Area between curves — horizontal and vertical slices: Question bank

4.2.13 · D5 · Maths › Calculus II — Integration › Area between curves — horizontal and vertical slices

True or false — justify

Naive

\int_{a}^{b} f - g

true area ke barabar hota hai jab bhi

f

aur

g

[a, b]

par kabhi cross nahi karte

True — agar wo cross nahi karte, toh ek curve poori tarah upar rehti hai, isliye

f - g

ek hi sign rakhta hai aur koi cancellation nahi hoti. Yahi exactly "top minus bottom" setup hai.

Agar

\int_{a}^{b} (f - g) d x = 0

, toh dono curves

[a, b]

par identical honi chahiye

False — iska matlab ye bhi ho sakta hai ki jahan

f > g

hai wahan ka positive area exactly cancel ho jata hai jahan

g > f

hai wahan ke negative area se. Curves cross kar sakti hain aur real, nonzero regions enclose kar sakti hain. Dekho Definite Integral as Riemann Sum ki kyon signed pieces cancel hote hain.

Vertical slices ki jagah horizontal slices choose karna usi region ke liye ek chota numerical area de sakta hai

False — region ka area ek fixed geometric fact hai. Slicing direction sirf ek bookkeeping choice hai; dono ko same number dena hi hoga (Worked Example 2 ne dono taraf se

1/6

nikala).

y = x

aur

y = x

ke beech ke region ke liye

[0, 1]

par, integrand

x - x

poori tarah non-negative hai

True —

(0, 1)

par hame

x \geq x

milta hai (jaise

x = 0.25

par,

0.5 > 0.25

), isliye top-minus-bottom order interval ke andar kabhi flip nahi hota.

d y

-integral ki limits hamesha equivalent

d x

-integral ki limits ke same numbers hoti hain

False —

d x

limits

x

-range hain,

d y

limits

y

-range hain; ye usually alag hote hain. Sirf coincidence se (ek square region) ye match karenge.

Do curves ke beech ka area hamesha

\int (bigger function) - \int (smaller function)

hota hai jo pure

[a, b]

par evaluate hota hai

False in general — ye tabhi kaam karta hai jab same function pure interval mein bada ho. Agar wo roles swap karein toh tumhe crossing point par split karna hoga aur har piece par re-order karna hoga.

Formula

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

require karta hai ki curves ko

y

ke terms mein

x

likha jaye

True — horizontal slices ki thickness

d y

hoti hai aur length

x

-direction mein measure hoti hai, isliye dono boundaries ko

x = R (y)

aur

x = L (y)

ke roop mein express karna hoga. Yahan Inverse Functions enter hota hai.

Spot the error

y = x^{2}

aur

y = 2 x

ke beech ka area

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

hai, isliye area

= \frac{4}{3}

hai." Trap kahan hai?

Integrand bottom-minus-top likha gaya hai (

x^{2} - 2 x

) kyunki line

2 x

actually

(0, 2)

par upar hai. Negative result ek red flag hai; end mein absolute value se salvage karna yahan sirf is liye kaam karta hai kyunki wo andar cross nahi karte, lekin ye aadat dangerous hai.

y = x^{2} - 1

aur

y = 0

ke liye

[- 2, 2]

par, area

= \int_{- 2}^{2} (x^{2} - 1) d x

." Flaw dhundho

Curves

x = \pm 1

par cross karti hain, isliye

[- 2, 2]

ke andar parabola

(- 1, 1)

par axis ke neeche hai aur bahar upar. Single integral negative aur positive parts ko cancel karne deta hai; tumhe

\pm 1

par split karna hoga aur har piece par top-minus-bottom use karna hoga.

x = y^{2}

aur

x = y + 2

ke beech ke region ko horizontally slice karne ke liye main

x = 0

x = 4

tak integrate karta hun." Kya galat hai?

d y

-integral mein

y

-limits use honi chahiye, aur curves

y = - 1

aur

y = 2

par milti hain.

x

-values ko

d y

-integral mein dalna thickness aur limits ko mismatch karta hai (ek core parent-note mistake).

x = y^{2}

vertical slicing ke liye bina kisi problem ke top-ya-bottom curve ho sakta hai." Iska critique karo

Kisi fixed

x > 0

ke liye,

x = y^{2}

y

-values deta hai (

\pm x

), isliye ye

x

ka single function nahi hai. Vertical slices ko ise

y = x

aur

y = - x

mein split karna hoga; horizontal slicing split ko bilkul avoid karti hai.

"Main koi bhi sign issue fix karne ke liye

∣ final answer ∣

lega." Ye fail kyun hota hai?

Cancellation integral ke andar hoti hai usse pehle ki tum koi final number dekho, isliye jis magnitude ka tum absolute value lete wo already bahut choti hai. Tumhe

∣ f - g ∣

piece by piece insert karna hoga, end mein nahi.

"Dono curves origin se guzarti hain, isliye lower limit definitely

x = 0

hai." Assumption pakdo

Kisi common point se guzarna usse wahan region bound karna nahi hai. Limits wo intersection points hain jo actually region ko band karti hain — check karo ki region actually us shared point par shuru hoti hai ya nahi, sirf itna nahi ki dono curves use visit karti hain. Limits Solving Quadratic & Polynomial Equations se aati hain.

Why questions

Geometrically, integration limits intersection points se kyun aate hain?

Intersection par top aur bottom curves milti hain, isliye slice ki height zero hoti hai — region "pinch shut" ho jata hai. Ye pinch points exactly enclosed area ke left/right (ya bottom/top) edges hain.

Slice height "top minus bottom" kyun hoti hai aur kabhi "bottom minus top" nahi?

Height ek positive vertical distance hai = (bada

y

) − (chota

y

). Ise top-minus-bottom likhna guarantee karta hai ek non-negative integrand taaki Riemann sum genuine areas add kare, signed ones nahi.

d x

d y

switch karna kabhi answer kyun nahi badalta?

Dono sirf same region ko thin rectangles se tile karne ke alag alag tarike hain; total tiled area region ki property hai, tile orientation ki nahi. Fundamental Theorem of Calculus dono sums ko same limit tak evaluate karta hai.

Jab vertical slices "hamesha exist karti hain" toh koi horizontal slices kyun prefer karega?

Jab top ya bottom boundary beech mein change ho jaye, vertical slicing mein multiple integrals chahiye; agar balki left/right boundaries

y

ke single clean functions hain, toh ek horizontal integral poora kaam karta hai (Worked Example 3).

Hume kabhi kabhi ek single pair of curves ke liye bhi interval split kyun karna padta hai?

Kyunki kaun si curve upar hai wo crossing par change ho sakti hai. Har sub-interval ka apna sahi top-minus-bottom order hota hai, aur splitting har integrand ko non-negative rakhti hai.

Koi ek interior point test karna kaafi kyun hota hai decide karne ke liye ki kaun si curve upar hai?

Do consecutive intersection points ke beech curves cross nahi karti, isliye unka order flip nahi ho sakta. Ek test point isliye pure sub-interval ke liye order fix kar deta hai.

Parent note kyun kehta hai "rectangle ka bottom sirf upar chala gaya"

\int f d x

\int (f - g) d x

jaate waqt?

Plain integral

f

aur line

y = 0

ke beech ka area hai. Baseline

0

ko curve

g

se replace karna rectangle ki bottom edge ko axis se utha kar

g

tak le jaata hai; slicing idea ke baare mein kuch aur nahi badlta.

Edge cases

Jab dono curves identical hain,

f = g

[a, b]

, tab area kya hota hai?

Zero — har slice ki height

f - g = 0

hai, isliye koi enclosed region nahi hai.

0

ka integral

0

hota hai.

Curves ek point par tangentially touch karti hain lekin cross nahi karti (jaise

y = x^{2}

aur

y = 0

sirf origin par) — kya wahan unke beech area hai?

Ek single touch point ki width zero hoti hai, isliye wo koi area contribute nahi karta; region trap karne ke liye curves ko ek interval par alag hona chahiye. Akela tangent point kuch enclose nahi karta.

Region ek taraf unbounded ho (jaise

y = e^{- x}

aur

y = 0

ke beech

x \geq 0

ke liye) — kya "intersection = limits" rule tab bhi apply hota hai?

Seedha nahi; right side par koi finite crossing nahi hai, isliye tum limit

b \to \infty

lete ho (ek improper integral). Area tab bhi finite ho sakta hai chahe interval infinite ho.

Do curves teen points par cross karti hain, do enclosed lobes dete hain — pure span par ek integral kya compute karta hai?

Ye do lobes ke areas ka signed difference compute karta hai, jo partially ya fully cancel ho sakta hai. Total area ke liye, har crossing par split karo aur har lobe par

∣ f - g ∣

alag alag sum karo.

Ek vertical line

x = k

ek boundary hai — kya ye ek "curve"

y = f (x)

hai?

Nahi, ek vertical line

x

ke function ke roop mein vertical-line test fail karti hai, lekin ye vertical slices ke liye ek perfectly good constant limit (

a

b

) hai, ya ek boundary

x = k

jo horizontal slices se naturally handle hoti hai.

Region

x

-axis ke baare mein symmetric hai (top curve

y = h (x)

, bottom

y = - h (x)

) — kya tum iska faayda utha sakte ho?

Haan; slice height

h - (- h) = 2 h (x)

hai, isliye

A = \int 2 h d x = 2 \int h d x

. Symmetry sirf axis ke upar ke area ko double karti hai, koi conceptual step nahi bachta lekin integrand simplify hota hai.

y

-value par horizontal slice ka kya hota hai jahan left aur right curves momentarily coincide karti hain?

Iski length

R (y) - L (y) = 0

hoti hai, isliye wo slice kuch contribute nahi karti — ye region ka ek top ya bottom pinch point mark karta hai aur typically ek

y

-limit par hota hai.

Recall Yahan har trap ki ek-line summary

Order matters (top−bottom, right−left), limits thickness variable se match karti hain, aur crossings ek split force karti hain — cancellation aur mismatched limits do tarike hain jisse almost sabka point jaata hai.

Connections

Parent: Area between curves
Definite Integral as Riemann Sum — kyun signed pieces cancel hote hain.
Fundamental Theorem of Calculus — dono slicing ko same value tak evaluate karta hai.
Inverse Functions — $y = f (x)$ ko $x = f^{- 1} (y)$ ke roop mein rewrite karna $d y$ -slices ke liye.
Solving Quadratic & Polynomial Equations — intersection limits dhundhna.
Volumes by Slicing & Disks — same slice-measure-integrate pattern ek dimension upar.