Class 12 Maths · Chapter 8
Short answer:
Is chapter (13 chapters wale 2026-27 rationalised syllabus me Chapter 8) me sirf 2 exercises hain — sabse chhota chapter integration ka. Ex 8.1 me area under simple curves (line, parabola, circle, ellipse — x-axis ya y-axis se bounded) aata hai, Ex 8.2 me area between two curves (dono curves ke intersection points nikal ke limits set karna) aur miscellaneous questions me triangle ka area integration se. Poora chapter Chapter 7 (Integrals) ka direct application hai — pehle definite integral solid honi chahiye.
Application of Integrals matlab curves ke beech ka area nikalna — bina graph paper ke, sirf integration se. Idea simple hai: chhoti-chhoti vertical (ya horizontal) strips lo, unka area (height × dx) integrate karo poori range me. Jahan do curves ke beech ka area chahiye, wahan pehle unka intersection point dhoondo — wahi integration ki limits banenge — phir upar wala curve minus neeche wala curve ka integral lo.
Chapter 8 Summary — 5 Minute Revision
1. Area under a simple curve
Curve y = f(x), x-axis, aur x = a, x = b ke beech ka area (jab f(x) ≥ 0 poori range me):
Area = ∫ab y dx = ∫ab f(x) dx
Agar curve x-axis ke neeche hai (y < 0), integral negative aayega — modulus lagao, kyunki area kabhi negative nahi hota.
Agar y-axis ke against area chahiye (x = g(y), y = c se y = d tak):
Area = ∫cd x dy
2. Area between two curves
Do curves y = f(x) (upar) aur y = g(x) (neeche), jo x = a aur x = b par intersect karte hain, unke beech ka area:
Area = ∫ab [f(x) − g(x)] dx (upper − lower)
Steps hamesha same rahenge:
- Dono curves ki equations solve karke intersection points nikalo — ye hi integration ki limits banenge.
- Range ke beech me koi ek point le ke check karo kaunsa curve upar hai aur kaunsa neeche.
- Upar wala minus neeche wala integrate karo.
3. Symmetry — kaam aasan banane ka shortcut
| Shape | Symmetry | Shortcut |
|---|---|---|
| Circle x² + y² = r² | Dono axes ke around symmetric | Quarter circle nikalo (first quadrant), × 4 |
| Ellipse x²/a² + y²/b² = 1 | Dono axes ke around symmetric | Quarter ellipse × 4 |
| Parabola y² = 4ax | x-axis ke around symmetric | Upar wala half nikalo, × 2 |
↔ Table ko side me swipe karein
Symmetry use karne se integration sirf first quadrant (positive x, positive y) tak simit ho jaati hai — kaafi calculation bach jaati hai.
4. Standard results (yaad rakho, derive bhi aana chahiye)
| Region | Area |
|---|---|
| Circle x² + y² = r² (poora) | π r² |
| Ellipse x²/a² + y²/b² = 1 (poori) | π a b |
| Parabola y² = 4ax aur latus rectum x = a ke beech | 8a²/3 |
↔ Table ko side me swipe karein
5. Sign / absolute value ka dhyan
Agar curve range ke beech me x-axis cross karti hai (jaise y = sin x, y = cos x, ya koi line jo negative se positive jaati hai), toh poori range ko us point pe todo jahan curve x-axis cross karti hai, har piece ka modulus lo, phir add karo. Ek hi integral me poora nikaloge toh positive aur negative parts cancel ho ke galat (chhota) answer aayega.
Poore Class 12 Maths ke handwritten colour notes
IITian & district toppers ke banaye short notes — revision-ready, diagram ke saath. Board se pehle poora syllabus 3 din me revise.
Exercise Questions — Solutions (Q1–Q15)
Q1. Curve y² = x aur lines x = 1, x = 4 ke beech (x-axis ke upar) area nikalo.
Curve: y = √x (upper branch, kyunki x-axis ke upar chahiye). Strips vertical lenge, x = 1 se x = 4 tak.
Area = ∫14 √x dx = [ (2/3) x3/2 ]14 = (2/3)(8 − 1) = 14/3 sq units
Q2. Curve y = x², lines x = 0, x = 2, aur x-axis ke beech ka area nikalo.
Parabola pura first quadrant me x-axis ke upar hai is range me, seedha integrate karo.
Area = ∫02 x² dx = [x³/3]02 = 8/3 sq units
Q3. Circle x² + y² = 4 ka poora area integration (symmetry) se nikalo.
Circle dono axes ke around symmetric hai — first quadrant ka quarter nikalo, phir ×4.
Quarter area = ∫02 √(4 − x²) dx = [ (x/2)√(4−x²) + 2 sin⁻¹(x/2) ]02 = (0 + 2·π/2) − 0 = π
Poora circle area = 4 × π = 4π sq units — jo πr² (r = 2) se match karta hai.
Q4. Parabola y² = 4x aur line y = x ke beech ghira hua area nikalo.
Intersection: y² = 4x aur y = x → x² = 4x → x(x − 4) = 0 → x = 0, 4. Yani points (0,0) aur (4,4).
Range 0 < x < 4 me check karo (x = 1 pe): parabola ka y = 2√1 = 2, line ka y = 1 — parabola upar hai.
Area = ∫04 (2√x − x) dx = [ (4/3)x3/2 − x²/2 ]04 = (4/3·8 − 8) − 0 = 32/3 − 8 = 8/3 sq units
Q5. Do parabolas y² = 4x aur x² = 4y ke beech ka area nikalo.
Intersection: x² = 4y me y = x²/4, isse y² = 4x me daalo: (x²/4)² = 4x → x⁴ = 64x → x⁴ − 64x = 0 → x(x³ − 64) = 0 → x = 0, 4. Points: (0,0) aur (4,4).
x = 1 pe check: y² = 4x se y = 2√1 = 2 (upar), x² = 4y se y = 1/4 (neeche). Toh y² = 4x wala curve upar hai.
Area = ∫04 (2√x − x²/4) dx = [ (4/3)x3/2 − x³/12 ]04 = (32/3 − 64/12) − 0 = 32/3 − 16/3 = 16/3 sq units
Q6. Parabola y = x² aur line y = 4 ke beech ghira hua area nikalo.
Intersection: x² = 4 → x = −2, 2. Range me line (y = 4) upar hai, parabola neeche.
Area = ∫−22 (4 − x²) dx = [4x − x³/3]−22 = (8 − 8/3) − (−8 + 8/3) = 16/3 · 2 ...
Directly: = (8 − 8/3) − (−8 + 8/3) = (16/3) − (−16/3) = 32/3 sq units. (Ya symmetry se: y ke around symmetric hai x = 0 ke, toh 2 × ∫₀² (4 − x²) dx = 2 × 16/3 = 32/3.)
Area = 32/3 sq units
Q7. Ellipse x²/16 + y²/9 = 1 ka poora area integration (symmetry) se nikalo.
a = 4, b = 3. Symmetric dono axes ke around — first quadrant ka quarter nikalo, ×4. Curve: y = (3/4)√(16 − x²).
Quarter area = (3/4) ∫04 √(16 − x²) dx = (3/4) [ (x/2)√(16−x²) + 8 sin⁻¹(x/4) ]04 = (3/4)(0 + 8·π/2) = 3π
Poora ellipse = 4 × 3π = 12π sq units — formula πab = π·4·3 = 12π se match.
Q8. Line y = 3x + 2, x-axis, aur x = −1, x = 1 ke beech ka area nikalo.
Pehle check karo line kahin x-axis cross toh nahi karti: 3x + 2 = 0 → x = −2/3, jo (−1, 1) ke andar hai! Isliye do parts me todna padega — modulus lagao.
Area = |∫−1−2/3 (3x+2) dx| + ∫−2/31 (3x+2) dx
Antiderivative F(x) = (3x²/2) + 2x. F(−1) = 1.5 − 2 = −0.5, F(−2/3) = 2/3 − 4/3 = −2/3, F(1) = 3.5.
Part 1 (curve neeche, x-axis ke neeche hai is chhote piece me) = |F(−2/3) − F(−1)| = |−2/3 − (−0.5)| = |−1/6| = 1/6.
Part 2 = F(1) − F(−2/3) = 3.5 − (−2/3) = 25/6.
Total Area = 1/6 + 25/6 = 26/6 = 13/3 sq units
Q9. Triangle jiske vertices A(1, 0), B(2, 2), C(3, 1) hain, uska area integration se nikalo.
Teeno sides ki equations: AB: y = 2x − 2 (x: 1→2); BC: y = −x + 4 (x: 2→3); AC: y = (x−1)/2 (x: 1→3, base).
Area(ABC) = ∫12(2x−2)dx + ∫23(−x+4)dx − ∫13 (x−1)/2 dx
∫₁²(2x−2)dx = [x²−2x]₁² = 0 − (−1) = 1
∫₂³(−x+4)dx = [−x²/2+4x]₂³ = 7.5 − 6 = 1.5
∫₁³ (x−1)/2 dx = [(x−1)²/4]₁³ = 1 − 0 = 1
Area = 1 + 1.5 − 1 = 3/2 sq units
Q10. y = sin x, x-axis, x = 0, x = π ke beech ka area nikalo.
sin x poori range [0, π] me x-axis ke upar (non-negative) hai, toh direct integrate karo — koi sign issue nahi.
Area = ∫0π sin x dx = [−cos x]0π = (1) − (−1) = 2 sq units
Q11. Semicircle y = √(9 − x²) (upper half, radius 3) aur x-axis ke beech ka area nikalo.
Yeh poore circle x² + y² = 9 ka upper half hai — is baar direct formula/symmetry use karo (integration full circle ka half).
Area = (1/2) π r² = (1/2) π (3)² = 9π/2 sq units
(Verify: ∫₋₃³ √(9−x²) dx = [(x/2)√(9−x²) + (9/2) sin⁻¹(x/3)]₋₃³ = (9π/4) − (−9π/4) = 9π/2 ✓)
Q12. Circle x² + y² = 8x aur parabola y² = 4x ke beech, x-axis ke upar wala area nikalo.
Circle: (x−4)² + y² = 16 (center (4,0), radius 4). Intersection nikalo: y² = 4x ko circle me daalo → x² − 8x + 4x = 0 → x² − 4x = 0 → x = 0, 4. x = 4 pe y² = 16, y = 4.
x = 0 se x = 4 tak parabola circle ke andar/neeche hai (region parabola se bounded), x = 4 se x = 8 tak region circle se bounded.
Area = ∫04 2√x dx + ∫48 √(16 − (x−4)²) dx
Part 1 = [(4/3)x3/2]₀⁴ = 32/3.
Part 2: u = x−4, u: 0→4. ∫₀⁴ √(16−u²) du = [(u/2)√(16−u²) + 8 sin⁻¹(u/4)]₀⁴ = (0 + 8·π/2) − 0 = 4π.
Total Area = 32/3 + 4π sq units
Q13. y = cos x, x-axis, x = 0 aur x = 2π ke beech (poora absolute) area nikalo.
cos x [0, π/2] me +ve, [π/2, 3π/2] me −ve, [3π/2, 2π] me +ve hai — teen pieces me todo, har piece me modulus lo.
Area = |∫0π/2 cos x dx| + |∫π/23π/2 cos x dx| + |∫3π/22π cos x dx|
= |1 − 0| + |(−1) − 1| + |0 − (−1)| = 1 + 2 + 1 = 4 sq units.
Total Area = 4 sq units (yaad rahe: seedha ∫₀^{2π} cos x dx karne se 0 aayega — galat, kyunki +ve/−ve cancel ho jaate hain)
Q14. Parabola y² = 4ax aur uski latus rectum (x = a) ke beech ghira area nikalo.
Parabola x-axis ke around symmetric hai — upper half (y = 2√(ax), x: 0→a) nikalo, phir ×2.
Upper half area = ∫0a 2√(a) √x dx = 2√a · [(2/3)x3/2]0a = 2√a · (2/3) a3/2 = (4/3) a²
Symmetry se poora area (upar + neeche wala half) = 2 × (4/3)a².
Total Area = 8a²/3 sq units
Q15. Region {(x, y) : x² ≤ y ≤ x} ka area nikalo (parabola y = x² aur line y = x ke beech).
Intersection: x² = x → x(x−1) = 0 → x = 0, 1. Range (0,1) me line (y = x) parabola (y = x²) se upar hai — jaise x = 0.5 pe x = 0.5 > x² = 0.25.
Area = ∫01 (x − x²) dx = [x²/2 − x³/3]01 = 1/2 − 1/3 = 1/6 sq units
Important Equations — Ek Nazar Me
| Situation | Formula |
|---|---|
| Area under curve (x-axis, x = a to b) | ∫ab y dx |
| Area under curve (y-axis, y = c to d) | ∫cd x dy |
| Area between two curves (upper f, lower g) | ∫ab [f(x) − g(x)] dx |
| Curve neeche x-axis ke (y < 0) | Modulus lo — |∫ y dx| |
| Circle x² + y² = r² (poora) | π r² (quarter × 4 se derive) |
| Ellipse x²/a² + y²/b² = 1 (poori) | π a b (quarter × 4 se derive) |
| Semicircle (radius r) | (1/2) π r² |
| Parabola y² = 4ax aur latus rectum x = a | 8a²/3 |
| Curve jo range me x-axis cross kare | Intersection point pe todo, har piece ka modulus lo, add karo |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Intersection points nikale bina limits laga dena. Do curves ke beech area chahiye toh sabse pehle unhe equal karke intersection x (ya y) values nikalo — wahi integration ki limits hongi. Random ya diye hue x = 0, x = 2 jaise limits use karna galat hoga jab tak curves khud wahin intersect na karein.
- Upper aur lower curve ulta subtract karna. Area = (upar wala) − (neeche wala) hona chahiye. Agar galti se neeche − upar kar diya, negative answer aayega — phir bina samjhe modulus laga ke 'sahi' kar lena galat habit hai; pehle range ke beech ek point check karke sahi order pata karo.
- Curve x-axis ke neeche ho aur modulus na lagana. Agar y < 0 ho poori range ya kuch range me, uska integral negative aayega. Area kabhi negative nahi hota — |integral| lo. Poori range me sign check karna zaroori hai (jaise y = sin x ke case me [π, 2π] wala part negative hoga).
- Circle/ellipse ka area seedha bina symmetry ke lambi calculation se nikalna. Poora circle/ellipse integrate karne ki jagah quarter (first quadrant) nikalo aur ×4 karo — sign errors aur calculation mistakes kaafi kam ho jaati hain. Symmetry use na karna time waste aur error-prone dono hai.
- Region positive aur negative x dono me phailta ho toh single integral laga dena. Jaise y = x² aur y = 4 ka region x = −2 se x = 2 tak hai — yahan seedha ek hi integral sahi hai (function continuous, sign nahi badalta), lekin jab curve khud x-axis cross karti ho (jaise line jo negative se positive jaati hai), tab bina tode ek integral lagana galat answer dega — pehle check karo curve apni value change (sign) toh nahi kar rahi.
- ∫y dx aur ∫x dy me confuse ho jaana. x-axis ke against (vertical strips) area chahiye toh ∫y dx karo, y-axis ke against (horizontal strips) chahiye toh ∫x dy. Curve ka equation jis form me di hai usi ke hisab se sahi variable chuno — warna x aur y ulta daal ke poora integral hi galat ho jaata hai.
Board-Style Important Questions
- 2 marks: Curve y² = x aur lines x = 1, x = 4 ke beech (x-axis ke upar) area nikalo.
- 2 marks: Parabola y = x² aur line y = 4 ke beech ghira hua area nikalo.
- 3 marks: Circle x² + y² = 4 ka area integration aur symmetry use karke nikalo.
- 3 marks: Parabola y² = 4x aur line y = x ke beech ghira hua area nikalo, intersection points dikha ke.
- 5 marks: Circle x² + y² = 8x aur parabola y² = 4x ke beech, x-axis ke upar wala area nikalo.
- 5 marks: Ellipse x²/a² + y²/b² = 1 ka poora area integration se derive karo, symmetry use karte hue.
Aksar Poochhe Jaane Wale Sawaal
Application of Integrals me kitne exercises hain?
Sirf 2 — Ex 8.1 (area under simple curves) aur Ex 8.2 (area between two curves), plus miscellaneous questions. Yeh 13 chapters wale rationalised syllabus ka sabse chhota chapter hai.
Area between two curves ke liye limits kaise decide karein?
Dono curves ki equations solve karo (equal karke) — jo x (ya y) values common solution me milengi, wahi integration ki lower aur upper limits hongi. Diagram nahi bhi hai toh bhi ye method kaam karta hai.
Agar curve x-axis ke neeche ho toh kya karna hai?
Integral negative aayega kyunki y ki value negative hai us range me. Area hamesha positive hota hai, isliye modulus (absolute value) lagao — |∫ y dx|.
Circle ya ellipse ka poora area nikalne ka fastest tarika kya hai?
Symmetry use karo. Circle/ellipse dono axes ke around symmetric hote hain, toh sirf first quadrant (quarter) ka integral nikalo aur usse 4 se multiply kar do — poora circle/ellipse baar-baar integrate karne ki zarurat nahi.
y = f(x) aur x = g(y) me area formula alag kyun hai?
Formula strip ke orientation pe depend karta hai. Vertical strips (dx wide) lo toh height y hoti hai, integral ∫y dx. Horizontal strips (dy wide) lo toh width x hoti hai, integral ∫x dy. Dono same area denge, bas curve ka form dekh ke jo aasan ho wahi choose karo.
Triangle ka area integration se kyun nikalna sikhaya jaata hai, jab formula seedha hai?
Yeh concept clarify karta hai ki koi bhi polygon ya region — chahe uske sides straight lines hi kyun na hon — integration se area nikala ja sakta hai. Har side ki equation nikal ke, upar wale lines minus neeche wale lines integrate karke poore region ka area milta hai, jo shoelace formula se bhi match karta hai.
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