MHT-CET Maths · Definite Integration

Modulus and Greatest-Integer Integrands — Split the Interval

A modulus changes formula where its inside changes sign, and [x] changes value at every integer — so the interval is split at those points and each piece is integrated with its own formula.

All Definite Integration notes MHT-CET Maths strategy

Why this matters

15 PYQs at 27% HARD — the gentlest page in the chapter, and the most mechanical: find the break points, split, integrate each piece. The MODERATE tag is where the marks are lost, not the HARD one: a modulus integrated as if it were the bare expression, or a greatest-integer function evaluated at the wrong endpoint, produces a confident wrong answer that is always among the options. Two stems ask for the integral of an expression that is piecewise CONSTANT by an inverse-trig identity, and one of them carries an official key that ignores the sign of x.

Concept 1 of 3

Modulus: Split Where the Inside Changes Sign

Intuition

|2x - 5|

5 - 2x

until

x = \frac52

and

2x - 5

after it. The graph is a V; the integral is the area of two triangles, computed one at a time.

Definition

Solve $\text{inside} = 0$ to find the break points inside the interval; on each piece replace $|u|$ by $u$ or $-u$ according to the sign there.
$\int_0^4|2x - 5|\,dx = \int_0^{5/2}(5 - 2x)\,dx + \int_{5/2}^{4}(2x - 5)\,dx = \dfrac{25}{4} + \dfrac94 = \dfrac{17}{2}$ .
Quadratic inside: $x^2 - x - 2 = (x - 2)(x + 1)$ is negative between the roots; on $[-2, 2]$ split at $-1$ and flip the sign on $[-1, 2]$ .
Trigonometric inside: $\sin x - \cos x$ changes sign at $\frac{\pi}{4}$ ; $\sin x - \dfrac{2x}{\pi}$ at $\frac{\pi}{2}$ (where $\sin x = 1 = \frac{2x}{\pi}$ ). Sketch both curves to see which is on top on each side.
A factor outside the modulus, as in $x\left|x - \frac12\right|$ , rides along: split at $\frac12$ and integrate $x(\frac12 - x)$ then $x(x - \frac12)$ .

Modulus splitting

\int_a^b|u(x)|\,dx = \int_a^c\pm u\,dx + \int_c^b\mp u\,dx \ \text{ where } u(c) = 0,\ \text{sign chosen to make each piece } \ge 0

Worked example

Evaluate

\int_0^3|x - 1|\,dx

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1Definite IntegrationEASY

\displaystyle\int_{0}^{4}|2x-5|\,dx=

[Q145 · 12th May Shift 1 · 2024]

Integrating the bare expression

\int_0^4(2x - 5)\,dx = -4

, but

\int_0^4|2x - 5|\,dx = \frac{17}{2}

. A modulus integral is never negative and is never the signed integral of the inside; if you did not split, you answered a different question.

Drill 5 more on modulus: split where the inside changes sign

Concept 2 of 3

Greatest Integer: Piecewise Constant, So Sum the Pieces

Intuition

[x]

0

[0, 1)

1

[1, 2)

2

[2, 3)

. Its integral is a staircase of rectangles; multiplied by another function, each rectangle becomes that function's integral times a constant.

Definition

Split at every integer inside the interval; on each piece $[x]$ is a constant, so $\int[x]\,g(x)\,dx = (\text{that integer})\int g(x)\,dx$ over the piece.
$\int_{0.2}^{3.5}[x]\,dx = 0\cdot0.8 + 1\cdot1 + 2\cdot1 + 3\cdot0.5 = 4.5$ . Watch the fractional ends: the first piece is $[0.2, 1)$ , the last $[3, 3.5]$ .
$\int_0^5x^2[x]\,dx = \int_1^2x^2 + 2\int_2^3x^2 + 3\int_3^4x^2 + 4\int_4^5x^2 = \dfrac{7 + 38 + 111 + 244}{3} = \dfrac{400}{3}$ .
$\int_1^4\log[x]\,dx = 0 + \log 2 + \log 3 = \log 6$ .
Composite $[x^2]$ on $[0, 2]$ : it steps at $x = 1, \sqrt2, \sqrt3$ , so the integral is $1(\sqrt2 - 1) + 2(\sqrt3 - \sqrt2) + 3(2 - \sqrt3) = 5 - \sqrt2 - \sqrt3$ .
The same counting answers a sum of floors: $\left[\frac12 + \frac{i}{100}\right]$ is $0$ for $i \le 49$ and $1$ for $50 \le i \le 99$ , so the hundred-term sum is $50$ .

Integrating a staircase

\int_a^b[x]\,g(x)\,dx = \sum_{n}\; n\int_{\max(a,n)}^{\min(b,n+1)} g(x)\,dx \qquad \int_0^2[x^2]\,dx = 5 - \sqrt2 - \sqrt3

Worked example

Evaluate

\int_0^3 x[x]\,dx

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Definite IntegrationMODERATE

\int_{0.2}^{3.5} [x]\,dx =

(where

[x] =

greatest integer not greater than x)

[Q111 · 11th May Shift 1 · 2023]

Using the endpoint's floor for the whole last piece

[3, 3.5]

the floor is

3

throughout, contributing

3 \times 0.5 = 1.5

— not

3.5 \times

anything. And

[x]

[0.2, 1)

0

, so that piece contributes nothing however long it is.

Drill 6 more on greatest integer: piecewise constant, so sum the pieces

Concept 3 of 3

Piecewise Constant by an Identity: tan⁻¹u + tan⁻¹(1/u)

Intuition

\tan^{-1}\dfrac{x}{x^2 + 1} + \tan^{-1}\dfrac{x^2 + 1}{x}

\tan^{-1}u + \tan^{-1}\dfrac1u

, which is

\dfrac{\pi}{2}

when

u > 0

and

-\dfrac{\pi}{2}

when

u < 0

. The integrand is a constant on each side of

0

— no integration, just lengths times constants.

Definition

$\tan^{-1}u + \tan^{-1}\dfrac1u = \dfrac{\pi}{2}$ for $u > 0$ and $-\dfrac{\pi}{2}$ for $u < 0$ . With $u = \dfrac{x}{x^2 + 1}$ the sign is the sign of $x$ .
Strictly, on $[-1, 3]$ : $\int_{-1}^{0}\left(-\dfrac{\pi}{2}\right)dx + \int_0^3\dfrac{\pi}{2}\,dx = -\dfrac{\pi}{2} + \dfrac{3\pi}{2} = \pi$ .
The official MHT-CET key treats the sum as $\dfrac{\pi}{2}$ throughout and marks $\dfrac{\pi}{2}\times 4 = 2\pi$ — in both the $\tan^{-1}$ (2025) and $\cot^{-1}$ (2024) versions. On the paper, answer $2\pi$ .
The $\cot^{-1}$ version is subtler still: $\cot^{-1}$ has range $(0, \pi)$ , so for $u < 0$ the two terms sum to $\dfrac{3\pi}{2}$ , and the strict value on $[-1, 3]$ is $3\pi$ .
General lesson: when an integrand is an identity in disguise, evaluate the identity on each sign region before multiplying by the interval length.

The reciprocal arctangent identity

\tan^{-1}u + \tan^{-1}\frac{1}{u} = \begin{cases} \dfrac{\pi}{2}, & u > 0 \\[4pt] -\dfrac{\pi}{2}, & u < 0 \end{cases}

Worked example

Evaluate

\int_1^5\left[\tan^{-1}\dfrac{x}{x^2 + 4} + \tan^{-1}\dfrac{x^2 + 4}{x}\right]dx

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3Definite IntegrationMODERATE

\int_{-1}^{3} \left[\cot^{-1}\frac{x}{x^2+1} + \cot^{-1}\frac{x^2+1}{x}\right]dx =

[Q126 · 11th May Shift 1 · 2024]

The key that ignores the sign of x

Both sittings of this question mark

2\pi

, which assumes

\tan^{-1}u + \tan^{-1}(1/u) = \pi/2

even for negative

u

. Mathematically the answer on

[-1, 3]

\pi

; on the paper choose

2\pi

, and know why.

Drill 1 more on piecewise constant by an identity: tan⁻¹u + tan⁻¹(1/u)

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Modulus: Split Where the Inside Changes Sign
Modulus splitting
$\int_a^b|u(x)|\,dx = \int_a^c\pm u\,dx + \int_c^b\mp u\,dx \ \text{ where } u(c) = 0,\ \text{sign chosen to make each piece } \ge 0$
Greatest Integer: Piecewise Constant, So Sum the Pieces
Integrating a staircase
$\int_a^b[x]\,g(x)\,dx = \sum_{n}\; n\int_{\max(a,n)}^{\min(b,n+1)} g(x)\,dx \qquad \int_0^2[x^2]\,dx = 5 - \sqrt2 - \sqrt3$
Piecewise Constant by an Identity: tan⁻¹u + tan⁻¹(1/u)
The reciprocal arctangent identity
$\tan^{-1}u + \tan^{-1}\frac{1}{u} = \begin{cases} \dfrac{\pi}{2}, & u > 0 \\[4pt] -\dfrac{\pi}{2}, & u < 0 \end{cases}$

Watch out for (3)

Integrating the bare expression
→ Modulus: Split Where the Inside Changes Sign
Using the endpoint's floor for the whole last piece
→ Greatest Integer: Piecewise Constant, So Sum the Pieces
The key that ignores the sign of x
→ Piecewise Constant by an Identity: tan⁻¹u + tan⁻¹(1/u)

Drill every past-year question on this subtopic

15 questions from the bank — paginated, with cart and Word-export support.

Drill the 15 questions

Drill every past-year question on this subtopic

Related notes