Exercises — Area between curves — horizontal and vertical slices

HUM KYA KARTE HAIN: pehle dekhte hain kahan milte hain, phir ek interior point test karte hain. Dono ko equal karo: $x+1=x^2-2x+1\Rightarrow 0=x^2-3x\Rightarrow x(x-3)=0$, toh woh $x=0$ aur $x=3$ pe milte hain (isme Solving Quadratic & Polynomial Equations ka use hota hai). Ek point test karo $x=1$ pe (interior): line deta hai $1+1=2$; parabola deta hai $1-2+1=0$. Kyunki $2>0$ hai, line $y=x+1$ poore interval $[0,3]$ mein upar hai. Ek point kyun kaafi hai: curves sirf dono endpoints pe cross karti hain, isliye beech mein order (kaun upar hai) badal nahi sakta. Answer: top $=y=x+1$, bottom $=y=x^2-2x+1$, $x=0,3$ pe milte hain.

Picture dekho (s02): parabola $x=y^2$ rightward kholti hai. Ek fixed height $y$ ke liye, region left pe parabola ($x=y^2$) se right pe line ($x=4$) tak jaata hai. Har $y$ ke liye yeh ek right curve aur ek left curve hai — ek clean horizontal slice. Ek vertical slice fixed $x$ pe curve ko $y=+\sqrt{x}$ aur $y=-\sqrt{x}$ dono jagah hit karega — top aur bottom ek hi curve ki do branches hain, $f(x)$ ke roop mein likhna awkward hai. $y$-limits: parabola $x=4$ se milti hai jab $y^2=4$, yaani $y=\pm 2$. Toh $c=-2,d=2$. Answer: horizontal slices; $y\in [-2,2]$, integrand $R(y)-L(y)=4-y^2$.

Intersections: $4-x^2=0\Rightarrow x^2=4\Rightarrow x=\pm 2$. Toh $a=-2,b=2$. Top/bottom: $x=0$ pe, curve $=4>0=$ axis, toh parabola top hai, $y=0$ bottom hai. Top minus bottom integrate karo (vertical slices, thickness $dx$): $A={\int }_{-2}^2((4-x^2)-0)dx=[4x-\frac{x^3}{3}{]}_{-2}^2.$ $x=2$ pe: $8-\frac{8}{3}=\frac{16}{3}$. $x=-2$ pe: $-8+\frac{8}{3}=-\frac{16}{3}$. $A=\frac{16}{3}-(-\frac{16}{3})=\frac{32}{3}.$ Yeh antiderivative kyun: $\frac{d}{dx}(4x)=4$ aur $\frac{d}{dx}(x^3/3)=x^2$. ✔

Intersections: $x^3=x\Rightarrow x^3-x=0\Rightarrow x(x-1)(x+1)=0$, toh $x=-1,0,1$. Interval ke andar $x=0$ pe ek crossing hai — yeh warning sign hai. Order flip hota hai! $x=0.5$ test karo: $x=0.5$, $x^3=0.125$, toh line top hai $(0,1)$ pe. $x=-0.5$ test karo: $x=-0.5$, $x^3=-0.125$, toh $x^3>x$ yahan, cube top hai $(-1,0)$ pe. Crossing pe split karo aur har piece pe (top$-$bottom) lo: $A={\int }_{-1}^0(x^3-x)dx+{\int }_0^1(x-x^3)dx.$ Pehla: $[\frac{x^4}{4}-\frac{x^2}{2}{]}_{-1}^0=0-(\frac{1}{4}-\frac{1}{2})=\frac{1}{4}$. Doosra: $[\frac{x^2}{2}-\frac{x^4}{4}{]}_0^1=(\frac{1}{2}-\frac{1}{4})-0=\frac{1}{4}$. $A=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}.$

s02 dekho. Parabola $x=y^2$ ka nose $x=0$ pe hai aur $x=1$ se guzarta hai ($y=\pm 1$ pe) aur $x=4$ se ($y=\pm 2$ pe). Line isse $(1,-1)$ aur $(4,2)$ pe milti hai. Left se right scan karte hue, ek vertical strip ka top aur bottom $x=1$ pe change hota hai:

• $0\le x\le 1$ ke liye: strip lower parabola branch $y=-\sqrt{x}$ se upper branch $y=+\sqrt{x}$ tak jaata hai. Height $=\sqrt{x}-(-\sqrt{x})=2\sqrt{x}$.
• $1\le x\le 4$ ke liye: strip line $y=x-2$ (bottom) se $y=+\sqrt{x}$ (top) tak jaata hai. Height $=\sqrt{x}-(x-2)$. $A={\int }_0^{1}2\sqrt{x}dx+{\int }_1^4(\sqrt{x}-x+2)dx.$ Pehla: $[\frac{4}{3}{x}^{3/2}{]}_0^1=\frac{4}{3}$. Doosra: $[\frac{2}{3}{x}^{3/2}-\frac{x^2}{2}+2x{]}_1^4$. $4$ pe: $\frac{2}{3}\cdot 8-8+8=\frac{16}{3}$. $1$ pe: $\frac{2}{3}-\frac{1}{2}+2=\frac{13}{6}$. Difference $=\frac{16}{3}-\frac{13}{6}=\frac{32}{6}-\frac{13}{6}=\frac{19}{6}$. $A=\frac{4}{3}+\frac{19}{6}=\frac{8}{6}+\frac{19}{6}=\frac{27}{6}=\frac{9}{2}.✓$ Lesson: same answer, lekin do integrals aur ek branch-split ki zaroorat padi, jabki single $dy$ integral kaafi tha. Horizontal yahan jeetta hai.

Andar crossing hai? $\sin x=\cos x\Rightarrow \tan x=1\Rightarrow x=\frac{\pi }{4}$, jo $[0,\frac{\pi }{2}]$ ke andar hai. Wahan split karo. Har piece pe order: $x=0$ pe, $\cos 0=1>0=\sin 0$, toh $\cos$ top hai $[0,\frac{\pi }{4}]$ pe. $x=\frac{\pi }{2}$ pe, $\sin =1>0=\cos$, toh $\sin$ top hai $[\frac{\pi }{4},\frac{\pi }{2}]$ pe. $A={\int }_0^{\pi /4}(\cos x-\sin x)dx+{\int }_{\pi /4}^{\pi /2}(\sin x-\cos x)dx.$ Pehla: $[\sin x+\cos x{]}_0^{\pi /4}=(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2})-(0+1)=\sqrt{2}-1$. Doosra: $[-\cos x-\sin x{]}_{\pi /4}^{\pi /2}=(0-1)-(-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2})=-1+\sqrt{2}$. $A=(\sqrt{2}-1)+(\sqrt{2}-1)=2\sqrt{2}-2\approx \mathrm{0.828.}$

s03 dekho. Teeno edge lines nikalo.

• $(0,0)\to (2,1)$: slope $\frac{1}{2}$, toh $y=\frac{1}{2}x$.
• $(0,0)\to (1,3)$: slope $3$, toh $y=3x$.
• $(2,1)\to (1,3)$: slope $\frac{3-1}{1-2}=-2$, toh $y-1=-2(x-2)\Rightarrow y=-2x+5$. Left se right scan karte hue top edge $x=1$ pe change hoti hai (vertex $(1,3)$):
• $0\le x\le 1$: top $=y=3x$, bottom $=y=\frac{1}{2}x$. Height $=3x-\frac{1}{2}x=\frac{5}{2}x$.
• $1\le x\le 2$: top $=y=-2x+5$, bottom $=y=\frac{1}{2}x$. Height $=-2x+5-\frac{1}{2}x=-\frac{5}{2}x+5$. $A={\int }_0^1\frac{5}{2}xdx+{\int }_1^2(-\frac{5}{2}x+5)dx.$ Pehla: $[\frac{5}{4}{x}^2{]}_0^1=\frac{5}{4}$. Doosra: $[-\frac{5}{4}{x}^2+5x{]}_1^2=(-5+10)-(-\frac{5}{4}+5)=5-\frac{15}{4}=\frac{5}{4}$. $A=\frac{5}{4}+\frac{5}{4}=\frac{5}{2}.$ Shoelace check: $\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|=\frac{1}{2}|0(1-3)+2(3-0)+1(0-1)|=\frac{1}{2}|6-1|=\frac{5}{2}.$ ✔

Horizontal kyun: parabola already $x=f(y)$ ke form mein hai, toh $y$ mein slice karo — koi inversion zaroorat nahi. Intersections: $6y-y^2=y\Rightarrow 5y-y^2=0\Rightarrow y(5-y)=0$, toh $y=0,5$. Right/left: $y=1$ pe, parabola $=6-1=5$; line $=1$. Toh parabola right hai, line left hai. $A={\int }_0^5((6y-y^2)-y)dy={\int }_0^5(5y-y^2)dy=[\frac{5y^2}{2}-\frac{y^3}{3}{]}_0^5.$ $y=5$ pe: $\frac{5\cdot 25}{2}-\frac{125}{3}=\frac{125}{2}-\frac{125}{3}=\frac{375-250}{6}=\frac{125}{6}.$ $A=\frac{125}{6}\approx \mathrm{20.83.}$

Intersections: $x-x^2=mx\Rightarrow x-x^2-mx=0\Rightarrow x((1-m)-x)=0$, toh $x=0$ aur $x=1-m$. Top/bottom: small $x>0$ ke liye parabola $x-x^2\approx x$ line $mx$ se zyada hai (kyunki $m<1$), toh parabola top hai. $A(m)={\int }_0^{1-m}((x-x^2)-mx)dx={\int }_0^{1-m}((1-m)x-x^2)dx.$ $k=1-m$ lo: ${\int }_0^k(kx-x^2)dx=[\frac{k{x}^2}{2}-\frac{x^3}{3}{]}_0^k=\frac{k^3}{2}-\frac{k^3}{3}=\frac{k^3}{6}.$ $\frac{1}{6}$ ke equal karo: $\frac{k^3}{6}=\frac{1}{6}\Rightarrow k^3=1\Rightarrow k=1\Rightarrow 1-m=1\Rightarrow m=0.$ Yeh $m>0$ ko violate karta hai, isliye koi admissible $m$ area $\frac{1}{6}$ nahi deta; maximum area ($m\to 0^{+}$ pe) exactly $\frac{1}{6}$ hai aur yeh sirf approach hoti hai, $m>0$ ke liye kabhi reach nahi hoti. Answer: $A(m)=\frac{(1-m{)}^3}{6}$; area $=\frac{1}{6}$ ke liye $m=0$ chahiye, toh koi valid $m>0$ exist nahi karta — supremum $\frac{1}{6}$ attain nahi hota.

Vertical ($dx$). $[0,1]$ pe, $x=\frac{1}{4}$ pe: $\sqrt{x}=\frac{1}{2}>\frac{1}{4}=x$, toh $\sqrt{x}$ top hai. $A={\int }_0^1(\sqrt{x}-x)dx=[\frac{2}{3}{x}^{3/2}-\frac{x^2}{2}{]}_0^1=\frac{2}{3}-\frac{1}{2}=\frac{1}{6}.$ Horizontal ($dy$). Invert karo (Inverse Functions use hota hai): $y=\sqrt{x}\Rightarrow x=y^2$; $y=x\Rightarrow x=y$. $y\in [0,1]$ ke liye, $y=\frac{1}{2}$ pe: $x=y=\frac{1}{2}$ vs $x=y^2=\frac{1}{4}$, toh line $x=y$ right hai, $x=y^2$ left hai. $A={\int }_0^1(y-y^2)dy=[\frac{y^2}{2}-\frac{y^3}{3}{]}_0^1=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}.$ Yeh match kyun karna chahiye: dono integrals usi region ke unit-area pieces ke Riemann sums hain (Definite Integral as Riemann Sum); slice karne ki direction sirf sum ko reorder karti hai, aur ek convergent area sum ka finite reordering total change nahi kar sakta. Isliye $A=\frac{1}{6}$ dono taraf se. $■$

Intersections: $y^2=y^2-2y\Rightarrow 0=-2y\Rightarrow y=0$. Woh sirf ek point pe milte hain, isliye yeh curves apne aap ek bounded region enclose nahi karti — humein explicit $y$-bounds ya koi aur boundary chahiye. Dobara padho: "between their intersections" ek trap hai kyunki $y=0$ pe sirf ek tangential meeting hai. Gap ka careful check: $y\ne 0$ ke liye, $R(y)-L(y)=y^2-(y^2-2y)=2y$. Yeh $y>0$ ke liye positive aur $y<0$ ke liye negative hai, matlab $x=y^2$ sirf $y>0$ ke liye right hai; $y<0$ ke liye roles flip ho jaate hain. Sirf in do curves aur ek crossing ke saath, koi closed area exist nahi karta — answer hai "enclosed area $0$ / region bounded nahi." Lesson: integrate karne se pehle hamesha confirm karo ki curves actually ek region enclose karti hain (lens ke liye $\ge 2$ crossings chahiye).

Technique ka point ::: Pucho "jab main sweep karta hoon toh kya change hota hai?" — agar top/bottom change hota hai lekin right/left nahi, toh horizontally slice karo. Integrate karne se pehle ::: Confirm karo ki curves do points pe cross karti hain (real bounded region), phir order ke liye ek interior point test karo. Agar curves andar cross karein ::: Har crossing pe split karo; har piece pe (top$-$bottom) integrate karo — seedha through kabhi nahi jaao. Sanity check ::: Thickness ko limits se match karo ($dx\to x$-limits, $dy\to y$-limits) aur, jab possible ho, dono taraf se compute karo.