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When a definite integral contains |x − 1| or [x²], there is no formula to apply. You have to rewrite the function piece by piece, and the whole question turns on finding the right places to split. Get the split points right and the rest is routine. This post shows how to find them, with worked examples for modulus, greatest integer and fractional part functions.
Where is this on the syllabus?
The JEE (Main) 2026 syllabus lists "properties of definite integrals" and "evaluation of definite integrals" in Unit 8. The property behind every example here is NCERT's P2: you can split ∫ from a to b at any point c. The current NCERT Class 12 chapter on integrals uses this in its Example 41, which splits |x sin πx| at x = 1, and its exercises include ∫ from −5 to 5 of |x + 2| dx.
The greatest integer function [x], the greatest integer less than or equal to x, is defined in the NCERT Class 11 chapter on relations and functions. The JEE (Advanced) 2026 brochure lists it among the special functions you should know. Check the documents for your exam year.
Why does splitting work?
Think of a definite integral as adding up thin strips of area. A modulus or a floor function changes its formula at certain points, so you add up the strips in separate groups, each with its own simple formula. Orton (1983) interviewed 100 students aged 16 to 22 and found that items on integration as the limit of a sum were difficult even for good students. Keeping the "sum of strips" picture in mind makes splitting feel natural rather than like a trick.
How do you split a modulus?
Find where the expression inside equals zero. Between those points it has one sign, so |f(x)| is either f(x) or −f(x).
Worked example 1. Find ∫ from 0 to 3 of |x² − 3x + 2| dx.
x² − 3x + 2 = (x − 1)(x − 2), which is zero at 1 and 2. It is positive on (0, 1), negative on (1, 2) and positive on (2, 3).
From 0 to 1: ∫ (x² − 3x + 2) dx = 1/3 − 3/2 + 2 = 5/6.
From 1 to 2: ∫ −(x² − 3x + 2) dx. The antiderivative x³/3 − 3x²/2 + 2x is 2/3 at x = 2 and 5/6 at x = 1, so this piece is −(2/3 − 5/6) = 1/6.
From 2 to 3: the antiderivative is 3/2 at x = 3 and 2/3 at x = 2, so the piece is 5/6.
Total: 5/6 + 1/6 + 5/6 = 11/6.
Check. Every piece of a modulus integral must be positive. If one comes out negative, you used the wrong sign there.
How do you split a greatest integer function?
[f(x)] stays constant until f(x) reaches the next integer. So split where f(x) equals an integer.
Worked example 2. Find ∫ from 0 to 2 of [x²] dx.
As x goes from 0 to 2, x² goes from 0 to 4. It crosses 1, 2 and 3 at x = 1, √2 and √3. So [x²] is 0 on [0, 1), 1 on [1, √2), 2 on [√2, √3) and 3 on [√3, 2).
The integral is 0 + 1(√2 − 1) + 2(√3 − √2) + 3(2 − √3) = 5 − √2 − √3, about 1.854.
What about the fractional part?
The fractional part is {x} = x − [x]. It rises from 0 towards 1 on every unit interval, then drops back. On each interval from n to n + 1, its integral is the area of a triangle with base 1 and height 1, which is 1/2.
Worked example 3. ∫ from 0 to 5 of {x} dx = 5 × 1/2 = 5/2.
Worked example 4. ∫ from −1 to 1 of [x] dx. On [−1, 0), [x] = −1; on [0, 1), [x] = 0. The integral is −1 + 0 = −1. Note that [x] is not an odd function, so the shortcut "odd function over a symmetric interval gives zero" does not apply here.
Which mistakes cost the most marks?
Splitting [x²] at 1, 2, 3 instead of at 1, √2, √3.
Getting the sign wrong on the negative piece of a modulus.
Treating [−0.5] as 0. It is −1, because −1 is the greatest integer not more than −0.5.
Assuming [x] or |x| is odd or even without testing f(−x). |x| is even; [x] is neither.
Worrying about the endpoints. Single points do not change a definite integral, so it does not matter which piece an endpoint belongs to.
A practice routine
Split points only (10 minutes, three times a week). For ten integrals, list the split points and the formula on each piece. Do not integrate.
Sketch. Draw the step graph for each floor function. The integral is the area of the rectangles.
Sign check. For each modulus integral, confirm every piece is positive.
Combine with the symmetry properties in our post on definite integral properties.
If the step graph is hard to picture, the Fermi Simulation Lab has interactive simulations for ideas that are easier to see than read.
Today, find ∫ from 0 to 2 of [x² + 1] dx and check it against example 2.
