Exercises — Antiderivative — definition, family of solutions (+C)

Power Rule with $n=5$: power ko ek se badhao yaani $6$ kar do, phir us nayi power se divide karo. $\int x^{5}dx=\frac{x^6}{6}+C.$ Check (differentiate forward): $\frac{d}{dx}\left(\frac{x^6}{6}\right)=\frac{6x^5}{6}=x^5.✓$

Dono "known-slope" facts hain jo ulte padhe ja rahe hain. Kyunki $\frac{d}{dx}\sin x=\cos x$: $\int \cos xdx=\sin x+C$. Kyunki $\frac{d}{dx}{e}^x=e^x$ (exponential apna khud ka slope hai): $\int e^{x}dx=e^x+C$.

Woh kaun sa function hai jiska slope $1$ har jagah ho? Slope $1$ wali ek straight line: $x$. $\int 1dx=x+C,\frac{d}{dx}(x+C)=1.✓$ (Yeh Power Rule hai $n=0$ ke saath: $x^0=1\to \frac{x^1}{1}=x$.)

Term by term integrate karo (sum ka integral, integrals ka sum hota hai; constants factor out ho jaate hain). $\int 3x^{2}dx=3\cdot \frac{x^3}{3}=x^3,\int (-4x)dx=-4\cdot \frac{x^2}{2}=-2x^2,\int 7dx=7x.$ $\int \left(3x^2-4x+7\right)dx=x^3-2x^2+7x+C.$ Check: $\frac{d}{dx}(x^3-2x^2+7x+C)=3x^2-4x+7.✓$ (Poore expression ke liye ek $+C$, har term ke liye alag nahi.)

Root ko power ki tarah likhte hain: $\sqrt{x}=x^{1/2}$. Phir Power Rule with $n=\frac{1}{2}$. $\int x^{1/2}dx=\frac{x^{3/2}}{3/2}+C=\frac{2}{3}{x}^{3/2}+C.$ Check: $\frac{d}{dx}\left(\frac{2}{3}{x}^{3/2}\right)=\frac{2}{3}\cdot \frac{3}{2}{x}^{1/2}=x^{1/2}=\sqrt{x}.✓$

Negative power ki tarah likhte hain: $\frac{5}{x^2}=5x^{-2}$, toh $n=-2$ (jo $-1$ nahi hai, isliye Power Rule allowed hai). $\int 5x^{-2}dx=5\cdot \frac{x^{-1}}{-1}+C=-\frac{5}{x}+C.$ Check: $\frac{d}{dx}\left(-5x^{-1}\right)=-5\cdot (-1)x^{-2}=5x^{-2}=\frac{5}{x^2}.✓$

$\frac{1}{x}=x^{-1}$, toh $n=-1$. Power Rule $n+1=0$ se divide karta hai — division by zero, jo forbidden hai. Yahi woh ek exception hai. Correct antiderivative hai $\int \frac{1}{x}dx=\ln |x|+C,$ jo $x\ne 0$ par valid hai. Absolute value dono signs cover karta hai: $\frac{d}{dx}\ln |x|=\frac{1}{x}$ har $x\ne 0$ ke liye. Dekho Logarithmic & Exponential Integrals.

Nahi. "Do antiderivatives ek constant se alag hote hain" waala theorem ek connected interval chahta hai, kyunki yeh Mean Value Theorem par rely karta hai (zero slope ek single interval par ⇒ flat). Set $x\ne 0$ do intervals hain $(-\infty ,0)$ aur $(0,\infty )$ ek gap ke saath. Har piece apni height par ho sakti hai: $F(x)=\{\begin{array}{ll}\ln (-x)+C_1,& x<0,\\ \ln (x)+C_2,& x>0,\end{array}$ jahan $C_1,C_2$ independent hain. Har branch ko differentiate karne par $\frac{1}{x}$ milta hai, phir bhi $C_1\ne C_2$ allowed hai. Toh $\ln |x|+C$ us gap ke across jump waale solutions miss karta hai.

Single interval $(-\infty ,\infty )$ par: har jagah zero slope ⇒ $g(x)=C$, ek constant (MVT). $x\ne 0$ par (do intervals): $h$ har piece par constant hai lekin constants ka match karna zaroori nahi: $h(x)=\{\begin{array}{ll}{C}_1,& x<0,\\ C_2,& x>0.\end{array}$

Step 1 — general family. $\int (6x-2)dx=3x^2-2x+C.$ Step 2 — ek curve choose karne ke liye condition use karo. Point $(1,4)$ us par lie karna chahiye: $F(1)=3(1{)}^2-2(1)+C=3-2+C=1+C=4\Rightarrow C=3.$ $\boxed{F(x)=3x^2-2x+3.}$ Check: $F^{\prime }(x)=6x-2$ aur $F(1)=3-2+3=4.✓$

Position, velocity ka ek antiderivative hai (velocity, position ka slope hai). Integrate karo: $s(t)=\int (t^2-4t)dt=\frac{t^3}{3}-2t^2+C.$ $s(0)=5$ apply karo: $\frac{0}{3}-0+C=5\Rightarrow C=5$. $\boxed{s(t)=\frac{t^3}{3}-2t^2+5.}$ Sanity check ek point par: $s(3)=\frac{27}{3}-2(9)+5=9-18+5=-4$ m, aur $s^{\prime }(t)=t^2-4t=v(t).✓$

$\int (\cos x+e^x)dx=\sin x+e^x+C.$ Condition: $F(0)=\sin 0+e^0+C=0+1+C=0\Rightarrow C=-1$. $\boxed{F(x)=\sin x+e^x-1.}$ Check: $F^{\prime }(x)=\cos x+e^x$, $F(0)=0+1-1=0.✓$

$2x$ ka ek antiderivative $x^2+C$ hai. Fundamental Theorem of Calculus se, $[x^2+C]$ ko $1$ se $3$ tak evaluate karo: $(3^2+C)-(1^2+C)=9+C-1-C=8.$ $+C$ cancel ho jaata hai ($C-C=0$), isliye answer plain number $8$ hai — yahi wajah hai ki definite integrals kabhi constant carry nahi karte.

Maan lo $H(x)=G(x)-F(x)$. Toh $H^{\prime }(x)=G^{\prime }(x)-F^{\prime }(x)=0$ har $x\in I$ ke liye. Ek function jiska slope single interval par zero ho, woh rise ya fall nahi kar sakta, isliye woh constant hai — yeh precisely Mean Value Theorem guarantee karta hai. Isliye $H(x)=C$, yaani $G(x)=F(x)+C.✓$

Note karo $|x{|}^3=|x{|}^3$; kyunki $|x{|}^2=x^2$ aur cubing $x$ ka sign rakhhti hai, hume milta hai $|x{|}^3=x^2\cdot |x|$. Sign se split karo:

• $x>0$ ke liye: $|x{|}^3=x^3$, toh $F(x)=\frac{x^3}{3}$, $F^{\prime }(x)=x^2.✓$
• $x<0$ ke liye: $|x|=-x$, toh $|x{|}^3=(-x{)}^3=-x^3$, jis se $F(x)=-\frac{x^3}{3}$ aur $F^{\prime }(x)=-x^2\ne x^2.$ ✗

Toh $F$, $x<0$ par $x^2$ ka antiderivative nahi hai. $\mathbb{R}$ par sach mein general antiderivative simple wala $\frac{x^3}{3}+C$ hai (check: $x$ ke har sign ke liye derivative $x^2$ hai, including $x=0$).

Kyunki $\frac{d}{dx}\tan x={\sec }^{2}x$, dono $\tan x+C$ aur $\tan x+3$ ka derivative ${\sec }^{2}x$ hai, isliye dono ek valid antiderivative hain. Lekin sirf $\tan x+C$ general answer hai (poori family); $\tan x+3$ ek particular member hai ($C=3$). Ek specific number likhna $+C$ ki jagah silently infinitely many solutions discard karta hai.