JEE Mains Maths · Application of Integrals

Curves and Lines: Vertical Strips

Regions bounded by parabolas, cubics and lines, found by integrating the top curve minus the bottom curve between the points where they meet.

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Why this matters

Twenty PYQs, fourteen of them multiple choice, and three from 2026. Ten are a parabola against a line or against a second parabola; six have a top or bottom that changes partway — at a tangent, a second line or an axis — so the interval splits; four take y² = kx as y = √(kx) against a line or a cubic. Three ideas cover the page.

Concept 1 of 3: Top minus bottom between the meeting points

Equate the two curves to find where they meet; those x-values are the limits. Between them one curve stays on top, so the area is the integral of top minus bottom. For a parabola and a line, or two parabolas, top minus bottom is a quadratic that vanishes at both limits, and its integral has a closed form.

Definition

Solve $f(x)=g(x)$ for the limits $\alpha<\beta$ .
Test one point between them to see which curve is on top.
If top minus bottom is $a(x-\alpha)(\beta-x)$ with $a>0$ , the area is $\frac{a(\beta-\alpha)^3}{6}$ .

Area between two curves

A=\int_{\alpha}^{\beta}\big(f(x)-g(x)\big)\,dx,\qquad\int_{\alpha}^{\beta}a(x-\alpha)(\beta-x)\,dx=\frac{a(\beta-\alpha)^3}{6}

Worked example

Find the area enclosed by

y=x^2

and

y=2x+3

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 6 Apr 2026 Shift 2 · Q69Moderate

Example 1 · Application of Integrals · Curves and Lines: Vertical Strips

The area of the region

\left\{ (x,y):x^{2}- 8x \leq y \leq - x \right\}

is:

A third bound can cut the top off

When the region also has

y\le c

, the top is the lower of the two upper bounds. Find where the cap meets each curve and split the interval there before integrating.

Drill 9 more on top minus bottom between the meeting points

Concept 2 of 3: Split where the top or bottom changes

Some regions have three or more boundaries, so the top or the bottom switches partway across. List every x where two boundaries meet. Between consecutive ones the top and bottom are fixed, so integrate each piece and add. A tangent line is a common switching boundary: write it first, then find where it meets the axis or the other lines.

Definition

Tangent to $y=f(x)$ at $x=a$ : $y=f(a)+f'(a)(x-a)$ .
Split the interval at every x where the top or the bottom changes.
Pieces bounded only by straight lines are triangles: use $\frac12\times\text{base}\times\text{height}$ .

Split at the switch point c

A=\int_{a}^{c}\big(f_1-g\big)\,dx+\int_{c}^{b}\big(f_2-g\big)\,dx

Worked example

Find the area bounded by

y=x^2

, its tangent at

(1,1)

and the x-axis.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 24 Jan 2026 Shift 1 · Q69Moderate

Example 2 · Application of Integrals · Curves and Lines: Vertical Strips

Let

A_{1}

be the bounded area enclosed by the curves

y =x^{2}+ 2,x + y = 8

and

y

-axis that lies in the first quadrant. Let

A_{2}

be the bounded area enclosed by the curves

y =x^{2}+ 2,y^{2}= x,x = 2

, and

y

-axis that lies in the first quadrant. Then

A_{1}-A_{2}

is equal to

A tangent to a cubic crosses it again

Solve curve = tangent in full. The point of contact is a double root, and the other root is where the tangent crosses the cubic again — that is the second limit.

Drill 5 more on split where the top or bottom changes

Concept 3 of 3: Sideways parabolas as square roots

In the first quadrant,

y^2=kx

is the curve

y=\sqrt{kx}

, which can be integrated in x like any other. Against a line

y=mx

through the origin, the two meet at

x=\frac{k}{m^2}

, and the area has a closed form.

Definition

In the first quadrant, $y^2=kx$ is $y=\sqrt{kx}$ .
$\int_0^c\sqrt{kx}\,dx=\frac23c\sqrt{kc}$ : two-thirds of the rectangle it sits in.
$y^2=4ax$ meets $y=mx$ at $x=\frac{4a}{m^2}$ , and the area between them is $\frac{8a^2}{3m^3}$ .

Under a square-root curve

\int_0^{c}\sqrt{kx}\,dx=\frac23\,c\sqrt{kc}

Worked example

Find the area between

y^2=4x

and

y=x

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 30 January 2023 · Q83Moderate

Example 3 · Application of Integrals · Curves and Lines: Vertical Strips

Let

\alpha

be the area of the larger region bounded by the curve

y^{2}= 8x

and the lines

y = x

and

x = 2

, which lies in the first quadrant. Then the value of

3\alpha

is equal to

The lower branch

y^2=kx

also has the branch

y=-\sqrt{kx}

. If the region is not limited to

y\ge0

, include the part below the x-axis, or use symmetry and double.

Drill 3 more on sideways parabolas as square roots

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)

Top minus bottom between the meeting points
Area between two curves
$A=\int_{\alpha}^{\beta}\big(f(x)-g(x)\big)\,dx,\qquad\int_{\alpha}^{\beta}a(x-\alpha)(\beta-x)\,dx=\frac{a(\beta-\alpha)^3}{6}$
Split where the top or bottom changes
Split at the switch point c
$A=\int_{a}^{c}\big(f_1-g\big)\,dx+\int_{c}^{b}\big(f_2-g\big)\,dx$
Sideways parabolas as square roots
Under a square-root curve
$\int_0^{c}\sqrt{kx}\,dx=\frac23\,c\sqrt{kc}$

Watch out for (3)

A third bound can cut the top off
→ Top minus bottom between the meeting points
A tangent to a cubic crosses it again
→ Split where the top or bottom changes
The lower branch
→ Sideways parabolas as square roots

Test yourself on Application of Integrals

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