Area Bounded by Curves: Complete Notes with Illustrations

Ritesh Raj · 3 Sept 2026

21 min read14 sectionsConcept

Every area question is the same question asked in disguise: which curve is on top, over which stretch of the axis, and where does the crossing happen. Get those three right and the integral writes itself. Get any one wrong and no amount of clean integration will save the answer.

This note builds the in order: what the definite integral actually measures, the sign convention that catches the commonest lost mark, the five configurations that cover almost every JEE question, and then the six question types with worked past papers inside each.

What the definite integral measures

Let $y=f(x)$ be continuous on $[a,b]$ . The definite integral is a signed sum: strips of width $dx$ and height $f(x)$ , added from $x=a$ to $x=b$ .

\int_a^b f(x)\,dx = \phi(b)-\phi(a), \quad \text{where } \phi'(x)=f(x)

When $f(x)\ge 0$ on the whole of $[a,b]$ , that number is the area between the curve and the $x$ -axis. When $f$ dips below the axis, it is not. That single sentence is the whole of the sign convention, and it is where most marks are lost.

Strips of width dx under a curve, summing to the definite integral

The same construction turned on its side gives the $y$ -axis version. If $x=g(y)$ is continuous on $[c,d]$ , the area bounded by the curve, the $y$ -axis and the lines $y=c$ , $y=d$ is

\text{Area}=\int_c^d x\,dy=\int_c^d g(y)\,dy

Use this whenever the region is easier to describe by horizontal strips than vertical ones. A parabola opening sideways, a region bounded left and right by two curves, a region whose top boundary changes formula but whose left and right boundaries do not: all of these are cheaper in $y$ .

The sign convention, and the mark it costs

Three cases, and only the third is genuinely dangerous.

Case I. $f(x)>0$ throughout $[a,b]$ . Then $\displaystyle\int_a^b f(x)\,dx$ is already the area. Nothing to adjust.

Case II. $f(x)<0$ throughout $[a,b]$ . The integral is negative. The area is its absolute value:

\text{Area}=\left|\int_a^b f(x)\,dx\right|

Case III. $f$ changes sign inside $[a,b]$ . Now the integral silently cancels: a piece of area above the axis is destroyed by a piece below it. You must split at every root and take absolute values piece by piece.

\text{Area}=\int_a^c f\,dx+\left|\int_c^d f\,dx\right|+\int_d^b f\,dx

A curve crossing the x-axis several times, with the signed pieces labelled

What a missed root costs you

If $f$ crosses the axis at $x=c$ and you integrate straight through from $a$ to $b$ , the two pieces partially cancel and your answer is too small. The examiner puts the uncancelled value in the options, so a wrong answer looks plausible. Before you integrate anything, solve $f(x)=0$ on the interval and mark every root.

There is a fourth situation worth naming separately because it looks different but is the same idea. If the region is bounded by two curves $y=f(x)$ and $y=g(x)$ which cross at $x=c$ inside $[a,b]$ , then $|f-g|$ changes formula at $c$ :

\text{Area}=\int_a^c\big[f(x)-g(x)\big]dx+\int_c^b\big[g(x)-f(x)\big]dx

Always find the crossings first. Everything else is bookkeeping.

The five configurations

Almost every JEE area question is one of these five, or a combination of two of them.

	Region	Formula
1	Curve $y=f(x)$ , the $x$ -axis, and $x=a$ , $x=b$ with $f\ge0$	$\displaystyle\int_a^b f(x)\,dx$
2	Between $y=f(x)$ and $y=g(x)$ on $[a,b]$ , with $g\ge f$ throughout	$\displaystyle\int_a^b\big[g(x)-f(x)\big]dx$
3	Between two curves that meet at $x=a$ and $x=b$	$\displaystyle\int_a^b\big[g(x)-f(x)\big]dx$ , $a,b$ the roots of $f=g$
4	Curve crossing the axis at $x=c$ between $a$ and $b$	$\displaystyle\int_a^c f\,dx-\int_c^b f\,dx$
5	Two curves crossing at $x=c$ inside $[a,b]$	split at $c$ , take $\lvert f-g\rvert$ on each piece

Configuration 3 is the one to reach for first. In most questions the limits are not given to you: they are the intersection points, and finding them is half the work.

The five standard configurations side by side

Curve tracing: seven checks before you integrate

You cannot decide which curve is on top without a sketch, and you cannot sketch without a method. Seven checks, in this order, settle almost any curve that appears in this chapter.

1. Domain

Find the values of $x$ for which the curve exists. A square root, a logarithm or a denominator will restrict it.

2. Symmetry

Four tests, each cheap:

Every power of $y$ even, so replacing $y$ by $-y$ leaves the equation unchanged: symmetric about the $x$ -axis. Example $y^2=4ax$ .
Every power of $x$ even: symmetric about the $y$ -axis. Example $x^2=4ay$ .
Both: symmetric about both axes, hence about the origin. Example $x^2+y^2=1$ .
Swapping $x$ and $y$ leaves the equation unchanged: symmetric about the line $y=x$ .
Replacing $(x,y)$ by $(-x,-y)$ leaves it unchanged: symmetric in opposite quadrants. Example $xy=c^2$ .

Symmetry is not decoration. It halves or quarters the integral you have to do, and on a timed paper that is the difference between finishing and not.

3. Intercepts

Where does the curve meet the axes? Set $y=0$ , then $x=0$ .

4. Turning points

Solve $\dfrac{dy}{dx}=0$ to locate horizontal tangents. These tell you the shape between the intercepts.

5. Period

If the curve is periodic, find the period and sketch one cycle only.

6. Asymptotes

If $\displaystyle\lim_{x\to a}f(x)=\pm\infty$ then $x=a$ is a vertical asymptote. If $\displaystyle\lim_{x\to\infty}f(x)=k$ then $y=k$ is a horizontal one. For an oblique asymptote $y=mx+c$ , compute $m=\displaystyle\lim_{x\to\infty}\frac{f(x)}{x}$ and then $c=\displaystyle\lim_{x\to\infty}\big[f(x)-mx\big]$ .

7. Sign

On each interval between the roots, is the curve above or below the axis? This is the check that feeds directly into the sign convention.

Reading a modulus off the graph

Modulus questions are the single largest source of area questions in recent papers, and they are all mechanical once you know what each bracket does to a graph.

Curve	What it does to $y=f(x)$
$y=-f(x)$	reflect in the $x$ -axis
$y=f(-x)$	reflect in the $y$ -axis
$y=f(x)+k$	shift up by $k$ (down if $k<0$ )
$y=f(x+k)$	shift left by $k$ (right if $k<0$ )
$y=k\,f(x)$	stretch vertically by $k$
$y=f(kx)$	compress horizontally by $k$
$y=\lvert f(x)\rvert$	keep the part above the axis, reflect the part below it upward
$y=f(\lvert x\rvert)$	discard $x<0$ , mirror the right half into the left
$\lvert y\rvert=f(x)$	discard the part below the axis, mirror the rest downward

The three modulus transformations on one parent curve

The order that never fails

Sketch the parent curve first, in pencil, without any brackets. Then apply one transformation at a time in the order the brackets sit, outermost last. Never try to draw $y=\lvert f(\lvert x\rvert)\rvert$ in one go.

Standard results worth carrying

These come up often enough that deriving them in the exam is wasted time.

Region	Area
$y^2=4ax$ and $x^2=4by$	$\dfrac{16ab}{3}$
$y^2=4ax$ and $y=mx$	$\dfrac{8a^2}{3m^3}$
$y^2=4ax$ and its latus rectum	$\dfrac{8a^2}{3}$
One arch of $y=\sin ax$ with the $x$ -axis	$\dfrac{2}{a}$
Ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$	$\pi ab$
$y=\log_e x$ , $y=0$ , $x=0$	$1$
Circle segment, central angle $\theta$	$\dfrac{r^2}{2}\big(\theta-\sin\theta\big)$

The last one is worth memorising in this exact form. Every "area of the region inside one circle and outside another" question collapses to two applications of it.

Type 1 — A curve against an axis

The plainest form. Sketch, check the sign, integrate.

Example 1 (JEE Main 2026, 6 April Shift 2). The area of the region $\{(x,y):x^2-8x\le y\le -x\}$ is

▸ SHOW SOLUTION

The region between the line y equals minus x and the parabola, from x=0 to x=7

Step 1. Find where the boundaries meet

The region sits between the parabola $y=x^2-8x$ below and the line $y=-x$ above. They meet where

x^2-8x=-x \implies x^2-7x=0 \implies x=0,\ 7

Step 2. Confirm which is on top

At $x=1$ the parabola gives $-7$ and the line gives $-1$ . The line is above, as the inequality already told us.

Step 3. Integrate

A=\int_0^7\Big[(-x)-(x^2-8x)\Big]dx=\int_0^7\big(7x-x^2\big)dx

A=\left[\frac{7x^2}{2}-\frac{x^3}{3}\right]_0^7=\frac{343}{2}-\frac{343}{3}=\frac{343}{6}

Answer: (1) $\dfrac{343}{6}$

Note that the whole region lies below the $x$ -axis, and it did not matter in the slightest. Because we integrated the difference of two boundaries rather than a single curve against the axis, the sign convention never came into play. That is the standard reason to prefer the difference form.

Example 2 (JEE Main 2026, 23 January Shift 1). Let the area of the region bounded by $y=\max\{\sin x,\cos x\}$ , the lines $x=0$ , $x=\tfrac{3\pi}{2}$ and the $x$ -axis be $A$ . Then $A+A^2$ is equal to ____.

▸ SHOW SOLUTION

y equals max of sin x and cos x, with the negative pieces beyond pi shaded separately

Step 1. Find where the maximum switches

$\sin x=\cos x$ at $x=\tfrac{\pi}{4}$ and $x=\tfrac{5\pi}{4}$ in $[0,\tfrac{3\pi}{2}]$ . Testing a point in each piece:

on $\left[0,\tfrac{\pi}{4}\right]$ the maximum is $\cos x$
on $\left[\tfrac{\pi}{4},\tfrac{5\pi}{4}\right]$ the maximum is $\sin x$
on $\left[\tfrac{5\pi}{4},\tfrac{3\pi}{2}\right]$ the maximum is $\cos x$

Step 2. Check the sign on each piece

This is the step that decides the question. On $[0,\pi]$ the maximum is positive. Beyond $x=\pi$ both $\sin x$ and $\cos x$ are negative, so the maximum is negative there too, and those pieces must be taken in absolute value.

Step 3. Integrate piece by piece

\int_0^{\pi/4}\cos x\,dx=\frac{1}{\sqrt2},\qquad \int_{\pi/4}^{\pi}\sin x\,dx=1+\frac{1}{\sqrt2}

\left|\int_{\pi}^{5\pi/4}\sin x\,dx\right|=1-\frac{1}{\sqrt2}

\left|\int_{5\pi/4}^{3\pi/2}\cos x\,dx\right|=1-\frac{1}{\sqrt2}

Step 4. Add

A=\frac{1}{\sqrt2}+1+\frac{1}{\sqrt2}+1-\frac{1}{\sqrt2}+1-\frac{1}{\sqrt2}=3

A+A^2=3+9=12

Answer: $12$

Every one of the four $\tfrac{1}{\sqrt2}$ terms cancels, which is the examiner's signature that you have split at the right places. If you stop at $x=\pi$ and forget the two negative pieces you get $A=1+\sqrt2$ and $A+A^2=4+3\sqrt2$ , which is exactly the trap.

Type 2 — Between two curves

Find the intersections, decide which is on top, integrate the difference. If the curves cross inside the interval, split there.

Example 3 (JEE Main 2026, 4 April Shift 2). The area of the region bounded by the curves $x+3y^2=0$ and $x+4y^2=1$ is

▸ SHOW SOLUTION

The lens between the two sideways parabolas, measured in horizontal strips

Step 1. Write both as $x$ in terms of $y$

x=-3y^2 \qquad\text{and}\qquad x=1-4y^2

Both are parabolas opening in the $-x$ direction, so horizontal strips are the natural choice.

Step 2. Find the intersections

-3y^2=1-4y^2 \implies y^2=1 \implies y=\pm1

Step 3. Decide which is on the right

At $y=0$ the first gives $x=0$ and the second gives $x=1$ , so $x=1-4y^2$ is to the right.

Step 4. Integrate in $y$

A=\int_{-1}^{1}\Big[(1-4y^2)-(-3y^2)\Big]dy=\int_{-1}^{1}\big(1-y^2\big)dy

A=2\int_0^1\big(1-y^2\big)dy=2\left(1-\frac13\right)=\frac43

Answer: (3) $\dfrac43$

Written in $x$ this needs two integrals and a square root. Written in $y$ it is four lines. Whenever both boundaries are naturally $x=g(y)$ , integrate in $y$ .

Example 4 (JEE Main 2026, 24 January Shift 1). Let $A_1$ be the bounded area enclosed by $y=x^2+2$ , $x+y=8$ and the $y$ -axis in the first quadrant. Let $A_2$ be the bounded area enclosed by $y=x^2+2$ , $y^2=x$ , $x=2$ and the $y$ -axis in the first quadrant. Then $A_1-A_2$ is equal to

▸ SHOW SOLUTION

The two regions A1 and A2 side by side

Step 1. Locate $A_1$

x^2+2=8-x \implies x^2+x-6=0

x=2 \quad\text{or}\quad x=-3

In the first quadrant the relevant root is $x=2$ , and the line is above the parabola there.

A_1=\int_0^2\Big[(8-x)-(x^2+2)\Big]dx

A_1=\int_0^2\big(6-x-x^2\big)dx

A_1=\left[6x-\frac{x^2}{2}-\frac{x^3}{3}\right]_0^2=12-2-\frac83=\frac{22}{3}

Step 2. Locate $A_2$

Here the upper boundary is $y=x^2+2$ and the lower is $y=\sqrt x$ , from $x=0$ to $x=2$ .

A_2=\int_0^2\Big[(x^2+2)-\sqrt x\Big]dx

=\left[\frac{x^3}{3}+2x-\frac{2}{3}x^{3/2}\right]_0^2

A_2=\frac83+4-\frac{2}{3}\cdot2\sqrt2=\frac{20}{3}-\frac{4\sqrt2}{3}

Step 3. Subtract

A_1-A_2=\frac{22}{3}-\frac{20}{3}+\frac{4\sqrt2}{3}=\frac{2}{3}+\frac{4\sqrt2}{3}

Answer: (1) $\dfrac{2}{3}+\dfrac{4\sqrt2}{3}$

Two separate regions, no shared boundary, no clever trick. The only thing to be careful about is that $A_2$ is bounded by $x=2$ on the right, not by an intersection.

Example 5 (JEE Main 2026, 28 January Shift 2). Let $P_1:y=4x^2$ and $P_2:y=x^2+27$ . If the area enclosed between $P_1$ and $P_2$ is six times the area enclosed between the line $y=\alpha x$ ( $\alpha>0$ ) and $P_1$ , then $\alpha$ equals

▸ SHOW SOLUTION

The area between the two parabolas, and the area between the line and P1

Step 1. The area between the two parabolas

4x^2=x^2+27 \implies x^2=9 \implies x=\pm3

A_1=\int_{-3}^{3}\Big[(x^2+27)-4x^2\Big]dx=\int_{-3}^{3}\big(27-3x^2\big)dx

A_1=2\left[27x-x^3\right]_0^3=2\big(81-27\big)=108

Step 2. The area between the line and $P_1$

\alpha x=4x^2 \implies x=0 \text{ or } x=\frac{\alpha}{4}

A_2=\int_0^{\alpha/4}\big(\alpha x-4x^2\big)dx=\left[\frac{\alpha x^2}{2}-\frac{4x^3}{3}\right]_0^{\alpha/4}

A_2=\frac{\alpha^3}{32}-\frac{\alpha^3}{48}=\frac{\alpha^3}{96}

Step 3. Impose the condition

108=6\cdot\frac{\alpha^3}{96} \implies \alpha^3=1728 \implies \alpha=12

Answer: (3) $12$

The standard result $\dfrac{8a^2}{3m^3}$ for $y^2=4ax$ against $y=mx$ has a twin for the vertical parabola: the area between $y=kx^2$ and $y=mx$ is $\dfrac{m^3}{6k^2}$ . Here $k=4$ , $m=\alpha$ , giving $\dfrac{\alpha^3}{96}$ directly.

Type 3 — Regions given by inequalities

A set such as $\{(x,y):\ldots\}$ is not harder, it is just less generous. Translate every inequality into "above", "below", "left of" or "right of", sketch, and read the limits off the picture.

Example 6 (JEE Main 2026, 6 April Shift 1). The area of the region $\{(x,y):0\le y\le 6-x,\ y^2\ge 4x-3,\ x\ge 0\}$ is

▸ SHOW SOLUTION

The region in horizontal strips, split at y=3 where the line meets the parabola

Step 1. Translate each inequality

$0\le y$ puts the region above the $x$ -axis. $y\le 6-x$ puts it below the line. $y^2\ge 4x-3$ rearranges to $x\le\dfrac{y^2+3}{4}$ , so the region is to the left of that parabola. And $x\ge0$ .

Step 2. Find the corner

The line and the parabola meet where $x=6-y$ and $x=\dfrac{y^2+3}{4}$ :

y^2+3=4(6-y) \implies y^2+4y-21=0

y=3 \quad\text{or}\quad y=-7

With $y\ge0$ the corner is at $y=3$ , $x=3$ .

Step 3. Integrate in horizontal strips

For $0\le y\le3$ the right-hand boundary is the parabola; for $3\le y\le6$ it is the line.

A=\int_0^3\frac{y^2+3}{4}\,dy+\int_3^6\big(6-y\big)dy

\int_0^3\frac{y^2+3}{4}\,dy=\frac14\left[\frac{y^3}{3}+3y\right]_0^3=\frac14\big(9+9\big)=\frac92

\int_3^6(6-y)\,dy=\left[6y-\frac{y^2}{2}\right]_3^6=18-\frac{27}{2}=\frac92

A=\frac92+\frac92=9

Answer: (2) $9$

Example 7 (JEE Main 2026, 28 January Shift 1). The area of the region $R=\{(x,y):xy\le 8,\ 1\le y\le x^2,\ x\ge0\}$ is

▸ SHOW SOLUTION

The region bounded by y=1, the parabola and the hyperbola, ceiling changing at x=2

Step 1. Read the boundaries

$y\ge1$ and $y\le x^2$ together force $x\ge1$ . The condition $xy\le8$ means $y\le\dfrac8x$ .

Step 2. Find where the top boundary changes

x^2=\frac8x \implies x^3=8 \implies x=2

So for $1\le x\le2$ the ceiling is $y=x^2$ , and for $x\ge2$ it is $y=\dfrac8x$ . That second branch meets $y=1$ at $x=8$ .

Step 3. Integrate

A=\int_1^2\big(x^2-1\big)dx+\int_2^8\left(\frac8x-1\right)dx

\int_1^2\big(x^2-1\big)dx=\left[\frac{x^3}{3}-x\right]_1^2=\frac83-2-\frac13+1=\frac43

\int_2^8\left(\frac8x-1\right)dx=\Big[8\log_e x-x\Big]_2^8

=8\log_e 4-6=16\log_e 2-6

A=\frac43+16\log_e2-6=16\log_e2-\frac{14}{3}

A=\frac23\big(24\log_e2-7\big)

Answer: (3) $\dfrac23\big(24\log_e 2-7\big)$

The identical region with $8$ replaced by $27$ and $y=1$ appeared again on 5 April Shift 1, where the answer is $54\log_e3-\dfrac{52}{3}$ . Same three boundaries, same three steps.

Example 8 (JEE Main 2026, 21 January Shift 2). If the area of the region $\{(x,y):1-2x\le y\le 4-x^2,\ x\ge0,\ y\ge0\}$ is $\dfrac{\alpha}{\beta}$ with $\gcd(\alpha,\beta)=1$ , then $\alpha+\beta$ is

▸ SHOW SOLUTION

The region whose floor switches from the line to the x-axis at x=one half

Step 1. Sketch the constraints

The ceiling is the downward parabola $y=4-x^2$ , which meets the $x$ -axis at $x=2$ . The floor is the larger of $y=1-2x$ and $y=0$ . The line $y=1-2x$ is positive only for $x<\tfrac12$ .

Step 2. Split at $x=\tfrac12$

For $0\le x\le\tfrac12$ the floor is the line; beyond that it is the axis.

A=\int_0^{1/2}\Big[(4-x^2)-(1-2x)\Big]dx

\qquad+\int_{1/2}^{2}\big(4-x^2\big)dx

Step 3. Evaluate

\int_0^{1/2}\big(3+2x-x^2\big)dx=\left[3x+x^2-\frac{x^3}{3}\right]_0^{1/2}

=\frac32+\frac14-\frac{1}{24}=\frac{41}{24}

\int_{1/2}^{2}\big(4-x^2\big)dx=\left[4x-\frac{x^3}{3}\right]_{1/2}^{2}

=\frac{16}{3}-2+\frac{1}{24}=\frac{27}{8}

A=\frac{41}{24}+\frac{27}{8}=\frac{41+81}{24}=\frac{122}{24}=\frac{61}{12}

Step 4. Read off the answer

$\gcd(61,12)=1$ , so $\alpha+\beta=61+12=73$ .

Answer: (1) $73$

Forgetting that the floor changes at $x=\tfrac12$ , and integrating $(4-x^2)-(1-2x)$ all the way to $x=2$ , gives a different and wrong value. The constraint $y\ge0$ is not decoration.

Type 4 — Modulus, max, min and the greatest integer

The graph does the work. Convert every bracket into a piecewise formula, sketch, then integrate piece by piece.

Example 9 (JEE Main 2026, 4 April Shift 1). The area of the region $\{(x,y):y\le\pi-|x|,\ y\le|x\sin x|,\ y\ge0\}$ is

▸ SHOW SOLUTION

The region below both ceilings, symmetric about the y-axis

Step 1. Use the symmetry

Both $\pi-|x|$ and $|x\sin x|$ are even, so the region is symmetric about the $y$ -axis. Work on $x\ge0$ and double.

Step 2. Find which ceiling is lower

On $x\ge0$ the two ceilings are $\pi-x$ and $|x\sin x|$ . At $x=\pi$ both are $0$ , and for $0<x<\pi$ we have $\sin x>0$ , so $|x\sin x|=x\sin x$ . Comparing at $x=\tfrac{\pi}{2}$ : $\pi-\tfrac{\pi}{2}=\tfrac{\pi}{2}\approx1.571$ and $\tfrac{\pi}{2}\sin\tfrac{\pi}{2}=\tfrac{\pi}{2}$ . They are equal there. In fact $x\sin x\le\pi-x$ on $[0,\pi]$ , so the binding ceiling is $y=x\sin x$ throughout.

Step 3. Integrate

A=2\int_0^{\pi}x\sin x\,dx

By parts, with $u=x$ and $dv=\sin x\,dx$ :

\int_0^{\pi}x\sin x\,dx=\Big[-x\cos x\Big]_0^{\pi}+\int_0^{\pi}\cos x\,dx

=\pi+0=\pi

That gives $A=2\pi$ , which is not among the options, so the ceiling must switch somewhere. Checking $x=3$ : $\pi-3\approx0.142$ while $3\sin3\approx0.423$ . So beyond some point the line is the lower ceiling.

Step 4. Locate the switch honestly

$x\sin x=\pi-x$ has a root at $x=\tfrac{\pi}{2}$ , since $\tfrac{\pi}{2}\cdot1=\tfrac{\pi}{2}$ . For $x<\tfrac{\pi}{2}$ the product $x\sin x$ is the smaller; for $\tfrac{\pi}{2}<x<\pi$ the line $\pi-x$ is the smaller.

A=2\left[\int_0^{\pi/2}x\sin x\,dx+\int_{\pi/2}^{\pi}\big(\pi-x\big)dx\right]

\int_0^{\pi/2}x\sin x\,dx=\Big[-x\cos x+\sin x\Big]_0^{\pi/2}=1

\int_{\pi/2}^{\pi}\big(\pi-x\big)dx=\left[\pi x-\frac{x^2}{2}\right]_{\pi/2}^{\pi}=\frac{\pi^2}{2}-\frac{3\pi^2}{8}=\frac{\pi^2}{8}

A=2\left(1+\frac{\pi^2}{8}\right)=2+\frac{\pi^2}{4}

Answer: (2) $2+\dfrac{\pi^2}{4}$

The moment an answer is not in the options, stop and re-test the comparison at a second point. Two curves that agree at one interior point almost always cross there.

Example 10 (JEE Main 2026, 24 January Shift 2). Let $f(\alpha)$ be the area 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

▸ SHOW SOLUTION

The two regions for alpha=0 and alpha=1

Step 1. Simplify the top boundary for $\alpha=0$

y=|{-5}|-|1|+0=5-1=4

A horizontal line. The floor is $y=\sqrt x$ .

f(0)=\int_0^1\big(4-\sqrt x\big)dx=4-\frac23=\frac{10}{3}

Step 2. Simplify for $\alpha=1$

On $0\le x\le1$ we have $x-5<0$ and $1-x\ge0$ , so

y=(5-x)-(1-x)+x^2=4+x^2

f(1)=\int_0^1\Big[(4+x^2)-\sqrt x\Big]dx=4+\frac13-\frac23=\frac{11}{3}

Step 3. Add

f(0)+f(1)=\frac{10}{3}+\frac{11}{3}=7

Answer: (3) $7$

The two modulus terms cancel their $\alpha x$ parts on the whole interval, which is why the messy-looking boundary is a plain quadratic. Always resolve the moduli on the given interval before you draw anything.

Example 11 (JEE Main 2026, 21 January Shift 1). The area inside the ellipse $x^2+4y^2=4$ and outside the region bounded by $y=|x|-1$ and $y=1-|x|$ is

▸ SHOW SOLUTION

The ellipse with the square removed from its centre

Step 1. Identify both regions

The ellipse is $\dfrac{x^2}{4}+y^2=1$ , with semi-axes $2$ and $1$ , so its area is $\pi\cdot2\cdot1=2\pi$ .

The curves $y=|x|-1$ and $y=1-|x|$ are two V shapes, one opening up from $(0,-1)$ and one opening down from $(0,1)$ . Together they enclose a square standing on its corner, with vertices $(0,1)$ , $(1,0)$ , $(0,-1)$ , $(-1,0)$ .

Step 2. Confirm the square is inside the ellipse

Its vertices $(\pm1,0)$ and $(0,\pm1)$ all satisfy $x^2+4y^2\le4$ , with equality only at $(0,\pm1)$ . So the square sits inside.

Step 3. Subtract

The square has diagonals of length $2$ and $2$ , so its area is $\tfrac12\cdot2\cdot2=2$ .

A=2\pi-2=2(\pi-1)

Answer: (1) $2(\pi-1)$

No integration at all. Recognising the two V shapes as a square is the entire question.

Type 5 — A parameter to be found

The area is given and a constant is unknown. Set up the integral with the letter in it, evaluate, then solve.

Example 12 (JEE Main 2026, 22 January Shift 1). Let the line $x=-1$ divide the area of the region $\{(x,y):1+x^2\le y\le 3-x\}$ in the ratio $m:n$ with $\gcd(m,n)=1$ . Then $m+n$ is

▸ SHOW SOLUTION

The region between parabola and line, split by x equals minus one into parts m and n

Step 1. Find the extent of the region

1+x^2=3-x \implies x^2+x-2=0

x=-2 \quad\text{or}\quad x=1

Step 2. Integrate each part

The height is $(3-x)-(1+x^2)=2-x-x^2$ .

m\text{-part}=\int_{-2}^{-1}\big(2-x-x^2\big)dx

=\left[2x-\frac{x^2}{2}-\frac{x^3}{3}\right]_{-2}^{-1}

=\left(-2-\frac12+\frac13\right)-\left(-4-2+\frac83\right)

=-\frac{13}{6}+\frac{10}{3}=\frac{7}{6}

n\text{-part}=\int_{-1}^{1}\big(2-x-x^2\big)dx=\left[2x-\frac{x^2}{2}-\frac{x^3}{3}\right]_{-1}^{1}

=\left(2-\frac12-\frac13\right)-\left(-2-\frac12+\frac13\right)

=\frac76+\frac{13}{6}=\frac{10}{3}

Step 3. Form the ratio

\frac{m}{n}=\frac{7/6}{10/3}=\frac{7}{20}

Since $\gcd(7,20)=1$ , $m+n=27$ .

Answer: (4) $27$

Example 13 (JEE Main 2026, 5 April Shift 2). Let $f:\mathbb{R}\to\mathbb{R}$ satisfy $f(x)+3f\!\left(\tfrac{\pi}{2}-x\right)=\sin x$ , and let $\alpha$ be the maximum value of $f$ . If the area bounded by $g(x)=x^2$ and $h(x)=\beta x^3$ ( $\beta>0$ ) is $\alpha^2$ , then $30\beta^3$ is

▸ SHOW SOLUTION

The region between the square and the cubic

Step 1. Solve the functional equation

Replace $x$ by $\tfrac{\pi}{2}-x$ :

f\!\left(\tfrac{\pi}{2}-x\right)+3f(x)=\cos x

Multiply this by $3$ and subtract the original:

8f(x)=3\cos x-\sin x

f(x)=\frac{3\cos x-\sin x}{8}

Step 2. Take the maximum

\alpha=\frac{\sqrt{3^2+1^2}}{8}=\frac{\sqrt{10}}{8}, \qquad \alpha^2=\frac{10}{64}=\frac{5}{32}

Step 3. Find the area between the two curves

x^2=\beta x^3 \implies x=0 \text{ or } x=\frac1\beta

A=\int_0^{1/\beta}\big(x^2-\beta x^3\big)dx=\left[\frac{x^3}{3}-\frac{\beta x^4}{4}\right]_0^{1/\beta}

A=\frac{1}{3\beta^3}-\frac{1}{4\beta^3}=\frac{1}{12\beta^3}

Step 4. Solve

\frac{1}{12\beta^3}=\frac{5}{32} \implies \beta^3=\frac{32}{60}=\frac{8}{15} \implies 30\beta^3=16

Answer: $16$

The area formula $\dfrac{1}{12\beta^3}$ is worth keeping: the region between $y=x^2$ and $y=\beta x^3$ always has that area, whatever $\beta$ is.

Example 14 (JEE Main 2026, 8 April Shift 2). Let $f:(1,\infty)\to\mathbb{R}$ be $f(x)=\dfrac{x-1}{x+1}$ , with $f^{i+1}(x)=f\big(f^i(x)\big)$ and $f^1(x)=f(x)$ . If $g(x)+f^{26}(x)=0$ , the area bounded by $y=g(x)$ , $2y=2x-3$ , $y=0$ and $x=4$ is

▸ SHOW SOLUTION

The triangle under the line plus the region under the hyperbola up to x=4

Step 1. Iterate until it repeats

f^1(x)=\frac{x-1}{x+1},\qquad f^2(x)=\frac{\frac{x-1}{x+1}-1}{\frac{x-1}{x+1}+1}=\frac{-2}{2x}=-\frac1x

f^3(x)=\frac{-\frac1x-1}{-\frac1x+1}=\frac{1+x}{1-x},\qquad f^4(x)=x

So $f$ has period $4$ under composition.

Step 2. Reduce the index

$26=4\cdot6+2$ , so $f^{26}=f^2=-\dfrac1x$ , and therefore

g(x)=-f^{26}(x)=\frac1x

Step 3. Locate the region

The line $2y=2x-3$ is $y=x-\tfrac32$ . It meets $y=0$ at $x=\tfrac32$ and meets $y=\dfrac1x$ where

x-\frac32=\frac1x \implies 2x^2-3x-2=0 \implies x=2

taking the positive root. The region is bounded below by $y=0$ , on the left by the line, above by $y=\dfrac1x$ , and on the right by $x=4$ .

Step 4. Split at $x=2$

For $\tfrac32\le x\le2$ the ceiling is the line; for $2\le x\le4$ it is the hyperbola.

\int_{3/2}^{2}\left(x-\frac32\right)dx=\frac12\cdot\frac12\cdot\frac12=\frac18

\int_2^4\frac{dx}{x}=\Big[\log_e x\Big]_2^4=\log_e 2

A=\frac18+\log_e 2

Answer: (1) $\dfrac18+\log_e 2$

The first integral is a triangle with legs $\tfrac12$ and $\tfrac12$ , so it needs no calculus at all.

Type 6 — Conics

Circles, parabolas, ellipses and hyperbolas bring one extra tool: the segment formula. Whenever a chord cuts a circle, reach for it before you reach for an integral.

For a circle of radius $r$ and a chord subtending a central angle $\theta$ at the centre, the smaller piece has area

\text{segment}=\frac{r^2}{2}\big(\theta-\sin\theta\big)

A circular segment with its central angle marked

Example 15 (JEE Main 2026, 23 January Shift 2). The area of the region enclosed between the circles $x^2+y^2=4$ and $x^2+(y-2)^2=4$ is

▸ SHOW SOLUTION

Step 1. Set up

Both circles have radius $2$ ; their centres are $(0,0)$ and $(0,2)$ , a distance $d=2$ apart. The overlap is a lens made of two identical segments.

The lens where the two circles of radius 2 overlap

Step 2. Find the central angle

Half the chord subtends an angle $\phi$ at each centre with $\cos\phi=\dfrac{d/2}{r}=\dfrac12$ , so $\phi=\dfrac{\pi}{3}$ and the full central angle is $\theta=\dfrac{2\pi}{3}$ .

Step 3. Apply the segment formula twice

\text{one segment}=\frac{r^2}{2}\big(\theta-\sin\theta\big)

=2\left(\frac{2\pi}{3}-\frac{\sqrt3}{2}\right)=\frac{4\pi}{3}-\sqrt3

A=2\left(\frac{4\pi}{3}-\sqrt3\right)=\frac{8\pi}{3}-2\sqrt3=\frac23\big(4\pi-3\sqrt3\big)

Answer: (4) $\dfrac23\big(4\pi-3\sqrt3\big)$

Two circles of equal radius $r$ whose centres are $r$ apart always give this: the lens is $\dfrac{2r^2}{3}\big(2\pi-\dfrac{3\sqrt3}{2}\big)$ , and here $r=2$ .

Example 16 (JEE Main 2026, 22 January Shift 2). The area of the region $A=\{(x,y):4x^2+y^2\le8 \text{ and } y^2\le4x\}$ is

▸ SHOW SOLUTION

The region inside both the ellipse and the parabola

Step 1. Identify the curves

$4x^2+y^2=8$ is the ellipse $\dfrac{x^2}{2}+\dfrac{y^2}{8}=1$ , with semi-axes $\sqrt2$ along $x$ and $2\sqrt2$ along $y$ . The parabola is $y^2=4x$ .

Step 2. Find the intersection

Substituting $y^2=4x$ into the ellipse:

4x^2+4x=8 \implies x^2+x-2=0

x=1 \quad\text{or}\quad x=-2

Only $x=1$ is admissible, giving $y=\pm2$ .

Step 3. Integrate, using symmetry about the $x$ -axis

For $0\le x\le1$ the ceiling is the parabola $y=2\sqrt x$ ; for $1\le x\le\sqrt2$ it is the ellipse $y=\sqrt{8-4x^2}$ .

A=2\left[\int_0^1 2\sqrt x\,dx+\int_1^{\sqrt2}\sqrt{8-4x^2}\,dx\right]

\int_0^1 2\sqrt x\,dx=\frac43

\int_1^{\sqrt2}\sqrt{8-4x^2}\,dx=2\int_1^{\sqrt2}\sqrt{2-x^2}\,dx

=2\left[\frac{x\sqrt{2-x^2}}{2}+\sin^{-1}\frac{x}{\sqrt2}\right]_1^{\sqrt2}

=2\left[\frac{\pi}{2}-\frac12-\frac{\pi}{4}\right]=\frac{\pi}{2}-1

A=2\left[\frac43+\frac{\pi}{2}-1\right]=\frac83+\pi-2=\pi+\frac23

Answer: (2) $\pi+\dfrac23$

Example 17 (JEE Main 2026, 2 April Shift 2). If the area bounded by $16x^2-9y^2=144$ and $8x-3y=24$ is $A$ , then $3\big(A+6\log_e3\big)$ is ____.

▸ SHOW SOLUTION

Step 1. Put the hyperbola in standard form

\frac{x^2}{9}-\frac{y^2}{16}=1

Its right branch has vertex $(3,0)$ , and the line passes through $(3,0)$ as well, since $8(3)-3(0)=24$ .

Horizontal strips between the line and the right branch of the hyperbola

Step 2. Find the other intersection

Write both as $x$ in terms of $y$ :

x=3\sqrt{1+\frac{y^2}{16}} \qquad\text{and}\qquad x=3+\frac{3y}{8}

Squaring and simplifying gives $y=0$ and $y=\dfrac{16}{3}$ .

Step 3. Integrate in $y$

A=\int_0^{16/3}\left[\left(3+\frac{3y}{8}\right)-3\sqrt{1+\frac{y^2}{16}}\right]dy

\int_0^{16/3}\left(3+\frac{3y}{8}\right)dy=16+\frac{3}{16}\cdot\frac{256}{9}

=16+\frac{16}{3}=\frac{64}{3}

For the hyperbola part, put $y=4\tan\theta$ :

3\int_0^{16/3}\sqrt{1+\frac{y^2}{16}}\,dy=\frac34\int_0^{16/3}\sqrt{16+y^2}\,dy

=\frac{3}{16}\left[y\sqrt{16+y^2}+16\log_e\!\left(y+\sqrt{16+y^2}\right)\right]_0^{16/3}

Evaluating gives $\dfrac{64}{3}-8+6\log_e3$ , so

A=8-6\log_e 3

Step 4. Finish

3\big(A+6\log_e3\big)=3\cdot8=24

Answer: $24$

The logarithm was always going to cancel: the question hands you $+6\log_e3$ precisely so that it does. When a numerical-answer question adds a transcendental term to your unknown, it is telling you what your answer contains.

The decision chart

Six questions, asked in this order, settle which method to use.

Decision chart for choosing the integration variable and the split points

Are both boundaries naturally $x=g(y)$ ? If yes, integrate in $y$ . Sideways parabolas and left-right regions belong here.
Is there a modulus, a max, a min or a greatest integer? Resolve it into a piecewise formula on the given interval first. Sketch second. Integrate third.
Does a circle appear with a chord across it? Use the segment formula $\dfrac{r^2}{2}(\theta-\sin\theta)$ , not an integral.
Is the region symmetric? Halve or quarter the work. Check for even powers of $x$ or of $y$ .
Do the boundaries cross inside the interval? Split at every crossing and take the difference the right way round on each piece.
Is a single curve measured against an axis? Only then does the sign convention matter. Find every root first.

Quick recall

Situation	What to write
Curve above the axis on $[a,b]$	$\displaystyle\int_a^b f\,dx$
Curve below the axis	$\left\lvert\displaystyle\int_a^b f\,dx\right\rvert$
Curve crossing at $x=c$	$\displaystyle\int_a^c f\,dx-\displaystyle\int_c^b f\,dx$
Two curves, $g$ on top	$\displaystyle\int_a^b(g-f)\,dx$
Region easier in horizontal strips	$\displaystyle\int_c^d g(y)\,dy$
Circle cut by a chord, angle $\theta$	$\dfrac{r^2}{2}(\theta-\sin\theta)$
$y^2=4ax$ with $x^2=4by$	$\dfrac{16ab}{3}$
$y^2=4ax$ with $y=mx$	$\dfrac{8a^2}{3m^3}$
$y=kx^2$ with $y=mx$	$\dfrac{m^3}{6k^2}$
$y=x^2$ with $y=\beta x^3$	$\dfrac{1}{12\beta^3}$
Ellipse	$\pi ab$
One arch of $\sin ax$	$\dfrac2a$

The habit that earns the mark

Before writing a single integral sign, draw the region and shade it. Then write down, in words, which curve is the ceiling and which is the floor, and over what range of $x$ or $y$ . Nine out of ten wrong answers in this chapter come from a correct integral of the wrong region.

Every worked example above is a JEE Main 2026 question, re-derived from the statement rather than copied from a key. Practise the rest on doMath, where the same six types are graded and solved step by step.

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Reinforce this post with live problems on doMath.

Indefinite Integration Definite Integration Application of Integrals

Area Bounded by Curves: Complete Notes with Illustrations

What the definite integral measures

The sign convention, and the mark it costs

The five configurations

Curve tracing: seven checks before you integrate

1. Domain

2. Symmetry

3. Intercepts

4. Turning points

5. Period

6. Asymptotes

7. Sign

Reading a modulus off the graph

Standard results worth carrying

Type 1 — A curve against an axis

Step 1. Find where the boundaries meet

Step 2. Confirm which is on top

Step 3. Integrate

Step 1. Find where the maximum switches

Step 2. Check the sign on each piece

Step 3. Integrate piece by piece

Step 4. Add

Type 2 — Between two curves

Step 1. Write both as xxx in terms of yyy

Step 2. Find the intersections

Step 3. Decide which is on the right

Step 4. Integrate in yyy

Step 1. Locate A1A_1A1​

Step 2. Locate A2A_2A2​

Step 3. Subtract

Step 1. The area between the two parabolas

Step 2. The area between the line and P1P_1P1​

Step 3. Impose the condition

Type 3 — Regions given by inequalities

Step 1. Translate each inequality

Step 2. Find the corner

Step 3. Integrate in horizontal strips

Step 1. Read the boundaries

Step 2. Find where the top boundary changes

Step 3. Integrate

Step 1. Sketch the constraints

Step 2. Split at x=12x=\tfrac12x=21​

Step 3. Evaluate

Step 4. Read off the answer

Type 4 — Modulus, max, min and the greatest integer

Step 1. Use the symmetry

Step 2. Find which ceiling is lower

Step 3. Integrate

Step 4. Locate the switch honestly

Step 1. Simplify the top boundary for α=0\alpha=0α=0

Step 2. Simplify for α=1\alpha=1α=1

Step 3. Add

Step 1. Identify both regions

Step 2. Confirm the square is inside the ellipse

Step 3. Subtract

Type 5 — A parameter to be found

Step 1. Find the extent of the region

Step 2. Integrate each part

Step 3. Form the ratio

Step 1. Solve the functional equation

Step 2. Take the maximum

Step 3. Find the area between the two curves

Step 4. Solve

Step 1. Iterate until it repeats

Step 2. Reduce the index

Step 3. Locate the region

Step 4. Split at x=2x=2x=2

Type 6 — Conics

Step 1. Set up

Step 2. Find the central angle

Step 3. Apply the segment formula twice

Step 1. Identify the curves

Step 2. Find the intersection

Step 3. Integrate, using symmetry about the xxx-axis

Step 1. Put the hyperbola in standard form

Step 2. Find the other intersection

Step 3. Integrate in yyy

Step 4. Finish

The decision chart

Quick recall

Step 1. Write both as $x$ in terms of $y$

Step 4. Integrate in $y$

Step 1. Locate $A_1$

Step 2. Locate $A_2$

Step 2. The area between the line and $P_1$

Step 2. Split at $x=\tfrac12$

Step 1. Simplify the top boundary for $\alpha=0$

Step 2. Simplify for $\alpha=1$

Step 4. Split at $x=2$

Step 3. Integrate, using symmetry about the $x$ -axis

Step 3. Integrate in $y$