Definite integration JEE questions: properties, patterns and the mistakes that cost marks

The fundamental theorem

If F′ = f on [a, b], then ∫ₐᵇ f(x)dx = F(b) − F(a). The variable of integration is a dummy: ∫ₐᵇ f(x)dx and ∫ₐᵇ f(t)dt are the same number, which is why the answer never contains x.

The reflection or king property

∫ₐᵇ f(x)dx = ∫ₐᵇ f(a + b − x)dx. Adding the original integral to its reflection often produces a constant integrand, so 2I becomes trivial. This single property accounts for a large share of exam integrals.

Even and odd symmetry

Over [−a, a], an odd function integrates to zero and an even one to twice the half-interval integral. Testing f(−x) takes five seconds and can replace a page of work.

Substitution changes the limits

A definite integral is a number attached to an interval. When you substitute u = g(x), the limits must become g(a) and g(b), or you must return to x before evaluating. This is where most marks in the chapter are lost.

Periodicity

If f has period T, then ∫₀^{nT} f(x)dx = n∫₀^T f(x)dx, and the integral over any full period is the same wherever it starts. Questions with |sin x| or fractional parts are usually periodicity questions in disguise.

Integral as the limit of a sum

lim(n→∞) (1/n) Σ_{r=1}^{n} f(r/n) = ∫₀¹ f(x)dx. Recognising the 1/n factor and the r/n argument turns a forbidding limit into a one-line integral.

Fundamental theorem

∫ₐᵇ f(x) dx = F(b) − F(a), where F′ = f

The variable of integration is a dummy.

Reversal and splitting

∫ₐᵇ f = −∫ᵦᵃ f · ∫ₐᵇ f = ∫ₐᶜ f + ∫_c^b f

Split at any point where the definition changes.

King property

∫ₐᵇ f(x) dx = ∫ₐᵇ f(a + b − x) dx

With a = 0: ∫₀ᵃ f(x)dx = ∫₀ᵃ f(a − x)dx.

Even and odd functions

∫₋ₐᵃ f = 2∫₀ᵃ f if f is even · ∫₋ₐᵃ f = 0 if f is odd

Test f(−x) before integrating.

Double-interval property

∫₀^{2a} f = 2∫₀ᵃ f if f(2a − x) = f(x) · 0 if f(2a − x) = −f(x)

The symmetric-interval rule, shifted.

Periodic functions

∫₀^{nT} f = n ∫₀^T f

T is the period; n is a positive integer.

Leibniz rule

d/dx ∫_{u(x)}^{v(x)} f(t) dt = f(v)·v′ − f(u)·u′

Differentiating an integral with variable limits.

Integral as a limit of a sum

lim_{n→∞} (1/n) Σ_{r=1}^{n} f(r/n) = ∫₀¹ f(x) dx

Spot 1/n outside and r/n inside.

Equal trigonometric integrals

∫₀^{π/2} sinⁿx dx = ∫₀^{π/2} cosⁿx dx

A direct consequence of the king property.

Modulus integrals

∫ₐᵇ |f(x)| dx: split at every zero of f

For example ∫₀^{2π} |sin x| dx = 4.

King property · 3 marks · moderate

Evaluate I = ∫₀^{π/2} dx / (1 + tan x).

1. Apply the king property with a = π/2: I = ∫₀^{π/2} dx / (1 + tan(π/2 − x)) = ∫₀^{π/2} dx / (1 + cot x).
2. Rewrite the second form: 1/(1 + cot x) = tan x/(tan x + 1).
3. Add the two expressions for I: 2I = ∫₀^{π/2} [1/(1 + tan x) + tan x/(1 + tan x)] dx = ∫₀^{π/2} 1 dx.
4. 2I = π/2, so I = π/4.

Answer: I = π/4.

King property with a linear factor · 4 marks · moderate

Evaluate I = ∫₀^{π} x sin x dx using a property of definite integrals.

1. Apply the king property with a + b = π: I = ∫₀^{π} (π − x) sin(π − x) dx = ∫₀^{π} (π − x) sin x dx.
2. Add the two forms: 2I = ∫₀^{π} [x sin x + (π − x) sin x] dx = π ∫₀^{π} sin x dx.
3. ∫₀^{π} sin x dx = [−cos x]₀^{π} = (1) − (−1) = 2.
4. 2I = 2π, so I = π.
5. Check by parts: ∫ x sin x dx = −x cos x + sin x, which between 0 and π gives π — the same answer with more work.

Answer: I = π.

Limit as a sum · 3 marks · moderate

Evaluate lim(n→∞) (1/n)[ 1/(1 + 1/n) + 1/(1 + 2/n) + … + 1/(1 + n/n) ].

1. The bracket is Σ_{r=1}^{n} 1/(1 + r/n), and the whole expression is (1/n)Σ f(r/n) with f(x) = 1/(1 + x).
2. So the limit equals ∫₀¹ dx/(1 + x).
3. ∫₀¹ dx/(1 + x) = [ln(1 + x)]₀¹ = ln 2 − ln 1.
4. The limit is ln 2 ≈ 0.693.

Answer: ln 2.