King's Rule & Definite Integral Properties: JEE Main

Definite integral properties are among the highest-return topics in JEE Main Mathematics. A question that appears to require a complex antiderivative often simplifies to a trivial calculation using King's rule or symmetry. Students who master these properties can solve "hard" integration questions in 30 seconds. This guide covers all the key properties with the specific manipulation that unlocks each type.

The Standard Properties

Property 1 (Reversal): ∫_a^b f(x)dx = −∫_b^a f(x)dx. Property 2 (Splitting): ∫_a^b f(x)dx = ∫_a^c f(x)dx + ∫_c^b f(x)dx for any c. Property 3 (Translation): ∫_a^b f(x)dx = ∫_a^b f(a+b−x)dx. This is KING'S RULE — the most powerful property in the chapter. Property 4 (Shift): ∫_0^a f(x)dx = ∫_0^a f(a−x)dx (special case of King with b=a). Property 5 (Double period): ∫_0^{2a} f(x)dx = 2∫_0^a f(x)dx if f(2a−x) = f(x); = 0 if f(2a−x) = −f(x). Property 6 (Odd/Even): ∫_{-a}^a f(x)dx = 2∫_0^a f(x)dx if f is even; = 0 if f is odd. Property 7 (Periodicity): ∫_0^{nT} f(x)dx = n∫_0^T f(x)dx where T is the period of f.

The most important for JEE: Property 3 (King's rule) and Property 6 (odd/even). Master these two and you can handle 80% of the tricky definite integral questions. Take a free integration mock. See our definite integrals guide.

King's Rule in Action

King's rule: I = ∫_a^b f(x)dx = ∫_a^b f(a+b−x)dx. Adding both: 2I = ∫_a^b [f(x) + f(a+b−x)]dx. If f(x) + f(a+b−x) = constant C: 2I = C(b−a) → I = C(b−a)/2. Classic JEE question: I = ∫_0^{π/2} log(tanx) dx. Apply King: let x → π/2 − x: I = ∫_0^{π/2} log(tan(π/2−x)) dx = ∫_0^{π/2} log(cotx) dx = ∫_0^{π/2} −log(tanx) dx = −I. Therefore 2I = 0 → I = 0. In 10 seconds. Another: I = ∫_0^π x sinx/(1+cos²x) dx. King (a+b = π): numerator becomes (π−x)sin(π−x) = (π−x)sinx. Adding: 2I = π∫_0^π sinx/(1+cos²x) dx = π[−arctan(cosx)]_0^π = π[−arctan(−1)+arctan(1)] = π[π/4+π/4] = π²/2 → I = π²/4.

Odd and Even Function Symmetry

For ∫_{-a}^a f(x)dx: first check if f is odd (f(−x) = −f(x)) or even (f(−x) = f(x)). If odd: integral = 0 instantly (symmetry cancellation). If even: integral = 2∫_0^a f(x)dx (double the half). JEE question: ∫_{-π/2}^{π/2} [x² sinx + cosx] dx. Split: x² sinx is odd (x² is even, sinx is odd → product is odd) → its integral over symmetric interval = 0. cosx is even → integral = 2∫_0^{π/2} cosx dx = 2[sinx]_0^{π/2} = 2(1−0) = 2. Total = 0 + 2 = 2. JEE uses this in "identify odd/even, use symmetry" format consistently. Composite functions: f(−x) for f(g(x)) — check if g(−x) = g(x) (g even) or g(−x) = −g(x) (g odd), then compose with f's parity.

Periodic Functions and Repeated Integrals

For a periodic function f with period T: ∫_a^{a+T} f(x)dx = ∫_0^T f(x)dx (integral over one full period is independent of start). ∫_0^{nT} f(x)dx = n∫_0^T f(x)dx. JEE question: ∫_0^{100π} |sinx| dx. Period of |sinx| is π. ∫_0^π |sinx| dx = 2. So ∫_0^{100π} |sinx| dx = 100 × 2 = 200. Another type: ∫_0^n {x} dx where {x} = fractional part. {x} has period 1. ∫_0^1 {x} dx = ∫_0^1 x dx = 1/2. Therefore ∫_0^n {x} dx = n/2. The fractional-part integral appears every two to three JEE sessions. For comprehensive integration techniques see our integration guide and our integration by parts guide.