Integral Calculus – JEE Main Maths Formula Sheet & Class 12 Notes | All Formulas, Standard Integrals & Methods

Fundamental Integrals — Complete Table from the PDF

Why All Fundamental Integrals Are the Non-Negotiable Base of JEE Main Integral Calculus

Integration as antiderivative (from PDF — Integral Calculus): ∫f(x)dx = g(x) + c if and only if g'(x) = f(x). Here c is the constant of integration, f(x) is the integrand, and g(x) is the antiderivative. The integral is not unique due to c. Fundamental theorem connection: derivative of integral = integrand: d/dx[∫f(x)dx] = f(x).

Fundamental rules of integration (from PDF — Integral Calculus):

∫[f₁(x) + f₂(x) + … + fₙ(x)]dx = ∫f₁dx + ∫f₂dx + … + ∫fₙdx

∫k·f(x)dx = k·∫f(x)dx (k ≠ 0 constant)

If ∫f(x)dx = F(x) + c, then ∫f(ax+b)dx = F(ax+b)/a + c (a ≠ 0)

Fundamental algebraic integrals (from PDF — Integral Calculus JEE Main):

∫xⁿ dx = xⁿ⁺¹/(n+1) + c (n ≠ –1)

∫(ax+b)ⁿ dx = (ax+b)ⁿ⁺¹/[a(n+1)] + c (n ≠ –1, a ≠ 0)

∫(1/x)dx = ln|x| + c (x ≠ 0)

∫(1/(ax+b))dx = (1/a)·ln|ax+b| + c (a ≠ 0)

∫f'(x)/f(x) dx = ln|f(x)| + c — the "log formula" — used whenever the numerator is the derivative of the denominator

Fundamental trigonometric integrals (from PDF — Integral Calculus JEE Main):

∫sin x dx = –cos x + c

∫cos x dx = sin x + c

∫sec²x dx = tan x + c

∫cosec²x dx = –cot x + c

∫sec x tan x dx = sec x + c

∫cosec x cot x dx = –cosec x + c

∫tan x dx = ln|sec x| + c = –ln|cos x| + c

∫cot x dx = ln|sin x| + c = –ln|cosec x| + c

∫sec x dx = ln|sec x + tan x| + c = ln|tan(π/4 + x/2)| + c

∫cosec x dx = ln|cosec x – cot x| + c = –ln|cosec x + cot x| + c = ln|tan(x/2)| + c

Fundamental exponential integrals (from PDF — Integral Calculus JEE Main):

∫eˣ dx = eˣ + c

∫e^(ax+b) dx = e^(ax+b)/a + c (a ≠ 0)

∫aˣ dx = aˣ/(ln a) + c (a > 0, a ≠ 1)

∫a^(px+q) dx = a^(px+q)/(p·ln a) + c (p ≠ 0)

Fundamental inverse trig integrals (from PDF — Integral Calculus JEE Main):

∫dx/√(a²–x²) = sin⁻¹(x/a) + c = –cos⁻¹(x/a) + c (|x| < a, a > 0)

∫dx/(a²+x²) = (1/a)tan⁻¹(x/a) + c = –(1/a)cot⁻¹(x/a) + c (a ≠ 0)

∫dx/(|x|√(x²–a²)) = (1/a)sec⁻¹(x/a) + c = –(1/a)cosec⁻¹(x/a) + c (x > a > 0)

Download the Free PDF for Integral Calculus for the complete fundamental integral table with worked examples for JEE Main.

9 Standard Formulas — Direct from the PDF for JEE Main

Why These 9 Standard Integral Formulas Are the Most Directly Tested Integral Calculus Results

These 9 formulas from the PDF are used in both indefinite and definite integration whenever the integrand matches the given pattern. Every one appears in JEE Main either directly or after completing the square / decomposing the numerator.

All 9 Standard Integral Formulas (from PDF — Integral Calculus JEE Main):

1. ∫dx/(a²+x²) = (1/a)tan⁻¹(x/a) + c (a ≠ 0)

2. ∫dx/(a²–x²) = (1/2a)ln|(a+x)/(a–x)| + c (a ≠ 0)

3. ∫dx/(x²–a²) = (1/2a)ln|(x–a)/(x+a)| + c (a ≠ 0)

4. ∫dx/√(a²–x²) = sin⁻¹(x/a) + c (a > 0)

5. ∫dx/√(a²+x²) = ln|x + √(a²+x²)| + c (a > 0)

6. ∫dx/√(x²–a²) = ln|x + √(x²–a²)| + c (a > 0)

7. ∫√(a²–x²) dx = (x/2)√(a²–x²) + (a²/2)sin⁻¹(x/a) + c (a > 0)

8. ∫√(a²+x²) dx = (x/2)√(a²+x²) + (a²/2)ln|x + √(a²+x²)| + c (a > 0)

9. ∫√(x²–a²) dx = (x/2)√(x²–a²) – (a²/2)ln|x + √(x²–a²)| + c (a > 0)

How to apply the 9 standard formulas in Integral Calculus (JEE Main method): For integrals like ∫dx/(ax²+bx+c) or ∫dx/√(ax²+bx+c): complete the square to write ax²+bx+c = a[(x+b/2a)² ± (b²–4ac)/4a²], then substitute X = x+b/2a to match one of formulas 1–6. For ∫(px+q)/(ax²+bx+c)dx: write px+q = A·d(ax²+bx+c)/dx + B = A(2ax+b)+B, split into two integrals (log formula + standard formula). For ∫√(ax²+bx+c)dx: complete the square, substitute X = x+b/2a, match formulas 7–9. Download the Free PDF for Integral Calculus for worked examples of all 9 standard formulas for JEE Main.

Method of Substitution — All 10 Standard Substitution Types

Why Substitution Is the Most Versatile Integral Calculus Method for JEE Main

Substitution principle (from PDF — Integral Calculus): ∫f(x)dx = ∫f(φ(t))·φ'(t)dt where x = φ(t). The substitution converts the integral into a simpler form that matches a standard formula. After integration in t, substitute back in terms of x.

Key substitution patterns from the PDF — Integral Calculus:

(i) Integrand of form f(ax+b): substitute ax+b = t → dx = dt/a → ∫f(ax+b)dx = (1/a)∫f(t)dt = (1/a)F(ax+b) + c.

(ii) Integrand of form xⁿ⁻¹·f(xⁿ): substitute xⁿ = t → nxⁿ⁻¹dx = dt → ∫xⁿ⁻¹f(xⁿ)dx = (1/n)∫f(t)dt.

(iii) Integrand of form [f(x)]ⁿ·f'(x): substitute f(x) = t → f'(x)dx = dt → ∫[f(x)]ⁿ·f'(x)dx = tⁿ⁺¹/(n+1) + c = [f(x)]ⁿ⁺¹/(n+1) + c.

(iv) Integrand of form f'(x)/f(x): substitute f(x) = t → ∫f'(x)/f(x)dx = ∫dt/t = ln|t| + c = ln|f(x)| + c.

All 10 Standard Trigonometric Substitution Types (from PDF — Integral Calculus JEE Main):

(i) 1/√(a²–x²) or √(a²–x²): put x = a sinθ or x = a cosθ

(ii) 1/√(a²+x²) or √(a²+x²): put x = a tanθ or x = a cotθ

(iii) 1/√(x²–a²) or √(x²–a²): put x = a secθ or x = a cosecθ

(iv) (a–x)/(a+x) or √((a–x)/(a+x)): put x = a cos2θ

(v) √(x/(a–x)) or √(x/(a+x)): put x = a sin²θ or x = a cos²θ

(vi) √(x/(a+x)) or √(x/(a–x)) (second form): put x = a tan²θ or x = a cot²θ

(vii) √((x–b)/(a–x)) or √((a–x)(x–b)) for b < x < a: put x = a cos²θ + b sin²θ

(viii) √((x–a)/(b–x)) or √((x–a)(b–x)) for a < x < b: put x = a sec²θ – b tan²θ

(ix) 1/√((x–a)(x–b)): put x – a = t² or x – b = t²

(x) √(x–a)/(x+a) or √(x+a)/(x–a): put x = a secθ or x = a sec²θ

Substitution for ∫dx/√Y where X and Y are linear/quadratic (from PDF — Integral Calculus):

X linear, Y linear: z² = Y. X quadratic, Y linear: z² = Y. X linear, Y quadratic: z = 1/X. X quadratic, Y quadratic: z² = Y/X or z = 1/X. Download the Free PDF for Integral Calculus for worked examples of all substitution types for JEE Main.

Integration by Parts — ILATE Rule and Key Formulas

Why Integration by Parts Is the Second-Most-Used Integral Calculus Method in JEE Main

Integration by Parts formula (from PDF — Integral Calculus):

∫u·v dx = u·∫v dx – ∫[u' · ∫v dx] dx

= (first function) × (integral of second function) – ∫[(derivative of first function) × (integral of second function)]dx

ILATE Rule (from PDF — Integral Calculus) — choice of first function: Choose u (first function) in the order of priority: I → L → A → T → E: Inverse trig (sin⁻¹x, cos⁻¹x, tan⁻¹x, …); Logarithmic (logₑx, log₁₀x, …); Algebraic (xⁿ, polynomial); Trigonometric (sinx, cosx, tanx, …); Exponential (eˣ, aˣ).

ILATE application rules (from PDF — Integral Calculus JEE Main):

(iv) For ∫xⁿ·f(x)dx: take xⁿ as first function (algebraic before trig/exponential).

(vii) For ∫(log x)ⁿ dx: take 1 as second function, log x as first → ∫(logx)·1 dx = x·logx – ∫x·(1/x)dx = x·logx – x + c.

(ix) Can be used repeatedly when a single application produces another byparts integral.

(x) When both functions are trigonometric: take the one whose integral is simpler as second function.

(xi) When both functions are algebraic: take the one whose derivative is simpler as first function.

Special integral formulas from Integration by Parts (from PDF — Integral Calculus JEE Main):

(iv) ∫eˣ[f(x) + f'(x)]dx = eˣf(x) + c — the most important integration by parts shortcut

Proof: ∫eˣf(x)dx + ∫eˣf'(x)dx = eˣf(x) – ∫eˣf'(x)dx + ∫eˣf'(x)dx = eˣf(x) + c.

(v) ∫eˣ[xf(x) + f'(x)]dx... wait — exact from PDF: ∫φ(x)[xf(x) + f'(x)]φ'(x)dx = φ(x)f(x)·φ(x) + c... cleaner version: if ∫f(x)dx exists, ∫φ'(x)f(φ(x))dx = f(φ(x)) + c (chain rule reverse)

The eˣ formula from PDF exactly: ∫eˣ{f(x) + f'(x)}dx = eˣ·f(x) + c

This covers: ∫eˣ(sinx + cosx)dx = eˣ sinx + c [f(x)=sinx, f'(x)=cosx]. ∫eˣ(cosx – sinx)dx = eˣ cosx + c. ∫eˣ(tan x + sec²x)dx = eˣ tanx + c. ∫eˣ(1/x – 1/x²)dx = eˣ/x + c.

Multiple Integration by Parts (from PDF — Integral Calculus):

∫f(x)·g(x)dx = f(x)g₁(x) – f'(x)g₂(x) + f''(x)g₃(x) – …

where gₖ(x) = kth integral of g(x), fʳ(x) = rth derivative of f(x). Works when f(x) is a polynomial (terminates). Download the Free PDF for Integral Calculus for all integration by parts examples for JEE Main.

Method of Partial Fractions — All 5 Cases for Integral Calculus

Why Partial Fractions Are the Most Systematic Integral Calculus Method for JEE Main Rational Integrals

Partial fractions setup (from PDF — Integral Calculus): For ∫p(x)/g(x)dx, if deg(p) ≥ deg(g): first perform polynomial long division to write p(x)/g(x) = quotient + f(x)/g(x) where deg(f) < deg(g). Then apply partial fractions to f(x)/g(x).

All 5 Cases of Partial Fractions (from PDF — Integral Calculus JEE Main):

Case 1 — Non-repeated linear factors: g(x) = (x–α₁)(x–α₂)…(x–αₙ)

f(x)/g(x) = A₁/(x–α₁) + A₂/(x–α₂) + … + Aₙ/(x–αₙ) (n constants to find)

Case 2 — Repeated linear factors: g(x) = (x–α₁)²(x–α₃)…(x–αₙ)

f(x)/g(x) = A₁/(x–α₁) + A₂/(x–α₁)² + A₃/(x–α₃) + … + Aₙ/(x–αₙ)

Note: for (x–α)ʳ in denominator, write r partial fractions: A₁/(x–α) + A₂/(x–α)² + … + Aᵣ/(x–α)ʳ

Case 3 — Non-repeated irreducible quadratic factor: g(x) = (ax²+bx+c)(x–α₃)…(x–αₙ) where ax²+bx+c cannot be factorised over ℝ

f(x)/g(x) = (A₁x+A₂)/(ax²+bx+c) + A₃/(x–α₃) + … + Aₙ/(x–αₙ)

Case 4 — Repeated irreducible quadratic factor: g(x) = (ax²+bx+c)²(x–α₅)…(x–αₙ)

f(x)/g(x) = (A₁x+A₂)/(ax²+bx+c) + (A₃x+A₄)/(ax²+bx+c)² + A₅/(x–α₅) + …

Case 5 — Integrand contains only even powers of x (from PDF):

Step 1: Put x² = z in the integrand. Step 2: Resolve the resulting rational expression in z into partial fractions. Step 3: Substitute z = x² back and integrate.

Special integration for non-factorisable quadratic (from PDF — Integral Calculus):

For ∫dx/(ax²+bx+c): write ax²+bx+c = a[(x+b/2a)² + (4ac–b²)/4a²]. Let D = b²–4ac. If D < 0: complete the square, match formula 1 (arctan type). If D > 0: factor, use partial fractions (Case 1).

For ∫(px+q)dx/(ax²+bx+c): write px+q = A·(2ax+b) + B → A = p/2a, B = q–pb/2a. Then: A·∫(2ax+b)/(ax²+bx+c)dx + B·∫dx/(ax²+bx+c) = A·ln|ax²+bx+c| + B·(standard formula). Download the Free PDF for Integral Calculus for all partial fraction integration examples for JEE Main.

Integration of Trigonometric Functions and Special Integrals

Why Trig Integration Patterns and Special Integrals Are JEE Main Integral Calculus Shortcuts

Integration of sinᵐx cosⁿx dx (from PDF — Integral Calculus JEE Main):

(i) If m (power of sinx) is odd: substitute cosx = t → sinx dx = –dt → integrate.

(ii) If n (power of cosx) is odd: substitute sinx = t → cosx dx = dt → integrate.

(iii) If both m and n are odd: substitute either sinx = t or cosx = t.

(iv) If both m and n are even: use trigonometric identities (sin²x = (1–cos2x)/2; cos²x = (1+cos2x)/2; sin2x = 2sinx cosx) to reduce powers.

(v) If m and n are rational and (m+n) is a negative even integer: substitute cotx = t or tanx = t.

Useful trig identities for integration (from PDF — Integral Calculus):

sin²mx = (1–cos2mx)/2; cos²mx = (1+cos2mx)/2

sin mx·cos mx = sin(2mx)/2

sin³mx = (3sinmx – sin3mx)/4; cos³mx = (3cosmx + cos3mx)/4

tan²mx = sec²mx – 1; cot²mx = cosec²mx – 1

2cosAcosB = cos(A+B) + cos(A–B); 2sinAcosB = sin(A+B) + sin(A–B); 2sinAsinB = cos(A–B) – cos(A+B)

Integration of (p sinx + q cosx)/(a sinx + b cosx) dx (from PDF — Integral Calculus): Write numerator = A(denominator) + B(d/dx of denominator) = A(asinx+bcosx) + B(acosx–bsinx). Match coefficients: p = Aa – Bb; q = Ab + Ba. Then integral = Ax + B·ln|asinx+bcosx| + c.

Integration of 1/(a+b sinx), 1/(a+b cosx), 1/(a sinx + b cosx) type (from PDF — Integral Calculus): Divide numerator and denominator by cos²x, substitute tanx = t → sec²x dx = dt. Denominator becomes a quadratic in t → apply standard formulas 1–3.

Special integral — eˣ[f(x)+f'(x)] (from PDF — Integral Calculus): ∫eˣ{f(x) + f'(x)}dx = eˣ·f(x) + c. This single formula handles: ∫eˣ(1/x–1/x²)dx = eˣ/x + c; ∫eˣ(sinx+cosx)dx = eˣsinx + c; ∫eˣ(tanx+sec²x)dx = eˣtanx + c; ∫eˣ(1+logx)dx = eˣlogx + c.

Algebraic Twins (from PDF — Integral Calculus JEE Main):

∫(x²+1)/(x⁴+1)dx = ∫(1+1/x²)/[(x–1/x)²+2]dx [put x–1/x = t] = (1/√2)tan⁻¹((x–1/x)/√2) + c

∫(x²–1)/(x⁴+1)dx = ∫(1–1/x²)/[(x+1/x)²–2]dx [put x+1/x = t] = (1/2√2)ln|(x+1/x–√2)/(x+1/x+√2)| + c

∫1/(x⁴+1)dx = (1/2)[∫(x²+1)/(x⁴+1)dx – ∫(x²–1)/(x⁴+1)dx] (split and use both twins). Download the Free PDF for Integral Calculus for all trig and special integral examples.

Definite Integral — Fundamental Theorem and Limit of Sum

Why Definite Integration and Limit-of-Sum Are the Most Formula-Rich Integral Calculus Section

Fundamental Theorem of Calculus (from PDF — Integral Calculus): Let f(x) be continuous on [a, b] and ∫f(x)dx = F(x) + c. Then: ∫ₐᵇ f(x)dx = F(b) – F(a) = [F(x)]ₐᵇ. This definite integral equals the net algebraic area bounded by y=f(x), x=a, x=b, and the x-axis (positive above, negative below x-axis).

Definite Integral as Limit of Sum (from PDF — Integral Calculus JEE Main): ∫₀¹ f(x)dx = lim_{n→∞} (1/n)·Σᵣ₌₁ⁿ f(r/n). Working method: to convert the given limit of sum to a definite integral, replace r/n → x, 1/n → dx, lim_{n→∞}Σ → ∫, lower limit = lim(first r/n term), upper limit = lim(last r/n term).

Key series summation formulas from PDF — Integral Calculus:

Σᵣ₌₁ⁿ r = n(n+1)/2

Σᵣ₌₁ⁿ r² = n(n+1)(2n+1)/6

Σᵣ₌₁ⁿ r³ = [n(n+1)/2]²

Sum of n-term AP sin: sinα+sin(α+β)+…+sin(α+(n–1)β) = sin(α+(n–1)β/2)·sin(nβ/2)/sin(β/2)

Sum of n-term AP cos: cosα+cos(α+β)+…+cos(α+(n–1)β) = cos(α+(n–1)β/2)·sin(nβ/2)/sin(β/2)

1–1/2+1/3–1/4+… = log_e 2

1/1²–1/2²+1/3²–1/4²+… = π²/12

1/1²+1/2²+1/3²+1/4²+… = π²/6

1/1²–1/3²+1/5²–… = π²/8 (from PDF: 1–1/9+1/25–… = π²/8)

1/2²+1/4²+1/6²+… = π²/24

Also from PDF — complex form: e^(iθ) = cosθ + i sinθ; sinθ = (e^(iθ) – e^(–iθ))/(2i); cosθ = (e^(iθ) + e^(–iθ))/2. Hyperbolic: sinhθ = (eθ–e^(–θ))/2; coshθ = (eθ+e^(–θ))/2. Download the Free PDF for Integral Calculus for limit-of-sum examples and series formulas for JEE Main.

All 20 Properties of Definite Integrals — Direct from the PDF

Why These 20 Properties Are the Most Pattern-Exploiting Integral Calculus Formulas in JEE Main

The 20 properties of definite integrals allow evaluation and simplification of definite integrals without computing the antiderivative. JEE Main regularly tests properties 2, 3, 4, 5, 6, 7, 8 (the "king", even-odd, and periodic properties) directly.

Properties 1–8 (Standard Properties from PDF — Integral Calculus):

P1: ∫ₐᵇ f(x)dx = ∫ₐᵇ f(z)dz (dummy variable)

P2: ∫ₐᵇ f(x)dx = –∫ᵦₐ f(x)dx (swapping limits reverses sign)

P3: ∫ₐᵇ f(x)dx = ∫ₐᶜ f(x)dx + ∫ᶜᵇ f(x)dx for any c (additive interval)

P4: ∫₀ᵃ f(x)dx = ∫₀ᵃ f(a–x)dx — the "King Property" — replace x by (a–x)

P5: ∫₀^(2a) f(x)dx = 2∫₀ᵃ f(x)dx if f(2a–x) = f(x) (even about a); = 0 if f(2a–x) = –f(x) (odd about a)

P6: ∫₀^(2a) f(x)dx = ∫₀ᵃ f(x)dx + ∫₀ᵃ f(2a–x)dx (split [0,2a] and substitute x = 2a–y)

P7 (from P5+P6): If f(2a–x) = f(x) → ∫₀^(2a) f(x)dx = 2∫₀ᵃ f(x)dx. If f(2a–x) = –f(x) → ∫₀^(2a) f(x)dx = 0.

P8 (Periodic): If f is periodic with period T: ∫ₐ^(a+nT) f(x)dx = n·∫₀ᵀ f(x)dx; ∫_(mT)^(nT) f(x)dx = (n–m)·∫₀ᵀ f(x)dx

Properties 9–20 (Advanced Properties from PDF — Integral Calculus):

P9: If f(x) ≥ 0 on [a,b] then ∫ₐᵇ f(x)dx ≥ 0

P10: If f(x) is odd, then ∫₀ˣ f(t)dt is an even function of x

P11: If f(x) is even, then ∫₀ˣ f(t)dt is an odd function of x

Consequence of P10 & P11: If f(x) is even: ∫₋ₐᵃ f(x)dx = 2∫₀ᵃ f(x)dx. If f(x) is odd: ∫₋ₐᵃ f(x)dx = 0.

P12 (Leibnitz Rule): d/dx[∫_{g(x)}^{h(x)} f(t)dt] = f(h(x))·h'(x) – f(g(x))·g'(x)

P13: If f(x) is continuous on [a,b] and ∫ₐᵇ f(x)dx = 0, then f(c) = 0 for at least one c ∈ [a,b]

P14: If f(x) < g(x) for all x ∈ (a,b) then ∫ₐᵇ f(x)dx < ∫ₐᵇ g(x)dx

P15: |∫ₐᵇ f(x)dx| ≤ ∫ₐᵇ |f(x)|dx (triangle inequality for integrals)

P16 (Estimation): m(b–a) ≤ ∫ₐᵇ f(x)dx ≤ M(b–a) where m = min f, M = max f on [a,b]

P17 (Mean Value Theorem): If f continuous on [a,b]: ∃c∈(a,b) such that ∫ₐᵇ f(x)dx = f(c)·(b–a). The value f(c) = (1/(b–a))·∫ₐᵇ f(x)dx is the mean value of f on [a,b].

P18 (Generalised MVT): If f(x) and φ(x) are continuous on [a,b] and φ(x) has constant sign on [a,b]: ∃c∈(a,b) such that ∫ₐᵇ f(x)φ(x)dx = f(c)·∫ₐᵇ φ(x)dx.

P19: If f(x) is continuous on [a,b], then Φ(x) = ∫ₐˣ f(t)dt is differentiable on (a,b) with Φ'(x) = f(x).

P20 (Cauchy-Schwarz type): [∫ₐᵇ f(x)g(x)dx]² ≤ ∫ₐᵇ [f(x)]²dx · ∫ₐᵇ [g(x)]²dx

Download the Free PDF for Integral Calculus for all 20 properties with worked examples and JEE Main applications.

Area Under Curves — All Formulas for Integral Calculus

Why Area Calculation Is the Most Visually Applied Integral Calculus Topic in JEE Main

Area bounded by curve, x-axis, and vertical lines (from PDF — Integral Calculus): If y = f(x) is continuous on [a, b] and f(x) ≥ 0 on [a, b], then area bounded by y = f(x), x = a, x = b, and the x-axis is: A = ∫ₐᵇ f(x)dx = ∫ₐᵇ y dx.

If f(x) < 0 on [a, b] (curve below x-axis): A = |∫ₐᵇ f(x)dx| = ∫ₐᵇ |f(x)|dx.

If f(x) changes sign in [a, b] at x = c: A = |∫ₐᶜ f(x)dx| + |∫ᶜᵇ f(x)dx| (evaluate separately and add magnitudes).

Area bounded by curve, y-axis, and horizontal lines (from PDF — Integral Calculus): If x = f(y) is continuous on [c, d] (c ≤ y ≤ d): A = ∫ᶜᵈ f(y)dy = ∫ᶜᵈ x dy.

Area between two curves (from PDF — Integral Calculus JEE Main): If y = f(x) lies above y = g(x) on [a, b] (f(x) ≥ g(x) for all x ∈ [a, b]): A = ∫ₐᵇ [f(x) – g(x)]dx.

If the curves intersect at x = a and x = b (boundaries degenerate to intersection points): A = ∫ₐᵇ |f(x) – g(x)|dx.

If the upper and lower curves swap at an interior point x = c: A = ∫ₐᶜ [f(x)–g(x)]dx + ∫ᶜᵇ [g(x)–f(x)]dx (use modulus to ensure positive area).

Key remarks for area problems (from PDF — Integral Calculus):

1. If the whole region is below the x-axis, the integral comes out negative — take absolute value for area.

2. Always split the region at any x-intercept or curve intersection before integrating, so each sub-region has a definite sign.

3. The formula A = ∫ₐᵇ [f(x)–g(x)]dx assumes f(x) ≥ g(x) throughout [a, b] — verify this before applying.

4. For parametric curves x=φ(t), y=ψ(t): A = ∫_{t₁}^{t₂} ψ(t)·φ'(t)dt (change integration variable to t). Download the Free PDF for Integral Calculus for all area formula examples including standard curves for JEE Main.