How to Solve: Definite Integration (Properties, Wallis’ Formula, Reduction Formula, Area Under Curve) – IIT JEE Guide

How to Solve: Definite Integration (Properties, Wallis’ Formula, Reduction Formula, Area Under Curve) – IIT JEE Guide

Introduction

Mastering definite integration unlocks 10-15% of your IIT JEE Math score—from calculating rocket trajectories to finding areas under curves in physics problems. One wrong sign or missed property, and you lose 5-6 marks in a single question. This guide ensures you never make those mistakes again.

What You Need To Know First

Before diving in, you must be comfortable with: 1. Indefinite Integration – Basic antiderivatives (e.g., ∫xⁿ dx, ∫sin x dx, ∫eˣ dx). 2. Limits & Continuity – Understanding how functions behave at boundaries. 3. Trigonometric Identities – Double-angle, product-to-sum, and Pythagorean identities.

If any of these feel shaky, pause here and review them first.

Key Vocabulary

Formulas To Know

1. Basic Properties of Definite Integrals

2. Wallis’ Formula (For ∫₀ᵖⁱ/² sinⁿ x dx or ∫₀ᵖⁱ/² cosⁿ x dx)

Example of Double Factorial: - 5!! = 5 × 3 × 1 = 15 - 6!! = 6 × 4 × 2 = 48

3. Reduction Formulas (For Powers of Trig Functions)

4. Area Under Curve

Step-by-Step Method

How to Solve Any Definite Integral Problem (IIT JEE Style)

Follow these exact steps for every problem:

1. Check for Symmetry (Even/Odd Function)
2. If the integral is from -a to a, check if f(x) is even or odd.
3. If even: ∫₋ₐᵃ f(x) dx = 2∫₀ᵃ f(x) dx.

4. If odd: ∫₋ₐᵃ f(x) dx = 0 (instant answer!).

5. Simplify Using Properties

6. Break the integral into smaller parts using linearity.

7. Example: ∫₀ᵖⁱ (3sin x + 2cos x) dx = 3∫₀ᵖⁱ sin x dx + 2∫₀ᵖⁱ cos x dx.

8. Apply Substitution (If Needed)

9. If the integrand has a composite function (e.g., √(1-x²)), use substitution.

10. Example: ∫₀¹ x√(1-x²) dx → Let u = 1-x², du = -2x dx.

11. Use Reduction Formula (For Powers)

12. If the integrand is sinⁿ x, cosⁿ x, tanⁿ x, or secⁿ x, apply the reduction formula.

13. Example: ∫₀ᵖⁱ/² sin⁴ x dx → Use Iₙ = (n-1)/n Iₙ₋₂.

14. Apply Wallis’ Formula (For Trig Powers)

15. If the integral is from 0 to π/2 and involves sinⁿ x or cosⁿ x, use Wallis’ formula.

16. Example: ∫₀ᵖⁱ/² sin⁶ x dx = (5!! / 6!!) × π/2.

17. Evaluate the Integral

18. Find the antiderivative and apply the limits.

19. Example: ∫₀¹ x² dx = [x³/3]₀¹ = 1/3 - 0 = 1/3.

20. Check for Area Under Curve (If Required)

21. If the question asks for area, ensure the result is non-negative.
22. Example: ∫₀ᵖⁱ sin x dx = 2, but area = 2 (since sin x ≥ 0 in [0, π]).

Worked Examples

Example 1 – Basic (Symmetry + Linearity)

Problem: Evaluate ∫₋₁¹ (x³ + 2x²) dx.

Step-by-Step Solution: 1. Check for Symmetry:
- f(x) = x³ + 2x².
- x³ is odd (f(-x) = -f(x)).
- 2x² is even (f(-x) = f(x)). 2. Split the Integral:
∫₋₁¹ (x³ + 2x²) dx = ∫₋₁¹ x³ dx + ∫₋₁¹ 2x² dx. 3. Apply Symmetry:
- ∫₋₁¹ x³ dx = 0 (odd function).
- ∫₋₁¹ 2x² dx = 2 × 2∫₀¹ x² dx (even function). 4. Evaluate:
2 × 2∫₀¹ x² dx = 4 [x³/3]₀¹ = 4 (1/3 - 0) = 4/3.

What we did and why: - Used symmetry to simplify the integral. - Split the integral into odd + even parts for faster evaluation.

Example 2 – Medium (Reduction Formula + Wallis’ Formula)

Problem: Evaluate ∫₀ᵖⁱ/² sin⁴ x dx.

Step-by-Step Solution: 1. Recognize the Form:
- Integrand is sin⁴ x, a power of sine.
- Limits are 0 to π/2 → Wallis’ formula applies. 2. Apply Reduction Formula (or Wallis’):
- I₄ = ∫₀ᵖⁱ/² sin⁴ x dx.
- Using reduction: I₄ = (3/4) I₂.
- I₂ = ∫₀ᵖⁱ/² sin² x dx = (1/2) I₀.
- I₀ = ∫₀ᵖⁱ/² dx = π/2. 3. Substitute Back:
- I₂ = (1/2)(π/2) = π/4.
- I₄ = (3/4)(π/4) = 3π/16.
(Alternatively, use Wallis’ formula directly: I₄ = (3!! / 4!!) × π/2 = (3×1)/(4×2) × π/2 = 3π/16.)

What we did and why: - Used reduction formula to break down the power. - Recognized Wallis’ formula as a shortcut for trigonometric powers.

Example 3 – Exam-Style (Area Under Curve + Tricky Limits)

Problem: Find the area bounded by y = x² - 4 and the x-axis between x = -3 and x = 3.

Step-by-Step Solution: 1. Find Points of Intersection:
- y = x² - 4 = 0 → x = ±2.
- The curve crosses the x-axis at x = -2 and x = 2. 2. Split the Integral at Roots:
- Area = ∫₋₃⁻² |x² - 4| dx + ∫₋₂² |x² - 4| dx + ∫₂³ |x² - 4| dx. 3. Evaluate Each Part:
- Part 1 (x = -3 to -2): x² - 4 ≥ 0 → ∫₋₃⁻² (x² - 4) dx = [x³/3 - 4x]₋₃⁻² = (8/3 + 8) - (-9 + 12) = 8/3 + 8 + 9 - 12 = 11/3.
- Part 2 (x = -2 to 2): x² - 4 ≤ 0 → ∫₋₂² -(x² - 4) dx = 2∫₀² (4 - x²) dx = 2[4x - x³/3]₀² = 2(8 - 8/3) = 32/3.
- Part 3 (x = 2 to 3): x² - 4 ≥ 0 → ∫₂³ (x² - 4) dx = [x³/3 - 4x]₂³ = (9 - 12) - (8/3 - 8) = -3 + 16/3 = 7/3. 4. Total Area:
11/3 + 32/3 + 7/3 = 50/3.

What we did and why: - Split the integral at roots to handle negative areas. - Used absolute value to ensure area is non-negative. - Evaluated three separate integrals for accuracy.

Common Mistakes

Exam Traps

1-Minute Recap (Night Before Exam)

"Listen up—this is your 5-minute crash course for definite integration in IIT JEE.

1. Symmetry is your best friend. If the integral is from -a to a, check if the function is even or odd. Even? Double the integral from 0 to a. Odd? Zero. Done.
2. Wallis’ formula = free marks. If you see ∫₀ᵖⁱ/² sinⁿ x dx or ∫₀ᵖⁱ/² cosⁿ x dx, use Wallis’. Even power? Multiply by π/2. Odd power? No π/2.
3. Reduction formulas = recursion. For ∫sinⁿ x dx, ∫cosⁿ x dx, or ∫tanⁿ x dx, write Iₙ in terms of Iₙ₋₂. Keep reducing until you hit I₀ or I₁.
4. Area ≠ integral. If the question says area, always take the absolute value of the function. Split the integral at roots if needed.
5. Common traps:
6. Forgetting to flip the sign when reversing limits.
7. Misapplying Wallis’ for limits other than 0 to π/2.
8. Ignoring symmetry in even/odd functions.

Now go crush that exam. You’ve got this!