A definite integral ∫ₐᵇ f(x) dx is not just "an antiderivative with limits plugged in." Look at it this way: it represents the net signed area between the curve y = f(x) and the x-axis, from x = a to x = b. Signed — meaning area below the x-axis subtracts, area above adds.
Here's a useful mental image. Imagine pouring water into a container shaped like the region under a curve. The definite integral measures how much water fits — except that regions below the x-axis drain water out. The indefinite integral gives you the container's formula; the definite integral gives you the actual volume for specific walls.
Formally, the Fundamental Theorem of Calculus ties this to antiderivatives: if F'(x) = f(x), then
∫ₐᵇ f(x) dx = F(b) - F(a)
This is written as [F(x)]ₐᵇ. The constant of integration vanishes — it cancels when you subtract F(a) from F(b). That is why indefinite integration demands + C but definite integration does not.
For NDA, the raw computation of antiderivatives matters less than properties. The exam repeatedly tests whether you recognise a property that collapses a hard-looking integral into a one-liner. If you approach every definite integral by hunting for an antiderivative first, you will burn two to three minutes on questions designed to be solved in thirty seconds. The property-first mindset is the single biggest leverage point in this chapter.
The analogy that works best: properties of definite integrals are like coordinate transformations in geometry — they do not change the value, only the form. Your job is to choose the transformation that makes the form simple.
There are about a dozen standard properties. NDA tests four of them obsessively. Learn these four cold.
Property 1 — Reversal of Limits
∫ₐᵇ f(x) dx = −∫ᵦᵃ f(x) dx
Swapping limits flips the sign. Directly from F(b) − F(a) = −[F(a) − F(b)]. This is mostly used to clean up signs mid-solution.
Property 2 — Additivity over Intervals
∫ₐᵇ f(x) dx = ∫ₐᶜ f(x) dx + ∫ᶜᵇ f(x) dx
c need not lie between a and b. This is powerful when f(x) has different expressions on different sub-intervals (absolute value functions, step functions, piecewise definitions). You split the integral at every "break point" and handle each piece separately.
Property 3 — King's Property (the NDA favourite)
∫ₐᵇ f(x) dx = ∫ₐᵇ f(a + b − x) dx
This is the single most-tested property in NDA definite integration questions. The substitution x → a + b − x maps the interval [a, b] onto itself (check: when x = a, the new variable equals b, and vice versa — but after the substitution and reversing limits, the result is identical). The value does not change; only the integrand's form changes.
Special case: ∫₀ᵃ f(x) dx = ∫₀ᵃ f(a − x) dx. This is what most textbooks call "King's property" in the [0, a] context.
When to use it: The integrand has the form f(x) / [f(x) + f(a+b−x)] or any expression where applying King's creates a complementary fraction that adds to 1 with the original. Then you add the two forms of I, the numerators sum to the denominator, and the integral simplifies to (b−a)/2 or something similarly clean.
Property 4 — Even and Odd Functions on Symmetric Intervals
For ∫₋ₐᵃ f(x) dx:
f(−x) = f(x) (even): ∫₋ₐᵃ f(x) dx = 2∫₀ᵃ f(x) dxf(−x) = −f(x) (odd): ∫₋ₐᵃ f(x) dx = 0Whenever the limits are symmetric about 0, immediately check parity. An odd function on [−a, a] gives zero — no calculation needed.
Take I = ∫ₐᵇ f(x) dx. Substitute t = a + b − x, so dt = −dx. When x = a, t = b; when x = b, t = a. Then:
I = ∫ᵦᵃ f(a+b−t)(−dt) = ∫ₐᵇ f(a+b−t) dt
Since t is a dummy variable, rename it x:
I = ∫ₐᵇ f(a+b−x) dx
Now add the two expressions for I:
2I = ∫ₐᵇ [f(x) + f(a+b−x)] dx
If the integrand has the structure f(x) / [f(x) + f(a+b−x)], King's gives you the complementary integrand f(a+b−x) / [f(a+b−x) + f(x)], and the two integrands add to 1. So 2I = ∫ₐᵇ 1 dx = b − a, giving I = (b−a)/2.
This is the template for a whole class of NDA questions involving sinx/cosx combinations, eˣ/(eˣ + e^(a−x)), and similar structures.
When f(x) involves |x − c| or a greatest integer function [x], the strategy is:
[a, b] where the expression inside | | changes sign (the break points).[a, b] at each break point using Property 2.For [x] (greatest integer / floor function): [x] = n for x ∈ [n, n+1). Split the integration limits at every integer.
Do not confuse |∫ₐᵇ f(x) dx| with ∫ₐᵇ |f(x)| dx. The latter is always ≥ the former. If f(x) ≥ 0 throughout [a, b], then |f(x)| = f(x) and they are equal. This distinction appears directly in NDA PYQs.
If f has period T, then ∫₀^(nT) f(x) dx = n ∫₀ᵀ f(x) dx. Useful when limits are multiples of a period.
When you see a definite integral of the form ∫₀ᵃ f(x)/[f(x) + f(a−x)] dx, do not attempt to integrate directly. Write I, apply King's (replace x with a−x) to get a second expression for I, add the two — the ratio collapses to 1, and the answer is always a/2. Standard algebraic method: 3-4 minutes. King's pattern recognition: under 30 seconds. The tell is any fraction where the denominator is a sum of two complementary expressions.
When limits are [−a, a], before doing anything else, test f(−x). If f(−x) = −f(x), write down 0 and move on. This eliminates roughly one-third of symmetric-interval questions with a single substitution check. Standard method (computing antiderivative, substituting limits): 90 seconds. Parity check: 10 seconds. Example: ∫₋π^π sinx dx = 0 by inspection — sin(−x) = −sinx, odd function.
For ∫ₐᵇ |f(x)| dx, list every zero of f(x) in [a, b] — call them c₁, c₂, … — and split the integral there. On each sub-interval, replace |f(x)| with ±f(x) based on sign. Skipping this step and integrating |f(x)| directly as if it were f(x) produces a wrong answer with no obvious error flag. Step count without split: 4 steps, wrong answer. With split: 6 steps, correct answer. The extra two steps are mandatory — do not shortcut this.
In problems involving f(x) and f(a−x) symmetry, the phrase "rename the dummy variable" is your friend. After substituting t = a − x, the limits and dx transform, but the final integral is in t — and since t and x are just names, you can rename t back to x. This rename step confuses many students who think they have two different integrals. They are the same integral. Recognising this saves one full algebraic step and eliminates a common error in the adding-two-I procedure.
For any integral involving eˣ, e^(−x), e^(f(x)), or any exponential: these functions are always strictly positive. So |eˣ| = eˣ everywhere. Whenever a question asks you to compare ∫f(x)dx and ∫|f(x)|dx for an exponential integrand, they are equal — no computation needed. This kills a question type in under 15 seconds versus 60 seconds of setting up the absolute value split.
When you see a definite integration question in the exam hall, run this decision tree before writing anything:
Step 1 — Are the limits symmetric about 0? ([−a, a] form)
Yes → test parity of f(x). Odd → answer is 0. Even → double the [0, a] integral.
Step 2 — Does the integrand have the King's structure? (fraction with complementary sum in denominator, or a sum f(x) + f(a−x) pattern)
Yes → write I, apply x → a + b − x, add the two expressions, simplify to (b−a)/2 or similar.
Step 3 — Does the integrand involve |something| or [x]?
Yes → find break points in [a, b], split using additivity, handle each piece as a standard polynomial/trig integral.
Step 4 — Is the function purely positive (exponential, x², etc.)?
Yes → |f(x)| = f(x), drop the modulus immediately.
Step 5 — None of the above? Find the antiderivative and apply the fundamental theorem directly. This is the last resort, not the first move.
Total time budget: steps 1-4 should each take under 20 seconds. Step 5 may take 90-120 seconds. Getting stuck on step 5 for a question that belongs in step 2 is the most expensive mistake in this chapter.
Why this question: This is the canonical King's property question. It tests whether you reach for the substitution immediately or attempt partial fractions.
Solving path: Identify the King's structure — numerator is a sinx + b cosx, denominator has the form (a+b)(sinx + cosx). Call the integral I. Apply x → π/2 − x: sinx → cosx, cosx → sinx. Numerator becomes a cosx + b sinx, denominator stays (a+b)(cosx + sinx) = same denominator. Add the two forms of I: numerators sum to (a+b)(sinx+cosx). So 2I = ∫₀^(π/2) 1 dx = π/2. Hence I = π/4.
Why this question: Directly follows from the previous result. Tests whether you carry forward the established value of k without recomputing.
Solving path: From the previous question, ∫₀^(π/2) (a sinx + b cosx)/[(a+b)(sinx+cosx)] dx = k. Multiply both sides by (a+b): ∫₀^(π/2) (a sinx + b cosx)/(sinx+cosx) dx = k(a+b). By the same King's symmetry, ∫₀^(π/2) (a cosx + b sinx)/(sinx+cosx) dx produces the same value (the two integrands are King's reflections of each other and both equal (a+b)π/4). Answer: k(a+b).
Why this question: Introduces a function defined through a relation g(x) = f(x) + f(4−x). Tests whether you recognise the King's structure when the interval is [0, 4] rather than [0, π/2].
Solving path: g(x) = f(x) + f(4−x). Write I = ∫₀⁴ f(x)/g(x) dx. Apply King's with a = 0, b = 4: replace x with 4 − x. Numerator becomes f(4−x). For the denominator: g(4−x) = f(4−x) + f(4−(4−x)) = f(4−x) + f(x) = g(x). So the reflected integral is ∫₀⁴ f(4−x)/g(x) dx. Add: 2I = ∫₀⁴ [f(x) + f(4−x)]/g(x) dx = ∫₀⁴ g(x)/g(x) dx = ∫₀⁴ 1 dx = 4. Therefore I = 2. Note: the specific form of f(x) = sinx is a red herring — the answer holds for any f.
Why this question: Tests the distinction between ∫f(x)dx and ∫|f(x)|dx when the function is everywhere positive.
Solving path: f(x) = e^(−x). Since e^(−x) > 0 for all real x, we have |e^(−x)| = e^(−x). Therefore |f(x)| = f(x) at every point in [a, b]. It follows directly that q = ∫ₐᵇ |f(x)| dx = ∫ₐᵇ f(x) dx = p. So p = q. No computation of the actual integral is required.
Why this question: Combines two non-standard functions — absolute value and greatest integer — in the same integral. Tests the split-at-breakpoints technique.
Solving path: f(x) = |x−1|, g(x) = [x] (floor function), h(x) = f(x)g(x). On [−1, 0): [x] = −1 (floor of any number in [−1, 0) is −1). Also, x − 1 < 0 throughout, so |x−1| = 1 − x. Therefore h(x) = (1−x)(−1) = x − 1. Integrate: ∫₋₁⁰ (x−1) dx = [x²/2 − x]₋₁⁰ = (0−0) − (1/2 − (−1)) = 0 − (1/2 + 1) = −3/2. The upper limit is 0 (not 0⁺) — at x = 0, [0] = 0, but since the interval is [−1, 0] and the integrand's value at a single point does not affect the integral, the result is −3/2.
Skipping the parity check on symmetric intervals. Every time the limits are [−a, a], test f(−x) before anything else. Students who proceed directly to antiderivatives on ∫₋π^π (x³ + sinx cosx) dx waste two minutes on a zero-answer question.
Applying King's without verifying the complementary structure. King's is useful when the two forms of the integrand add to 1 (or to a simple constant). Blindly applying x → a+b−x to every integral and then not knowing what to do with 2I is wasted time. Check the structure first.
Forgetting that g(a−x) = g(x) needs to be separately verified. In the f(x)/g(x) class of problems, after applying King's you get f(a−x)/g(a−x). Students often assume g(a−x) = g(x) without verifying. Always substitute into g's definition to confirm.
Integrating |f(x)| as if it were f(x) without splitting. This produces a sign error that often gives a clean-looking but wrong numerical answer. The wrong answer frequently matches one of the distractors.
Using p = q only when the function is explicitly labelled "non-negative." The criterion is whether f(x) ≥ 0 throughout [a, b] — not whether it is labelled as such. Always check the function's sign behaviour on the given interval.
Treating the dummy variable rename as creating a new integral. After substituting t = a − x in King's property, the result is an integral in t over [a, b]. This is the same integral as the original, with x renamed. Students who write both ∫f(x)dx and ∫f(t)dt as if they are different quantities produce 2I = I, an impossible equation. The rename is cosmetic — both are I.