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JEE Mains Maths · Definite Integration

Greatest Integer, Modulus and Max–Min Integrands

Integrands that change formula part-way: the greatest integer function, the fractional part, the modulus, and the larger or smaller of two functions. Split the interval where the formula changes and add the pieces.

All Definite Integration notesJEE Mains Maths strategy

Why this matters

Forty-eight PYQs, a quarter of the chapter. Three quarters involve the greatest integer function or the fractional part. The method never changes: find where the formula changes, split there, and add. The errors come from a missed breaking point. Three ideas cover the page.

Concept 1 of 3: Greatest integer: split where the value jumps

[g(x)][g(x)] stays constant until g(x)g(x) crosses an integer. List the points in the interval where gg takes integer values, split there, and add (integer value) × (length of the piece). For [x][x] the pieces are unit intervals; for [x2][x^2] the breaks are at 1,2,3,2,…1,\sqrt2,\sqrt3,2,\dots. An integer can be taken out: [n+t]=n+[t][n+t]=n+[t].

Definition

  • ∫ab[g(x)] dx=∑(value)×(length where it holds)\int_a^b[g(x)]\,dx=\sum(\text{value})\times(\text{length where it holds}).
  • [x2]=k[x^2]=k on [k,k+1)[\sqrt k,\sqrt{k+1}) for x≥0x\ge0.
  • [x+n]=[x]+n[x+n]=[x]+n for integer nn; [x]+[−x]=−1[x]+[-x]=-1 for non-integer xx.

Greatest-integer integral

∫ab[g(x)] dx=∑kk⋅ℓ{x:[g(x)]=k}\int_a^b[g(x)]\,dx=\sum_k k\cdot\ell\{x: [g(x)]=k\}

Worked example

Evaluate ∫02[x2] dx\int_0^2[x^2]\,dx.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 22 Jan 2026 Shift 1 · Q65Moderate

Example 1 · Definite Integration · Greatest Integer, Modulus and Max-Min Integrands

The value of ∫−π2π2(1[x]+4)dx\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( \frac{1}{\lbrack x\rbrack + 4} \right)dx, where [•] denotes the greatest integer function, is

Negative numbers round down

[−0.3]=−1[-0.3]=-1, not 0. On an interval below zero, [x][x] is the next integer to the left.
Drill 30 more on greatest integer: split where the value jumps

Concept 2 of 3: Fractional part and repeating pieces

{x}=x−[x]\{x\}=x-[x] repeats every 1, so any function of {x}\{x\} has period 1. Integrate over [0,1][0,1] once and multiply by the number of whole periods. A limit that is not a whole number leaves a partial period at the end, worked out separately.

Definition

  • {x}=x−[x]∈[0,1)\{x\}=x-[x]\in[0,1), period 1.
  • ∫0ng({x}) dx=n∫01g(x) dx\int_0^ng(\{x\})\,dx=n\int_0^1g(x)\,dx for integer nn.
  • Partial period: ∫nn+tg({x}) dx=∫0tg(x) dx\int_n^{n+t}g(\{x\})\,dx=\int_0^tg(x)\,dx.

Whole periods of the fractional part

∫0ng({x}) dx=n∫01g(x) dx\int_0^{n} g(\{x\})\,dx=n\int_0^{1} g(x)\,dx

Worked example

Evaluate ∫05{x}2 dx\int_0^5\{x\}^2\,dx.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 25 · Q63Moderate

Example 2 · Definite Integration · Greatest Integer, Modulus and Max-Min Integrands

The value of ∑n=1100∫n−1nex−[x]dx\sum_{n= 1}^{100} \int_{n- 1}^{n} e^{x- \lbrack x\rbrack}dx, where [x]\lbrack x\rbrack is the greatest integer ≤x\leq x, is:

The partial period at the end

With an upper limit like 10.5, there are ten whole periods and half of one more. Add the half-period's integral; it is not half of a full one unless the integrand is constant.
Drill 4 more on fractional part and repeating pieces

Concept 3 of 3: Modulus and the larger or smaller of two functions

∣g(x)∣|g(x)| changes formula where g(x)=0g(x)=0: find every root inside the interval, split there, and change the sign on the pieces where g<0g<0. For max⁡{f,g}\max\{f,g\} or min⁡{f,g}\min\{f,g\}, find where the two graphs cross; on each piece one function is the larger throughout.

Definition

  • ∫ab∣g∣=∑±∫g\int_a^b|g|=\sum\pm\int g over the pieces between roots of gg.
  • max⁡{f,g}\max\{f,g\}: split at f=gf=g, take the larger on each piece.
  • 1−sin⁡2x=∣sin⁡x−cos⁡x∣\sqrt{1-\sin2x}=|\sin x-\cos x|: a hidden modulus.

Modulus split at the roots

∫ab∣g(x)∣ dx=∑pieces∣∫g(x) dx∣\int_a^b|g(x)|\,dx=\sum_{\text{pieces}}\left|\int g(x)\,dx\right|

Worked example

Evaluate ∫03∣x2−4∣ dx\int_0^3|x^2-4|\,dx.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 24 Jan 2023 · Q90Moderate

Example 3 · Definite Integration · Greatest Integer, Modulus and Max-Min Integrands

The value of 12∫03∣x2−3x+2∣dx12\int_{0}^{3} \left| x^{2}- 3x + 2 \right|dx is

Every root in the interval

A quadratic may have two roots inside the limits. Missing one leaves a piece with the wrong sign, and the error is exactly twice that piece's integral.
Drill 11 more on modulus and the larger or smaller of two functions

Summary — formulas & gotchas at a glance

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