Area between curves — horizontal and vertical slices

4.2.13 · Maths › Calculus II — Integration

WHY "between curves" ki zaroorat hai?

Vertical slices (width $d x$ )

Horizontal slices (height $d y$ )

Figure — Area between curves — horizontal and vertical slices

Worked Example 1 — vertical slices

$y = x^{2}$ aur $y = 2 x$ ke beech ka area nikalo.

Step — intersections nikalo. $x^{2} = 2 x$ set karo $\Rightarrow x^{2} - 2 x = 0 \Rightarrow x (x - 2) = 0$ , toh $x = 0, 2$ . Kyun? Limits $a, b$ exactly wahin hain jahan curves milti hain (region wahan pinch ho jaata hai).
Step — kaun upar hai? $x = 1$ test karo: line $2 x = 2$ , parabola $x^{2} = 1$ . Toh $2 x > x^{2}$ ; line top par hai. Point test kyun? Sochne se sasta hai; ek interior point poore interval par order fix kar deta hai (wo sirf endpoints par cross karte hain).
Step — top minus bottom integrate karo. $A = \int_{0}^{2} (2 x - x^{2}) d x = [x^{2} - \frac{x ^{3}}{3}]_{0}^{2} = (4 - \frac{8}{3}) - 0 = \frac{4}{3} .$ Ye antiderivative kyun? $\frac{d}{d x} (x^{2}) = 2 x$ , $\frac{d}{d x} (x^{3} /3) = x^{2}$ . ✔

Worked Example 2 — same region, dono tarike se

$y = x$ , $y = 0$ , aur $x = 4$ se bounded region (ek triangle... lekin chalte hain curve $y = x$ wala idea force karte hain). $y = x$ aur line $y = x$ ke beech ka region $0 \leq x \leq 1$ ke liye lo.

Intersections: $x = x \Rightarrow x = x^{2} \Rightarrow x (x - 1) = 0$ , toh $x = 0, 1$ .
$x = 0.25$ par top: $0.25 = 0.5$ vs $x = 0.25$ , toh $x$ top hai.
Vertical: $A = \int_{0}^{1} (x - x) d x = [\frac{2}{3} x^{3/2} - \frac{x ^{2}}{2}]_{0}^{1} = \frac{2}{3} - \frac{1}{2} = \frac{1}{6} .$
Horizontal (variables switch karo): curves invert karo. $y = x \Rightarrow x = y^{2}$ (ye right curve hai kyunki fixed $y$ ke liye, $y^{2} \leq y$ ... check karo). $y = x \Rightarrow x = y$ . $0 \leq y \leq 1$ ke liye, kaun right hai? $y = 0.5$ par: $x = y = 0.5$ vs $x = y^{2} = 0.25$ , toh $x = y$ right par hai, $x = y^{2}$ left par. $A = \int_{0}^{1} (y - y^{2}) d y = [\frac{y ^{2}}{2} - \frac{y ^{3}}{3}]_{0}^{1} = \frac{1}{2} - \frac{1}{3} = \frac{1}{6} . ✓$ Dono $1/6$ kyun dete hain? Ye same region hai; slicing direction kabhi area nahi badal sakti.

Worked Example 3 — horizontal slices bahut aasaan hain

$x = y^{2}$ aur $x = y + 2$ se bounded region.

$y$ mein intersections: $y^{2} = y + 2 \Rightarrow y^{2} - y - 2 = 0 \Rightarrow (y - 2) (y + 1) = 0$ , toh $y = - 1, 2$ .
Kaun curve right par hai? $y = 0$ par: $x = y + 2 = 2$ vs $x = y^{2} = 0$ . Toh $x = y + 2$ right par, $x = y^{2}$ left par. $A = \int_{- 1}^{2} ((y + 2) - y^{2}) d y = [\frac{y ^{2}}{2} + 2 y - \frac{y ^{3}}{3}]_{- 1}^{2} .$ $y = 2$ par: $2 + 4 - \frac{8}{3} = \frac{10}{3}$ . $y = - 1$ par: $\frac{1}{2} - 2 + \frac{1}{3} = - \frac{7}{6}$ . $A = \frac{10}{3} - (- \frac{7}{6}) = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2} .$ Yahan horizontal kyun? $x$ ke function ke roop mein parabola $x = y^{2}$ do $y$ -values deta hai; vertical slicing ke liye $x = 1$ par split karna padta do pieces mein. Horizontal slicing ye poori mushkil avoid kar deta hai.

Common mistakes (Steel-man them)

Recall Feynman: 12-saal ke bacche ko samjhao

Socho do tedhi-medhi lines ke beech ki jagah ek garden hai, aur tum uska area chahte ho. Garden ko super-patli sticks mein kaato. Har khadi stick ek tiny rectangle hai: uski height hai kitni door top line bottom line ke upar hai us jagah, aur uski width tiny hai. Sab sticks ke areas jodo — ye jodna hi integral hai. Agar garden tricky tarike se zyada tall hai wide se, toh sticks ko sideways lita do (height $d y$ ). Same garden, same area — bas tumne direction badal di jisme sticks rakhi.

Flashcards

Vertical-slice area formula

A = \int_{a}^{b} (f (x) - g (x)) d x

jahan

f

top curve hai,

g

bottom.

Horizontal-slice area formula

A = \int_{c}^{d} (R (y) - L (y)) d y

jahan

R

right curve hai,

L

left.

Integration limits kahan se aate hain?

Do curves ke intersection points se (

x

mein vertical ke liye,

y

mein horizontal ke liye).

Horizontal (

d y

) slices kab prefer karni chahiye?

Jab right/left curves constant hoon lekin top/bottom badal jaaye — jaise jab ek curve naturally

x = f (y)

ho (jaise

x = y^{2}

), toh vertical slicing ko splitting chahiye hogi.

Subtract kyun karte hain, end mein absolute value kyun nahi lete?

Agar curves cross karte hain, signed areas cancel ho jaate hain; crossings par split karke har piece mein top−bottom use karna padta hai.

Riemann sum se integrand derive karo

Patli rectangle = (top−bottom)·Δx; sum karo aur Δx→0 karne par

\int (f - g) d x

milta hai.

y = 2 x

aur

y = x^{2}

ke beech ka area

\int_{0}^{2} (2 x - x^{2}) d x = 4/3

x = y^{2}

aur

x = y + 2

ke beech ka area

\int_{- 1}^{2} ((y + 2) - y^{2}) d y = 9/2

Connections

Definite Integral as Riemann Sum — yahan har area formula ke peeche ka engine.
Fundamental Theorem of Calculus — limit-of-sums ko antiderivative evaluation mein convert karta hai.
Volumes by Slicing & Disks — same "slice, measure, integrate" pattern, ek dimension upar.
Inverse Functions — horizontal slices ke liye $y = f (x)$ ko $x = f^{- 1} (y)$ likhne ke liye chahiye.
Solving Quadratic & Polynomial Equations — intersection points nikalne mein use hota hai.