Application of Integrals: Class 12 Maths Chapter 8 Hinglish mein

Curves ke neeche ka area, ek-ek patli strip karke

Mathematics Part 2 · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

50 min
1 simulation
12 practice questions
14 MCQs

Is chapter mein aap kya seekhenge

Samjhao ki curved edge wale region ke liye integration kyun chahiye, aur uske area ko patli strips ke sum ki tarah set up karo
Curve y = f(x) aur x-axis ke beech, a se b tak ke interval par, area ko definite integral ki tarah likho aur evaluate karo
Horizontal strips se curve x = g(y) aur y-axis ke beech, y = c se y = d tak, area nikaalo
Jo curves x-axis ke neeche hain ya use cross karti hain, unhe zeros par split karke aur absolute values jod kar handle karo
Symmetry se standard form wale circles aur ellipses ka enclosed area nikaalo, aur πr² aur πab dobara paao
Har calculate kiye gaye area ko ek quick geometric estimate se check karo

NCERT Mathematics Class 12, Chapter 8, pp. 292–299 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

8.1 Introduction

Nashik ke paas ek kisaan ka plot hai: aage ek road, dono sides par seedhi fences, aur peeche ghoomti hui ek stream (chhoti nadi). Sahi maatra mein fertiliser order karne ke liye use plot ka area chahiye. Tape measure se teen seedhi sides turant naap jaati hain; problem stream hai.

School geometry seedhe tukdon se bani shapes handle karti hai: polygon ko triangles, rectangles ya trapezia mein kaato aur unke areas jod do. Circle hi ek aisi curved shape hai jiska ready formula hai. Stream jaisi boundary, ya parabola ke arc ke liye koi formula nahi hai, aur hum jo bhi polygon fit karein, woh kahin thoda bada aur kahin thoda chhota padta hai (Fig. 8.1).

3/3Solah patli strips curve se kahin behtar chipakti hain; aur patli strips aur bhi behtar karti hain.

Fig. 8.1Curved edge ka koi area formula nahi hai, lekin patli strips us tak jitna chahein utna paas pahunch sakti hain.

Ek region seedhe base par hai, uski vertical sides seedhi hain aur top lehradaar curved hai. Pehle step mein region shaded hai aur label hai: koi formula fit nahi hota. Doosre step mein upar ke do corners ko jodne wali chord ke neeche ek dashed trapezium bana hai; hump ke paas curve chord ke upar hai, yaani gap, aur dip ke paas trapezium curve ke upar nikal jaata hai, yaani extra area. Teesre step mein region solah patle rectangles mein kata hai jinke tops lagbhag curve ko follow karte hain, label hai: jitna patla, utna paas.

Pichhle chapter mein humne woh tool banaya jo humein chahiye. Definite integral patle rectangles ke sums ki limit hai, aur fundamental theorem se hum use antiderivative ki madad se evaluate kar sakte hain. Yahan hum us tool ko ulta use karte hain: koi region diya ho to hum use strips mein kaatte hain, ek strip ka area likhte hain, aur integrate karte hain.

Is chapter ki curves aasaan wali hain, har ek apne standard form mein: parabolas, ellipses, circles aur straight lines. Har integral ke saath hum ek rough geometric estimate bhi banate hain, taaki algebra mein koi galti turant pakdi jaaye.

Quick check

2 mein se 1

Inmein se kaun se areas school geometry ka formula integration ke bina de sakta hai?

Trapezium ke shape ka ek khetEk semicircular stage0 se 1 tak

y = x^{2}

ke neeche ka regionLehradaar kinaare wali ek jheel

Saare sahi options chuno

Answer dekho

Answer

(A) Trapezium ke shape ka ek khet
(B) Ek semicircular stage

Trapezium aur semicircle (

π r^{2}

ka aadha) ke formulas hain. Parabola aur lehradaar kinaara curved edges hain jinka koi ready formula nahi hai; integration se pehle wala

\frac{1}{3}

aata hai.

8.2 Simple Curves ke neeche Area

Is poore section mein region ek curve, ek coordinate axis aur doosre axis ke parallel do straight lines se ghira hota hai. Vertical lines $x = a$ aur $x = b$ ko ordinates kehte hain, aur inke beech band hue region ka size area under the curve hai. Pehle hum region ko vertically slice karte hain, phir horizontally.

Vertical strips

DerivationVertical strips ke sum ke roop mein area

Maan lo $f$ , $[a, b]$ par continuous hai aur $f (x) \geq 0$ hai. $y = f (x)$ ke neeche ke region ko vertical strips mein kaato. $x$ par $d x$ width ke chhote se stretch mein curve ki height mushkil se badalti hai, isliye strip lagbhag $y = f (x)$ height ka rectangle hai.

Iska area elementary area $d A = y d x$ hai. $x = a$ se $x = b$ tak strips jodne aur unki width ko zero tak ghatane par sum ek definite integral ban jaata hai.

\begin{aligned} A & = \int_{a}^{b} y d x \\ = \int_{a}^{b} f (x) d x \end{aligned}

(8.1)

jahan

$A$: $x = a$ se $x = b$ tak $y = f (x)$ ke neeche ke region ka area
$d x$: ek strip ki width

2/3 $d x$ width aur $y$ height wali strip ka area $d A = y d x$ hai.

Fig. 8.2Curve ke neeche area: ek strip ka area

y d x

hai; integral un sabko jodta hai.NCERT Fig. 8.1 · p. 292

Axes par ek curve left se right dheere-dheere upar jaati hai, label hai y equals f of x. Pehle step mein a se b tak curve aur x-axis ke beech ka region shaded hai. Doosre step mein ek patli vertical strip highlight hai: uski width dx aur height y mark hai, aur label par likha hai dA equals y dx. Teesre step mein region baarah patle rectangles se bhara hai, aur label par likha hai A equals a se b tak y dx ka integral.

Definition

Elementary area

Region ki ek patli strip ka area $d A$ : $d x$ width wali vertical strip ke liye $y d x$ , ya $d y$ thickness wali horizontal strip ke liye $x d y$ .

Example $y = x^{2}$ ke neeche, $x = 3$ par $0.01$ width wali strip ka area lagbhag $9 \times 0.01 = 0.09$ hai.

Kaam karne ka tareeka chhota hai. Region ka sketch banao aur limits padh lo, strip ki height ko $x$ ke function ki tarah likho, integrate karo, aur phir answer ko sketch mein dikhne wale kisi rectangle ya trapezium se compare karo.

Example

Parabola ke neeche, sanity check ke saath

$x = 0$ se $x = 2$ tak curve $y = x^{2} + 1$ aur $x$ -axis ke beech kitna area hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Horizontal strips

Kabhi-kabhi boundary ko $x = g (y)$ ki tarah likhna aasaan hota hai, aur region is curve, $y$ -axis aur horizontal lines $y = c$ aur $y = d$ ke beech hota hai (Fig. 8.3). Tab region ko across slice karo. Height $y$ par $d y$ thickness wali horizontal strip $y$ -axis se curve tak jaati hai, isliye uski length $x = g (y)$ hai aur area $x d y$ hai.

\begin{aligned} A & = \int_{c}^{d} x d y \\ = \int_{c}^{d} g (y) d y \end{aligned}

(8.2)

jahan

$A$: $y = c$ se $y = d$ tak $x = g (y)$ aur $y$ -axis ke beech ke region ka area

2/3Horizontal strip ki length $x = g (y)$ aur thickness $d y$ hai.

Fig. 8.3Jab

x

y

ke terms mein diya ho, to

x

length aur

d y

thickness wali horizontal strips kaam kar deti hain.NCERT Fig. 8.2 · p. 293

Axes par ek curve x equals g of y hai jo y badhne ke saath y-axis se door jaati hai. Pehle step mein height c se height d tak y-axis aur curve ke beech ka region shaded hai. Doosre step mein ek patli horizontal strip highlight hai, uski thickness dy aur length x mark hai, aur label hai dA equals x dy. Teesre step mein das horizontal rectangles region ko bharte hain aur label par likha hai A equals c se d tak x dy ka integral.

Kis taraf slice karna hai, yeh tumhari choice hai. Woh direction chuno jismein ek hi formula region ki har strip ko describe kare; agla example dikhata hai ki achhi choice kitna kaam bachati hai.

Example

Ek sideways parabola

Parabola $x = 4 - y^{2}$ aur $y$ -axis ke beech ghire region ka area nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

3/3Yeh box ka two-thirds bharta hai, isliye area $\frac{32}{3}$ hai.

Fig. 8.4Parabolic region apne bounding box ka theek two-thirds bharta hai:

\frac{2}{3} \times 16 = \frac{32}{3}

Axes par left ki taraf khulta ek parabola, jiski tip (4, 0) par hai aur jo y-axis ko 2 aur minus 2 par cross karta hai. Pehle step mein parabola aur y-axis ke beech ka region shaded hai aur dy thickness wali ek horizontal strip highlight hai, saath mein minus 2 se 2 tak 4 minus y squared dy ka integral. Doosre step mein ek dashed 4 by 4 box region ko gherta hai, jiska area 16 hai. Teesre step mein ek note kehta hai ki region box ka two-thirds bharta hai, isliye uska area 32 over 3 hai.

Quick check

3 mein se 1

Line

y = 2 x

x

-axis aur line

x = 3

kitna area gherti hain?

sq. units

Answer dekho

Answer 9 sq. units

Integrate karne par

{[x^{2}]}_{0}^{3} = 9

square units aata hai, jo base 3 aur height 6 wale triangle ke barabar hai.

Jab curve x-axis ke neeche chali jaaye

Formula (8.1) mein humne $f (x) \geq 0$ maana tha. Agar curve $[a, b]$ par $x$ -axis ke neeche hai, to har strip ki "height" $y$ negative hai, isliye har $y d x$ negative hai aur integral bhi. Area kabhi negative nahi hota: us case mein area integral ki absolute value hota hai.

3/3Area $A_{1} + ∣ A_{2} ∣$ hai, $A_{1} + A_{2}$ nahi.

Fig. 8.5Axis ke neeche integral negative hota hai; area tukdon ke sizes ko jodta hai.NCERT Figs 8.3, 8.4 · p. 293

Axes par ek curve jo a par x-axis ke upar shuru hoti hai, uthkar ek hump banati hai, c par axis ko cross karke neeche jaati hai aur line x equals b se pehle ek trough tak girti hai. Pehle step mein sirf a aur c ke beech ka hump plus sign ke saath shaded hai, label hai A1 greater than 0. Doosre step mein c aur b ke beech ka dip bhi shaded hai, minus sign ke saath hatched, label hai A2 less than 0: axis ke neeche integral negative hota hai. Teesre step mein label par likha hai Area equals A1 plus A2 ki absolute value.

Aksar curve $[a, b]$ ke andar kisi point $c$ par axis ko cross karti hai (Fig. 8.5). Tab $a$ se $b$ tak ek hi integral lene par neeche wala tukda upar wale tukde ka kuch hissa cancel kar deta hai, aur humein net value milti hai, area nahi. Isliye $c$ par split karo.

A = ∣ A_{1} ∣ + ∣ A_{2} ∣

(8.3)

jahan

$A_{1}$: $\int_{a}^{c} f (x) d x$ , pehle tukde par integral
$A_{2}$: $\int_{c}^{b} f (x) d x$ , doosre tukde par integral
$c$: woh point jahan curve $x$ -axis ko cross karti hai

Example

Axis ke neeche ek region

Curve $y = x^{2} - 2 x$ , $x$ -axis ke neeche jaati hai aur wapas upar aati hai. Axis ke saath yeh jo region gherti hai, uska area nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

InteractiveArea explorer

Is chapter ki saat curves wala ek interactive graph: x squared plus 1, 2x minus 3, x squared minus 2x, sine 2x, x cubed, x times x ki absolute value, aur sideways parabola x equals 4 minus y squared. Do limits vertical lines ki tarah dikhti hain (sideways parabola ke liye horizontal lines), jinke gol handles drag kiye ja sakte hain, aur do sliders se bhi unhe set kiya ja sakta hai. Curve aur axis ke beech ka region axis ke upar plain shaded hai aur neeche hatched. Ek status panel signed integral, area, aur curve ke zeros ke beech har tukde ka size batata hai. Buttons 4, 8, 16 ya 40 strips banate hain aur unka total area aur true area se uska fark dikhate hain. Jab tak learner kuch na kare, kuch nahi hilta.

Fig. 8.6Ek curve chuno, phir limits ko drag karo (ya sliders use karo) aur signed integral ko area se compare karo. Strips on karo aur dekho ki unka total kaise area ke paas pahunchta hai.

Quick check

3 mein se 1

x = 0

x = 6

tak line

y = x - 2

aur

x

-axis ke beech total kitna area hai?

sq. units

Answer dekho

Answer 10 sq. units

Line

x = 2

par cross karti hai. Axis ke neeche, 0 se 2 tak wale tukde ka area 2 hai (uska integral

- 2

hai); axis ke upar, 2 se 6 tak wale tukde ka area 8 hai. Area 10 hai. 0 se 6 tak ek hi integral sirf 6 deta.

Circles aur ellipses ke areas

Circle ya ellipse kisi ek function $y = f (x)$ ka graph nahi hai: extremes ke beech har $x$ se $y$ ki do values milti hain, ek axis ke upar aur ek neeche. Symmetry isko solve karti hai. Standard form mein dono curves dono axes ke about symmetric hain, isliye enclosed area pehle quadrant wale tukde ke area ka chaar guna hai, jahan hum $y \geq 0$ le sakte hain.

Example 1NCERT Example 1 · p. 294

Ek gol table top

Ek gol table top ka radius 6 dm hai. Uska centre origin par rakhein to uska rim $x^{2} + y^{2} = 36$ hai. Integration se uska area nikaalo, pehle vertical strips se aur phir horizontal strips se.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

2/4Vertical strip ki height $y = \sqrt{36 - x^{2}}$ hai.

Fig. 8.7Circle ka ek quarter, kisi bhi taraf slice karke, phir 4 se multiply.NCERT Figs 8.5, 8.6 · p. 294

Origin O par centre wala radius 6 ka circle, jo A(6, 0) aur B(0, 6) se guzarta hai. Pehle step mein quarter OAB shaded hai aur equation x squared plus y squared equals 36 dikhayi gayi hai. Doosre step mein quarter mein dx width aur y equals square root of 36 minus x squared height wali ek vertical strip highlight hai. Teesre step mein uski jagah dy thickness aur x equals square root of 36 minus y squared length wali horizontal strip highlight hai. Aakhri step mein chaaron quarters shaded hain, saath mein four times 9 pi equals 36 pi.

SimulationHar x ke liye do points

Dono axes par minus 8 se 8 tak ka ek coordinate plane, jismein fixed origin O aur ek draggable point P hai jo (3.6, 4.8) se shuru hota hai, O se chhe units door. Distance OP draw aur measure ki gayi hai, aur x-axis mein P ki mirror image dikhayi gayi hai, taaki learner dekhe ki circle par har x ke liye ek point axis ke upar aur ek neeche hota hai.

Fig. 8.8P table ke rim

x^{2} + y^{2} = 36

par shuru hota hai.

x

-axis mein uski mirror image ka

x

same hai aur woh bhi rim par hai.

Try karo

P ko $(4.8, 3.6)$ , $(0, 6)$ aur $(6, 0)$ par drag karo. Kya $O P$ 6 hi rehta hai?
Mirror point ko dekho. Dono $x^{2} + y^{2} = 36$ ko satisfy karte hain; pehle quadrant mein strip ki height kaun sa deta hai?
P ko $(3, 5)$ par rakho. Kya woh rim ke andar hai ya bahar? $3^{2} + 5^{2}$ ko 36 se compare karo.

Try karke dekho

Gin kar π tak pahuncho

Tumhe chahiye

1 cm wala graph paper
ek compass
ek nukeeli pencil

Pehle guess karo

1 cm squares par 6 cm radius ka ek quarter circle bana hai. Yeh lagbhag kitne squares jitna area cover karta hai?

lagbhag 18lagbhag 28lagbhag 36

Compass ki nok grid ke ek corner par rakho aur ek quadrant mein 6 cm radius ka quarter circle banao.
Woh squares gino jo poori tarah arc ke andar hain.
Woh squares gino jinhe arc kaat kar guzarta hai.
Area ka estimate lo: poore squares plus kate squares ka aadha, aur use $9 π \approx 28.3$ se compare karo.

Table 8.1

Tum kya ginte ho	Tumhari value
Andar ke poore squares
Arc se kate squares
Estimate: poore + kate ka aadha

Tumhari entries apne aap save hoti hain.

Example 2NCERT Example 2 · p. 295

Ek elliptical lawn

Park ka ek lawn 10 m lamba aur 6 m chauda ellipse hai. Uska centre origin par aur lambi axis $x$ -axis ke along rakhein to uska edge $\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$ hai. Integration se uska area nikaalo, pehle vertical strips se aur phir horizontal strips se.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

2/4Har height ko $\frac{3}{5}$ se multiply karo: height 4 wala point 2.4 par aa jaata hai.

Fig. 8.9Har vertical strip ko

\frac{3}{5}

se dabane par circle ellipse ban jaata hai aur uska area

25 π

15 π

ho jaata hai.NCERT Figs 8.7, 8.8 · p. 295

Axes par origin par centre wala radius 5 ka dashed circle. Pehle step mein sirf circle, x squared plus y squared equals 25, dikhaya gaya hai. Doosre step mein uske andar semi-axes 5 aur 3 wala ellipse bana hai, aur x equals 3 aur minus 3 par arrows circle ke height 4 wale points ko height 2.4 tak neeche le aate hain: har height three-fifths se multiply hoti hai. Teesre step mein ellipse ka first-quadrant quarter shaded hai, saath mein ek vertical strip jiski height square root of 25 minus x squared ka three-fifths hai. Aakhri step mein poora ellipse shaded hai aur areas compare kiye gaye hain: circle 25 pi, ellipse 15 pi.

Figure answer $π a b$ ko samjhata hai. Circle $x^{2} + y^{2} = a^{2}$ ko factor $\frac{b}{a}$ se vertically dabao. Har vertical strip ki width wahi rehti hai jabki uski height $\frac{b}{a}$ se multiply hoti hai, isliye total area $\frac{b}{a} \times π a^{2} = π a b$ ho jaata hai. Numbers $a$ aur $b$ ellipse ke semi-axes hain.

Quick check

3 mein se 1

Ellipse

\frac{x^{2}}{16} + \frac{y^{2}}{49} = 1

ka enclosed area hai

11 π

784 π

65 π

28 π

Answer dekho

Answer (D) $28 π$

Semi-axes

a = 4

aur

b = 7

hain, isliye area

π a b = 28 π

hai. 16 ko 49 se multiply karne par

784 π

aata hai; jodne par

11 π

65 π

Practice · 5 questions

Practice 8.1: answers ke saath

Short answerMediumNCERT Exercise 8.1, Q1
Integration se ellipse $\frac{x^{2}}{36} + \frac{y^{2}}{16} = 1$ ka enclosed area nikaalo.
Solution dikhao
Answer $24 π \approx 75.4$ square units
Pehle quadrant mein $y = \frac{2}{3} \sqrt{36 - x^{2}}$ hai, aur ellipse dono axes ke about symmetric hai.
Isliye area integral $\int_{0}^{6} \sqrt{36 - x^{2}} d x$ ka $\frac{8}{3}$ guna hai.
Woh integral radius 6 ke quarter-circle ka area hai, yaani $9 π$ . Standard formula bhi yahi deta hai: $x = 6$ par bracket $18 \cdot \frac{π}{2}$ hai, aur $x = 0$ par 0.
Isliye $A = \frac{8}{3} \times 9 π = 24 π$ . Check: $π a b = π \cdot 6 \cdot 4$ .
Short answerMediumNCERT Exercise 8.1, Q2
Integration se curve $4 x^{2} + y^{2} = 36$ ka enclosed area nikaalo.
Solution dikhao
Answer $18 π \approx 56.5$ square units
36 se divide karne par $\frac{x^{2}}{9} + \frac{y^{2}}{36} = 1$ milta hai: ek ellipse jiske semi-axes $x$ ke along 3 aur $y$ ke along 6 hain.
Is lambe (tall) ellipse ke liye horizontal strips theek rehti hain. Pehle quadrant mein $x = \frac{1}{2} \sqrt{36 - y^{2}}$ hai.
Isliye area $\int_{0}^{6} \sqrt{36 - y^{2}} d y$ ka do guna hai. Woh integral ek quarter-circle ka area hai, $9 π$ , isliye $A = 18 π$ .
Check: $π a b = π \cdot 3 \cdot 6 = 18 π$ (lagbhag 56.5), jo axes par bane rhombus (36) aur rectangle (72) ke beech hai.
Quick answerEasyNCERT Exercise 8.1, Q3
Second quadrant ka jo hissa circle $x^{2} + y^{2} = 9$ ke andar hai, uska area hai
$\frac{3 π}{4}$ $\frac{9 π}{4}$ $\frac{9 π}{2}$ $9 π$
Answer dekho
Answer (B) $\frac{9 π}{4}$
Region radius 3 wale circle ka ek quarter hai, isliye uska area $\frac{9 π}{4}$ hai (integral ki tarah, second quadrant par $\int_{- 3}^{0} \sqrt{9 - x^{2}} d x$ ). Aadha circle $\frac{9 π}{2}$ hota, aur $\frac{3 π}{4}$ mein $r^{2}$ ki jagah $r$ use hua hai.
Quick answerMediumNCERT Exercise 8.1, Q4
Parabola $y^{2} = 9 x$ , $y$ -axis aur line $y = 6$ jo region gherte hain, uska area hai
7224168
Answer dekho
Answer (D) 8
Horizontal strips $\int_{0}^{6} \frac{y^{2}}{9} d y$ deti hain, jo ${[\frac{y^{3}}{27}]}_{0}^{6} = 8$ square units aata hai. $\frac{1}{9}$ chhod dene par 72 aata hai; 24 poora 4 by 6 rectangle hai; 16 curve ki doosri taraf, $x$ -axis ke paas wala region hai ( $16 + 8 = 24$ ).
Quick answerEasy
Ellipse $\frac{x^{2}}{49} + \frac{y^{2}}{4} = 1$ ka enclosed area do decimal places tak nikaalo.
sq. units
Answer dekho
Answer 43.98 sq. units
Semi-axes 7 aur 2 hain, isliye area $14 π \approx 43.98$ hai.

Miscellaneous Examples

Neeche ke dono examples apni limits ke beech $x$ -axis ko cross karte hain, isliye har baar tareeka wahi hai: zeros dhoondho, split karo, har tukde ko integrate karo, sizes jodo, aur kisi naapne layak shape se check karo.

Example 3NCERT Example 3 · p. 296

Axis ko cross karti ek line

$x = 0$ aur $x = 4$ ke beech line $y = 2 x - 3$ aur $x$ -axis se ghira total area nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

3/3Area $= \frac{9}{4} + \frac{25}{4} = \frac{17}{2}$ .

Fig. 8.10Do triangles: axis ke neeche

\frac{9}{4}

aur upar

\frac{25}{4}

.NCERT Fig. 8.9 · p. 297

Axes par line y equals 2x minus 3, jo x-axis ko three-halves par cross karti hai; ek dashed vertical line x equals 4 mark karti hai, jahan line height 5 par hai. Pehle step mein sirf line aur limits dikhti hain. Doosre step mein axis ke neeche, 0 se three-halves tak ka triangle minus sign ke saath hatched hai aur axis ke upar, three-halves se 4 tak ka triangle plus sign ke saath shaded hai. Teesre step mein triangles par nine-quarters aur twenty-five quarters likha hai, aur ek note area ko unka sum, seventeen halves, batata hai.

Example 4NCERT Example 4 · p. 297

Sign badalti ek wave

$x = 0$ se $x = \frac{3 π}{4}$ tak wave $y = \sin 2 x$ aur $x$ -axis ke beech ghira total area nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

3/3Area $= 1 + \frac{1}{2} = \frac{3}{2}$ .

Fig. 8.11Area 1 ka ek hump aur area

\frac{1}{2}

ka aadha dip: area

\frac{3}{2}

hai.NCERT Fig. 8.10 · p. 297

Axes par 0 se pi tak y equals sine 2x ki ek poori wave, jo pi over 4 par 1 tak uthti hai aur 3 pi over 4 par minus 1 tak girti hai, jahan ek dashed vertical line right limit mark karti hai. Pehle step mein sirf curve aur limit dikhti hai. Doosre step mein 0 se pi over 2 tak ka hump plus sign ke saath shaded hai aur pi over 2 se 3 pi over 4 tak dip ka hissa minus sign ke saath hatched hai. Teesre step mein tukdon par 1 aur one-half likha hai aur area three-halves diya gaya hai.

Dono examples ko area explorer mein try karo: right wali limit ko khiskao aur dekho ki jaise hi koi tukda axis ke neeche jaata hai, signed integral aur area alag ho jaate hain.

Quick check

3 mein se 1

Har region ko (interval par curve aur

x

-axis ke beech) uske area se match karo.

Left mein koi item tap karo, phir right mein uska match.

Answer dekho

Answer

$[- 1, 1]$ par $y = x$ : $1$
$[0, π]$ par $y = \sin x$ : $2$
$[- 1, 2]$ par $y = x^{2}$ : $3$
$[- 1, 1]$ par $y = x^{3}$ : $\frac{1}{2}$

y = x

\frac{1}{2}

ke do triangles.

\sin x

[- \cos x]_{0}^{π} = 2

x^{2}

axis ke upar hi rehta hai:

\frac{1}{3} + \frac{8}{3} = 3

x^{3}

\frac{1}{4}

ke do tukde.

History seNCERT Historical Note · p. 298

Exhaustion se ∫ tak

Greek geometers curved areas method of exhaustion se nikaalte the: region ko jaane-pehchaane area wale polygons se bharo, phir aur chhote polygons jodte raho jab tak bacha hua hissa jitna chaho utna chhota na ho jaaye. Eudoxus ne chauthi sadi BCE mein is method ko rigorous banaya, aur Archimedes ne ise circle aur parabola ke liye use kiya.

Kerala mein Sangamagrama ke Madhava (lagbhag 1400) aur unke followers bahut saare chhote-chhote tukde jod kar π ki infinite series tak pahunche; yeh soch integration ke kaafi kareeb thi.

1660s mein Isaac Newton ne rates wale apne rule ko ulta chala kar areas nikaale. Alag se, Gottfried Leibniz ne area ko infinitely patli strips ke sum ki tarah likha aur sign ∫ chuna, jo summa (sum) ka khincha hua S hai; unhone 1680s mein apna calculus journal Acta Eruditorum mein publish kiya. Integral naam 1690s mein Bernoulli brothers se aaya. Dono ne dekha ki areas nikaalna, rates nikaalne ka ulta hai.

Unki "infinitely small" strips ne baad ke mathematicians ko pareshaan kiya. 1820s mein Augustin-Louis Cauchy ne integral ko sums ki limit ki tarah define kiya, wahi definition jo tumne pichhle chapter mein dekhi.

Recap

Summary aur key points

Curved edges wale areas ka koi geometry formula nahi hota; hum unhe patli strips mein kaat kar integrate karte hain (Section 8.1).
$y = f (x)$ ke neeche $d x$ width wali vertical strip ka area $d A = y d x$ hai (Derivation).
Agar $[a, b]$ par $f \geq 0$ hai, to $y = f (x)$ ke neeche ka area $\int_{a}^{b} f (x) d x$ hai ((8.1)).
Agar $[c, d]$ par $g \geq 0$ hai, to $x = g (y)$ aur $y$ -axis ke beech ka area $\int_{c}^{d} g (y) d y$ hai ((8.2)).
$x$ -axis ke neeche integral negative hota hai; area uski absolute value hai (Jab curve x-axis ke neeche chali jaaye).
Jab curve axis ko cross kare, to zeros par split karo aur tukdon ki absolute values jodo ((8.3)).
Dono axes ke about symmetry ki wajah se hum circle ya ellipse ka area first-quadrant integral ke chaar gune ki tarah nikaal sakte hain (Circles aur ellipses ke areas).
Radius $r$ wala circle $π r^{2}$ area gherta hai (Example 1); ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , $π a b$ area gherta hai (Example 2).
Answer ko hamesha sketch mein dikhne wale kisi rectangle, triangle ya trapezium se compare karo (Example).

Flash cards · 7

Yaad rakhne wale formulas

1 / 7

Chapter 8 par Miscellaneous Exercise

Apna NCERT exercise question dhoondo

Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook mein	Yahan practice question
NCERT Exercise 8.1, Q1	Practice 8.1 · Q1
NCERT Exercise 8.1, Q2	Practice 8.1 · Q2
NCERT Exercise 8.1, Q3	Practice 8.1 · Q3
NCERT Exercise 8.1, Q4	Practice 8.1 · Q4
NCERT Miscellaneous Exercise 8, Q1	Miscellaneous exercise · Q1
NCERT Miscellaneous Exercise 8, Q2	Miscellaneous exercise · Q2
NCERT Miscellaneous Exercise 8, Q3	Miscellaneous exercise · Q3
NCERT Miscellaneous Exercise 8, Q4	Miscellaneous exercise · Q4
NCERT Miscellaneous Exercise 8, Q5	Miscellaneous exercise · Q5

Quick check: MCQs

8.1 Introduction2 questions
Horizontal strips3 questions
Jab curve x-axis ke neeche chali jaaye3 questions
Circles aur ellipses ke areas3 questions
Miscellaneous Examples3 questions

Glossary · 10

Key terms (English + Hindi)

area under a curveवक्र के अंतर्गत क्षेत्रफलvakra ke antargat kshetraphal: Ek curve, $x$ -axis aur do ordinates ke beech ke region ka area; jab $f \geq 0$ ho, to yeh $\int_{a}^{b} f (x) d x$ ke barabar hai.Chapter mein pehli baar yahan aaya →

definite integralनिश्चित समाकलनnishchit samaakalan: Patle rectangles ke sums ki limit, $\int_{a}^{b} f (x) d x$ , jise antiderivative $F$ se $F (b) - F (a)$ ki tarah evaluate karte hain.Chapter mein pehli baar yahan aaya →

elementary areaप्रारंभिक क्षेत्रफलpraarambhik kshetraphal: Region ki ek patli strip ka area: vertical strip ke liye $y d x$ , horizontal ke liye $x d y$ .Chapter mein pehli baar yahan aaya →
ellipseदीर्घवृत्तdeerghvritt: Standard form mein curve $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , yaani ek axis ke along khincha ya daba hua circle; yeh $π a b$ area gherta hai.Chapter mein pehli baar yahan aaya →

method of exhaustionनिःशेषण विधिnihsheshan vidhi: Curved areas nikaalne ka Greek tareeka: region ko aur-aur chhote polygons se bharte jao jab tak bacha hua hissa negligible na ho jaaye.Chapter mein pehli baar yahan aaya →

ordinateकोटिkoti: Yahan, $x = a$ jaisi vertical line jo curve ke neeche ke region ko ek side se band karti hai.Chapter mein pehli baar yahan aaya →

parabolaपरवलयparvalay: $y = x^{2}$ ya $y^{2} = 4 a x$ jaisi curve; Archimedes ne dikhaya ki parabolic region apne bounding box ka $\frac{2}{3}$ bharta hai.Chapter mein pehli baar yahan aaya →

quadrantचतुर्थांशchaturthaansh: Un chaar regions mein se ek jinmein axes plane ko baant dete hain; first quadrant mein $x \geq 0$ aur $y \geq 0$ .Chapter mein pehli baar yahan aaya →

semi-axisअर्ध-अक्षardh-aksh: Ellipse ki axis ka aadha: dono axes ke along centre se curve tak ki distances $a$ aur $b$ .Chapter mein pehli baar yahan aaya →
symmetryसममितिsammiti: Koi curve kisi axis ke about symmetric hai jab us axis mein reflect karne par wahi curve mile; tab dono sides ke areas barabar hote hain.Chapter mein pehli baar yahan aaya →

Application of Integrals: Class 12 Maths Chapter 8 Hinglish mein

Is chapter mein aap kya seekhenge

8.1 Introduction

8.2 Simple Curves ke neeche Area

Vertical strips

Horizontal strips

Jab curve x-axis ke neeche chali jaaye

Circles aur ellipses ke areas

Practice 8.1: answers ke saath

Miscellaneous Examples

Summary aur key points

Chapter 8 par Miscellaneous Exercise

Practice questions answers ke saath (NCERT “Miscellaneous exercise”)

Apna NCERT exercise question dhoondo

Quick check: MCQs

Key terms (English + Hindi)

Tum chapter ke end tak pahunch gaye