JEE Mathematics • Official PYQs

Integral Calculus (Definite Integrals & Area)

Practice JEE Main Definite Integrals & Area Under Curves PYQs online. Master King’s Property, Leibniz Rule, periodic integrals & 4-step solutions.

📝 15 Solved PYQs
📅 Exam Years: 2019–2024
📊 Weightage: 8%–10% (2–3 Questions per shift)
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🎯 High-Yield Syllabus Breakdown

These high-frequency concepts account for over 90% of all questions asked in Integral Calculus (Definite Integrals & Area):

⚡ Essential Formulas & Speed Shortcuts

Formulas and time-saving shortcuts to solve Definite Integrals & Area questions in under 45 seconds:

Speed Trick #1:

King’s Property shortcut: If f(x) + f(a + b - x) = c (constant), then ∫_a^b f(x) dx = c · (b - a) / 2.

Speed Trick #2:

Leibniz Rule: d/dx [∫_u(x)^v(x) f(t) dt] = f(v(x))·v’(x) - f(u(x))·u’(x).

Speed Trick #3:

Area between parabola y² = 4ax and line y = mx is 8a² / (3m³).

📖 Sample Solved Exam Questions

Preview authentic questions from this chapter with full step-by-step solutions:

Question 1 of 3 JEE Main 2024 (Online) 27th Jan Shift-1
The value of the definite integral ∫₀^(π/2) [sin³ x / (sin³ x + cos³ x)] dx is:
(A) π/4
(B) π/2
(C) π/8
(D) 1
💡 View Step-by-Step Solution & Answer

Correct Answer: π/4

  1. Step 1: Let I = ∫₀^(π/2) [sin³ x / (sin³ x + cos³ x)] dx.
  2. Step 2: Apply King's Property: ∫₀^a f(x) dx = ∫₀^a f(a - x) dx. Here a = π/2.
  3. Step 3: Replace x with π/2 - x: I = ∫₀^(π/2) [cos³ x / (cos³ x + sin³ x)] dx.
  4. Step 4: Add the two expressions for I: 2I = ∫₀^(π/2) [(sin³ x + cos³ x)/(sin³ x + cos³ x)] dx = ∫₀^(π/2) 1 dx = π/2. Thus I = π/4.

⚡ Speed Tip: King's Property Symmetry: For ∫₀^(π/2) f(sin x)/[f(sin x) + f(cos x)] dx, the integral is always (b - a)/2 = π/4.

Question 2 of 3 JEE Main 2023 (Online) 24th Jan Shift-2
The area bounded by the curves y = x² and y = 2x is:
(A) 4/3
(B) 2/3
(C) 1/3
(D) 8/3
💡 View Step-by-Step Solution & Answer

Correct Answer: 4/3

  1. Step 1: Find intersection points: x² = 2x ⇒ x(x - 2) = 0 ⇒ x = 0 and x = 2.
  2. Step 2: On the interval [0, 2], the line y = 2x lies above the parabola y = x².
  3. Step 3: Area = ∫₀² (2x - x²) dx = [x² - x³/3]₀².
  4. Step 4: Evaluate at limits: (2² - 2³/3) - 0 = (4 - 8/3) = (12 - 8)/3 = 4/3.

⚡ Speed Tip: Archimedes Area Formula: Area between parabola y² = 4ax and line y = mx is 8a²/(3m³). For x² = 2x, [x² - x³/3]₀² = 4/3.

Question 3 of 3 JEE Main 2024 (Online) 29th Jan Shift-1
If f(x) = ∫₀ˣ t sin t dt, then f'(π/2) is equal to:
(A) π/2
(B) 1
(C) 0
(D) π
💡 View Step-by-Step Solution & Answer

Correct Answer: π/2

  1. Step 1: Apply Leibniz Rule for differentiation under the integral sign: d/dx ∫_{u(x)}^{v(x)} g(t) dt = g(v(x)) · v'(x) - g(u(x)) · u'(x).
  2. Step 2: Here upper limit v(x) = x with v'(x) = 1, and lower limit is constant 0 with u'(x) = 0.
  3. Step 3: Therefore, f'(x) = (x sin x) · (1) - 0 = x sin x.
  4. Step 4: Evaluate at x = π/2: f'(π/2) = (π/2) · sin(π/2) = (π/2) · 1 = π/2.

⚡ Speed Tip: Leibniz Rule: d/dx ∫₀ˣ g(t) dt = g(x). Simply evaluate x sin x at π/2 to get π/2.

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❓ Frequently Asked Questions (FAQs)

How does King’s Rule simplify trigonometric integrals?

Adding the original integral I to King’s transformed integral gives 2I = ∫ (numerator / denominator) dx = ∫ 1 dx = (b - a), yielding I = (b - a)/2 immediately.

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