JEE Mains Maths · Application of Integrals

Modulus Curves

Regions bounded by curves with a modulus, split where the expression inside changes sign so that each piece is an ordinary line or parabola.

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Why this matters

Sixteen PYQs, ten of them multiple choice, and one from 2026. Eight have V shapes from the modulus of a linear term, against a line, a parabola or √x; eight take the modulus of a quadratic or of a product such as x|x − 3|, which folds part of a parabola up. Two ideas cover the page.

Concept 1 of 2: Moduli of linear terms

|x-a|

has a corner at

x=a

; a sum of such terms is a broken line with a corner at each one. Split at the corners and each side is a straight line. Against another line the region is a polygon, so triangles and trapezia give the area without integrating; against a curve, integrate top minus bottom on each side.

Definition

$|x-a|=x-a$ for $x\ge a$ , and $a-x$ for $x<a$ .
A sum of linear moduli is a broken line with corners at those points.
For a nested modulus such as $\big||x|-1\big|$ , draw the inside first, then reflect the parts below the axis.

Split at the corner

|x-a|=\begin{cases}x-a,&x\ge a\\a-x,&x<a\end{cases}

Worked example

Find the area bounded by

y=|x-2|

and

y=3

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 24 Jan 2026 Shift 2 · Q62Moderate

Example 1 · Application of Integrals · Modulus Curves

Let

f(\alpha)

denote the area of the region in the first quadrant bounded by

x = 0,x = 1,y^{2}= x

and

y = |\alpha x - 5| - |1 - \alpha x| + \alpha x^{2}

. Then

(f(0) + f(1))

is equal to

Each side of a V against a curve

When a V meets a parabola or

\sqrt x

, its two arms meet the curve at different points. Solve each arm separately; one equation for both sides loses a limit.

Drill 7 more on moduli of linear terms

Concept 2 of 2: The modulus of a quadratic

|f(x)|

keeps the parts of

y=f(x)

above the axis and reflects the parts below it. So

|x^2-a^2|

is the upward parabola outside

[-a,a]

and the cap

a^2-x^2

inside it. For a product like

x|x-a|

, the sign changes only at

x=a

. Split at every zero of the quantity inside the modulus.

Definition

$|x^2-a^2|=x^2-a^2$ for $|x|\ge a$ , and $a^2-x^2$ for $|x|<a$ .
$x|x-a|=x(x-a)$ for $x\ge a$ , and $x(a-x)$ for $x<a$ .
$x|x|=x^2$ for $x\ge0$ , and $-x^2$ for $x<0$ .

Split at the zeros

\int_p^q|f(x)|\,dx=\sum_{\text{pieces}}\left|\int f(x)\,dx\right|

Worked example

Find the area between

y=x|x-2|

and the x-axis for

0\le x\le3

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 2 Apr 2025 · Q73Moderate

Example 2 · Application of Integrals · Modulus Curves

If the area of the region

\left\{ (x,y):\left| 4 - x^{2} \right| \leq y \leq x^{2},y \leq 4,x \geq 0 \right\}

\left( \frac{80\sqrt{2}}{\alpha} - \beta \right),\alpha,\beta \in N

, then

\alpha + \beta

is equal to.

Where the line cuts the cap

If the line

y=c

sits below the top of the folded cap, the cap pokes through and makes a second region above the line. Check whether c is above or below the cap's peak before deciding how many regions there are.

Drill 7 more on the modulus of a quadratic

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 (2)

Moduli of linear terms
Split at the corner
$|x-a|=\begin{cases}x-a,&x\ge a\\a-x,&x<a\end{cases}$
The modulus of a quadratic
Split at the zeros
$\int_p^q|f(x)|\,dx=\sum_{\text{pieces}}\left|\int f(x)\,dx\right|$

Watch out for (2)

Each side of a V against a curve
→ Moduli of linear terms
Where the line cuts the cap
→ The modulus of a quadratic

Test yourself on Application of Integrals

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