Antiderivative — definition, family of solutions (+C): Worked examples

4.2.1 · D3 · Maths › Calculus II — Integration › Antiderivative — definition, family of solutions (+C)

Scenario matrix

Har antiderivative problem jo tum encounter karoge, in cells mein se kisi ek mein fit hogi. Sabse right column us example ka naam deta hai jo us cell ko clear karta hai.

Cell	Kya cheez tricky banati hai	Example
A. Plain power, $n \geq 0$	bas power rule seedha apply karo	Ex 1
B. Negative power, $n \neq = - 1$	power rule hi lagega — lekin exponent ka sign flip hoga	Ex 2
C. Forbidden $n = - 1$	power rule zero se divide karta hai → logarithm	Ex 3
D. Disconnected domain	ek gap ka matlab hai do constants, ek nahi	Ex 3
E. Particular solution	ek initial condition ek hi curve ko pin karta hai	Ex 4
F. Sum / constant multiple	linearity: alag-alag todkar solve karo	Ex 5
G. Trig + degenerate check	sign traps; special angles par verify karo	Ex 6
H. Real-world word problem	velocity → position, units matter karte hain	Ex 7
I. Exam twist (pehle rewrite karo)	integrate hota nahi lagta jab tak algebra na karo	Ex 8

Hum A B C D E F G H I — har cell — cover karenge.

Ex 1 — Cell A: plain power

Worked example Statement

$\int x^{4} d x$ nikalo.

Forecast: aage padhne se pehle, upar wale power ka andaza lagao. $x^{5}$ ko differentiate karne par $5 x^{4}$ milta hai — toh kaun sa power aur kaun sa divisor hume wapas $x^{4}$ par le jaayega?

Step 1. Power ko ek se badhao: $x^{4} \to x^{5}$ . Yeh step kyun? Differentiation power ko ek se ghataata hai, toh ise reverse karne ke liye hume ek badhana hoga. Hum kuch aisa chahte hain jiski derivative $x^{4}$ ho, aur $x^{5}$ ek hi aisi power hai jiski derivative mein $x^{4}$ aata hai.

Step 2. Naye power se divide karo taaki differentiation jo factor bahar nikalti hai, woh cancel ho jaaye. $\frac{d}{d x} (\frac{x ^{5}}{5}) = \frac{5 x ^{4}}{5} = x^{4} .$ Yeh step kyun? $x^{5}$ ko differentiate karne par saamne ek stray $5$ aa jaata hai; $5$ se divide karne par woh neutralise ho jaata hai.

Answer: $\int x^{4} d x = \frac{x ^{5}}{5} + C .$

Verify: $\frac{d}{d x} (\frac{x ^{5}}{5} + C) = x^{4}$ . ✓

Yeh clean Power Rule (Integration) hai $n = 4$ ke saath: $\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + C$ .

Ex 2 — Cell B: negative power (lekin $- 1$ nahi)

Ex 3 — Cells C & D: forbidden power aur split domain

Worked example Statement

$\int \frac{1}{x} d x$ nikalo, aur poori general family do.

Forecast: $n = - 1$ ke saath power rule try karo. $n + 1$ se divide karte hi kya galat ho jaata hai?

Step 1. Power rule try karo: $n + 1 = - 1 + 1 = 0$ , toh formula demand karta hai $\frac{x ^{0}}{0}$ — zero se division. Yeh step kyun? Exactly woh jagah dekhna jahan rule fail hota hai, humein bataata hai ki hume alag tool chahiye. Yeh sawaal ki "kis function ki slope $\frac{1}{x}$ hai?" ka jawaab koi power nahi balki natural logarithm deta hai.

Step 2. Yaad karo $\frac{d}{d x} ln x = \frac{1}{x}$ for $x > 0$ . Lekin $\frac{1}{x}$ , $x < 0$ ke liye bhi exist karta hai, jahan $ln x$ undefined hai. Fix hai absolute value: $\frac{d}{d x} ln ∣ x ∣ = \frac{1}{x} (x \neq = 0) .$ Yeh step kyun? $∣ x ∣$ ek hi formula ko $x$ ke dono signs cover karne deta hai (figure dekho — do branches hain).

Step 3 (Cell D). Domain $x \neq = 0$ do disconnected intervals hai. "No-other-antiderivatives" theorem (jo Mean Value Theorem se aata hai) ko ek connected interval chahiye, toh har branch ko apna khud ka constant milta hai: $\int \frac{1}{x} d x = {ln (- x) + C_{1}, ln (x) + C_{2}, x < 0 x > 0.$ Yeh step kyun? Kuch bhi left branch ko right branch ki same height par force nahi karta — unke beech ek gap hai, toh woh independent hain.

Verify: $x > 0$ ke liye, $\frac{d}{d x} ln x = \frac{1}{x}$ ✓. $x < 0$ ke liye, $\frac{d}{d x} ln (- x) = \frac{- 1}{- x} = \frac{1}{x}$ ✓.

Ex 4 — Cell E: initial condition se particular solution

Ex 5 — Cell F: sum aur constant multiple (linearity)

Worked example Statement

$\int (3 x^{2} + 5 cos x - \frac{2}{x ^{2}}) d x$ nikalo.

Forecast: integration aapko har term ko alag-alag attack karne deta hai aur constants ko bahar nikalane deta hai. Check karne se pehle teeno pieces predict karo.

Step 1. Linearity se alag karo: $\int 3 x^{2} d x + \int 5 cos x d x - \int 2 x^{- 2} d x .$ Yeh step kyun? Differentiation linear hai (sum ki derivative, derivatives ka sum hoti hai), toh iska reverse bhi aisa hi hoga. Yeh ek mushkil-dikhne wale integral ko teen aasaan waalon mein badal deta hai.

Step 2. Har piece solve karo:

$\int 3 x^{2} d x = 3 \cdot \frac{x ^{3}}{3} = x^{3}$ .
$\int 5 cos x d x = 5 sin x$ (kyunki $\frac{d}{d x} sin x = cos x$ ).
$\int 2 x^{- 2} d x = 2 \cdot \frac{x ^{- 1}}{- 1} = - 2 x^{- 1} = - \frac{2}{x}$ ; leading minus ke saath yeh $+ \frac{2}{x}$ ban jaata hai.

Step 3. Ek single $+ C$ ke saath dobara combine karo (ek constant poore sum ko cover karta hai, connected domain par): $\int (3 x^{2} + 5 cos x - \frac{2}{x ^{2}}) d x = x^{3} + 5 sin x + \frac{2}{x} + C .$

Verify: $\frac{d}{d x} (x^{3} + 5 sin x + \frac{2}{x} + C) = 3 x^{2} + 5 cos x - \frac{2}{x ^{2}}$ ✓.

Ex 6 — Cell G: trig with degenerate-angle sanity check

Worked example Statement

$f (x) = sin x$ ka antiderivative $F$ nikalo jahan $F (π /2) = 0$ .

Forecast: kya differentiate hoke $sin x$ deta hai? Sign ke saath careful raho — $\frac{d}{d x} cos x = - sin x$ hota hai, $+ sin x$ nahi.

Step 1. Family: $\int sin x d x = - cos x + C$ . Yeh step kyun? Kyunki $\frac{d}{d x} cos x = - sin x$ hai, hume ek leading minus chahiye taaki do negatives cancel hokar $+ sin x$ de sakein. Yahi sign classic trap hai.

Step 2. $F (π /2) = 0$ apply karo (ek degenerate/special angle jahan $cos (π /2) = 0$ hota hai): $- cos (\frac{π}{2}) + C = - 0 + C = 0 \Rightarrow C = 0.$ Yeh step kyun? Special angle choose karne se arithmetic exact rehti hai aur yeh sanity check bhi ban jaata hai.

Answer: $F (x) = - cos x .$

Verify: $\frac{d}{d x} (- cos x) = sin x$ ✓, aur $F (π /2) = - cos (π /2) = 0$ ✓. Ek aur angle bhi test karo: $F (0) = - cos 0 = - 1$ , aur wahan slope hai $sin 0 = 0$ — curve $x = 0$ par momentarily flat hai. ✓

Ex 7 — Cell H: real-world word problem (velocity → position)

Worked example Statement

Ek ball upar phenki jaati hai. Uski velocity (metres per second) hai $v (t) = 20 - 10 t$ . $t = 0$ par uski height $s (0) = 2$ m hai. Height $s (t)$ nikalo, aur peak par ball kahan hai.

Forecast: velocity, position ki slope hai, toh position, velocity ka ek antiderivative hai. Andaza lagao: velocity kab zero hogi (peak par)?

Step 1. Position, velocity ka antiderivative hai: $s (t) = \int (20 - 10 t) d t = 20 t - 10 \cdot \frac{t ^{2}}{2} + C = 20 t - 5 t^{2} + C .$ Yeh step kyun? $s^{'} (t) = v (t)$ velocity ki definition se hai, toh $v$ se $s$ recover karna exactly antidifferentiation hai. Units: $[20 t]$ hai $\frac{m}{s} \cdot s = m$ ✓; $[5 t^{2}]$ hai $\frac{m}{s ^{2}} \cdot s^{2} = m$ ✓.

Step 2. $s (0) = 2$ use karo: $20 (0) - 5 (0)^{2} + C = 2 \Rightarrow C = 2$ . Toh $s (t) = 20 t - 5 t^{2} + 2.$ Yeh step kyun? Initial height woh ek piece of info hai jo $C$ fix karti hai — ball ki launch height.

Step 3. Peak wahan hai jahan velocity $= 0$ : $20 - 10 t = 0 \Rightarrow t = 2$ s. Wahan height: $s (2) = 20 (2) - 5 (2)^{2} + 2 = 40 - 20 + 2 = 22 m .$ Yeh step kyun? Top par, ball momentarily uthna band kar deti hai ( $v = 0$ ), toh $s$ ki slope zero hoti hai — ek maximum.

Verify: $s^{'} (t) = 20 - 10 t = v (t)$ ✓; $s (0) = 2$ ✓; peak height $22$ m at $t = 2$ ✓.

Ex 8 — Cell I: exam twist (integrate karne se pehle rewrite karo)

Recall Quick self-test (answers chhupa lo)

Kaun sa cell power rule fail karta hai, aur kya replace karta hai? ::: Cell C, $n = - 1$ ; replaced by $ln ∣ x ∣ + C$ . Ex 4 mein $C$ kya hai? ::: $C = 12$ . Ex 7 mein peak height aur time? ::: $22$ m at $t = 2$ s. Ex 3 mein do constants kyun hain? ::: Domain $x \neq = 0$ do disconnected intervals hai; har ek ko apna constant milta hai. Ex 8 ke quotient par pehla move? ::: Integrate karne se pehle fraction ko $x + x^{- 1/2}$ mein split karo.

Connections

Power Rule (Integration) — Ex 1, 2, 8 aur Ex 3 mein iska $n = - 1$ failure.
Logarithmic & Exponential Integrals — Ex 3 ka $ln ∣ x ∣$ .
Mean Value Theorem — kyun ek interval ka matlab ek constant hai (Ex 3, 5).
Differential Equations — Initial Value Problems — Ex 4, 6, 7 mein $C$ fix karna.
Fundamental Theorem of Calculus / Definite Integral — jahan yeh families $+ C$ ko cancel hone deti hain.