish में
Chain Rule — har "perfect match" aur "hidden inside" example is rule ka ulta hai.
Antiderivatives & Indefinite Integrals — kyun constants (variables nahi) integral se bahar jaate hain; power aur log rules ka source.
Definite Integral & Fundamental Theorem of Calculus — limit translation aur swap rule $\int_{a}^{b} = - \int_{b}^{a}$ justify karta hai (Ex 5, 6, 9).
Integration by Parts — iske liye jaao jab koi inner-derivative pair nahi ho.
Trigonometric Substitution — $a^{2} \pm x^{2}$ forms ke liye specialised sub.

Worked examples — U-substitution — technique, change of limits for definite integrals

Scenario matrix

Ex 1 — Cell A: perfect match

Ex 2 — Cell B1: positive constant fix karo

Ex 3 — Cell B2: NEGATIVE constant mismatch (minus carry karo)

Ex 4 — Cell C: ek stray variable jo cancel nahi hoga

Ex 5 — Cell D: definite integral jahan limits CROSS karti hain (decreasing inside)

Ex 6 — Cell E: degenerate limits ( $u (a) = u (b)$ ), aur iska geometry kyun hai

Ex 7 — Cell F1: trig inside

Ex 8 — Cell F2: $g^{'} / g$ logarithm pattern

Ex 9 — Cell G: ek real-world word problem (build then substitute)

Ex 10 — Cell H: exam twist (rewrite karo $u$ dikhne se pehle)

Matrix fill karna — ek quick recap

Connections