Intermediate

24 min read

Area between Two Curves

Master Area between Two Curves for JEE Main and Advanced. Covers upper minus lower curve formula, finding intersection points, testing which curve is on top, splitting when curves cross, horizontal integration wrt y, standard examples line and parabola, trig curves sin and cos, symmetry tricks, 5-step checklist with 7 fully verified JEE-level practice questions.

New

When two curves enclose a finite region between them, the area of that region is the integral of the vertical distance between the curves — the upper curve minus the lower curve. This simple idea extends naturally to more complex scenarios: curves that cross each other (requiring the region to be split), regions bounded horizontally (requiring integration w.r.t. $y$ ), and combinations of curves and lines. In JEE, area-between-curves questions test three skills simultaneously: finding intersection points correctly, identifying which curve is on top at any given $x$ , and setting up the integral correctly. Every step matters — an error in any one step gives a completely wrong final answer.

1. The Core Formula JEE Main & Advanced

If $f (x) \geq g (x)$ for all $x \in [a, b]$ , the area of the region between the curves $y = f (x)$ (upper) and $y = g (x)$ (lower) is:

$A = \int_{a}^{b} [f (x) - g (x)] d x$

Key insight: This formula works even when both curves are below the $x$ -axis, or both above, or one on each side — as long as $f (x) \geq g (x)$ throughout $[a, b]$ . The individual signs don't matter; only the difference determines the area.

When Curves Cross — Splitting the Region

If the curves intersect at $x = c$ within $[a, b]$ , with $f > g$ on $[a, c]$ and $g > f$ on $[c, b]$ :

$A = \int_{a}^{c} [f (x) - g (x)] d x + \int_{c}^{b} [g (x) - f (x)] d x = \int_{a}^{b} | f (x) - g (x) | d x$

2. Standard Examples — Line and Parabola JEE Main & Advanced

Example 1 — Area between $y = x$ and $y = x^{2}$

Step 1 — Intersections: $x = x^{2} \Rightarrow x (x - 1) = 0 \Rightarrow x = 0, 1$ .

Step 2 — Which is on top? At $x = 0.5$ : $y = 0.5$ (line) vs $y = 0.25$ (parabola). So $x > x^{2}$ on $(0, 1)$ .

Step 3 — Integrate:

$A = \int_{0}^{1} (x - x^{2}) d x = {[\frac{x^{2}}{2} - \frac{x^{3}}{3}]}_{0}^{1} = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}$

Example 2 — Area between $y = \sqrt{x}$ and $y = x^{2}$

Step 1 — Intersections: $\sqrt{x} = x^{2} \Rightarrow x = x^{4} \Rightarrow x^{4} - x = 0 \Rightarrow x (x^{3} - 1) = 0 \Rightarrow x = 0, 1$ .

Step 2 — Which is on top? At $x = 0.25$ : $\sqrt{0.25} = 0.5 > (0.25)^{2} = 0.0625$ . So $\sqrt{x} > x^{2}$ on $(0, 1)$ .

Step 3 — Integrate:

$A = \int_{0}^{1} (\sqrt{x} - x^{2}) d x = {[\frac{2 x^{3 / 2}}{3} - \frac{x^{3}}{3}]}_{0}^{1} = \frac{2}{3} - \frac{1}{3} = \frac{1}{3}$

Example 3 — Area between parabola and line: $y = x^{2} - 4$ and $y = x + 2$

Step 1 — Intersections: $x^{2} - 4 = x + 2 \Rightarrow x^{2} - x - 6 = 0 \Rightarrow (x - 3) (x + 2) = 0 \Rightarrow x = - 2, 3$ .

Step 2: At $x = 0$ : line gives $y = 2$ , parabola gives $y = - 4$ . So line is above on $(- 2, 3)$ .

Step 3:

$A = \int_{- 2}^{3} [(x + 2) - (x^{2} - 4)] d x = \int_{- 2}^{3} (- x^{2} + x + 6) d x$

$= {[- \frac{x^{3}}{3} + \frac{x^{2}}{2} + 6 x]}_{- 2}^{3} = (- 9 + \frac{9}{2} + 18) - (\frac{8}{3} + 2 - 12) = \frac{27}{2} - (- \frac{22}{3}) = \frac{125}{6}$

3. Horizontal Slicing — Integrating w.r.t. $y$ JEE Main & Advanced

When curves are more naturally expressed as $x = f (y)$ , integrate w.r.t. $y$ :

$A = \int_{c}^{d} [f_{right} (y) - f_{left} (y)] d y$

Example — Area between $y^{2} = x$ and $x = y + 2$

Intersections: $y^{2} = y + 2 \Rightarrow y^{2} - y - 2 = 0 \Rightarrow (y - 2) (y + 1) = 0 \Rightarrow y = - 1, 2$ .

Right curve: $x = y + 2$ (line); Left curve: $x = y^{2}$ (parabola). Line is to the right of parabola on $(- 1, 2)$ .

$A = \int_{- 1}^{2} [(y + 2) - y^{2}] d y = {[\frac{y^{2}}{2} + 2 y - \frac{y^{3}}{3}]}_{- 1}^{2}$

$= (2 + 4 - \frac{8}{3}) - (\frac{1}{2} - 2 + \frac{1}{3}) = \frac{10}{3} - (- \frac{7}{6}) = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2}$

4. Area between Trig Curves JEE Main & Advanced

Area between $y = \sin x$ and $y = \cos x$ on $[0, π / 4]$

On $[0, π / 4]$ : $\cos x \geq \sin x$ (since $\cos 0 = 1 > 0 = \sin 0$ , and they meet at $π / 4$ ).

$A = \int_{0}^{π / 4} (\cos x - \sin x) d x = [\sin x + \cos x]_{0}^{π / 4}$

$= (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) - (0 + 1) = \sqrt{2} - 1$

Area between $y = \sin x$ and $y = \cos x$ on $[0, π]$

They cross at $x = π / 4$ . On $[0, π / 4]$ : $\cos > \sin$ ; on $[π / 4, π]$ : $\sin > \cos$ . Split:

$A = \int_{0}^{π / 4} (\cos x - \sin x) d x + \int_{π / 4}^{π} (\sin x - \cos x) d x = (\sqrt{2} - 1) + (1 + \sqrt{2}) = 2 \sqrt{2}$

5. Useful Symmetry Tricks JEE Main & Advanced

Symmetric about $y$ -axis: If the region is symmetric about $y$ -axis, compute for $x \geq 0$ and double it.
Symmetric about $x$ -axis: If the region is symmetric about $x$ -axis (e.g., $y^{2} = f (x)$ ), compute the upper half and double.
$y = x$ symmetry: The area between $y = \sqrt{x}$ and $y = x^{2}$ equals the area between $y = x$ and $y = x^{2}$ (since $\sqrt{x}$ is the inverse of $x^{2}$ — they're reflections in $y = x$ ), which is confirmed: both give $\frac{1}{3}$ ... actually $y = x$ gives $\frac{1}{6}$ , not $\frac{1}{3}$ . The two areas are different.

Area between $y = x^{2}$ and $y = 4$

Intersections: $x^{2} = 4 \Rightarrow x = \pm 2$ . The line $y = 4$ is above the parabola on $(- 2, 2)$ .

$A = \int_{- 2}^{2} (4 - x^{2}) d x = 2 \int_{0}^{2} (4 - x^{2}) d x$ (by symmetry) $= 2 {[4 x - \frac{x^{3}}{3}]}_{0}^{2} = 2 (8 - \frac{8}{3}) = 2 \cdot \frac{16}{3} = \frac{32}{3}$

Area Between Curves — 5-Step Checklist

Step	Action	Common error
1. Sketch	Draw both curves roughly to see the enclosed region	Skipping this leads to wrong top/bottom identification
2. Intersect	Solve $f (x) = g (x)$ to find limits $a$ and $b$	Arithmetic errors here propagate to the final answer
3. Test	Plug in a test point between $a$ and $b$ — which curve gives the larger $y$ ?	Getting top and bottom swapped gives a negative area
4. Split	If curves cross within $[a, b]$ , find crossing point and split integral	Not splitting causes partial cancellation — area too small
5. Integrate	$\int_{a}^{b} [upper - lower] d x$ ; always positive	Negative answer = upper/lower swapped; recheck Step 3

JEE Power Tips — Avoid These Common Mistakes

Always find intersection points first — do not assume limits from the question. "Find area between $y = x^{2}$ and $y = 2 x$ " — the limits $x = 0$ and $x = 2$ must come from solving $x^{2} = 2 x$ . If not stated, you must derive them.
Test which curve is on top with a specific number — don't guess. Between $y = \sqrt{x}$ and $y = x^{2}$ , plug in $x = 0.25$ : $\sqrt{0.25} = 0.5 > 0.0625 = (0.25)^{2}$ . So $\sqrt{x}$ is on top. Guessing without testing loses marks.
$y^{2} = f (x)$ regions need the factor of 2. The parabola $y^{2} = 4 x$ is symmetric about the $x$ -axis — upper half is $y = 2 \sqrt{x}$ , lower is $y = - 2 \sqrt{x}$ . Area $= 2 \int_{0}^{h} 2 \sqrt{x} d x$ . Forgetting the factor of 2 halves the answer.
Integrate w.r.t. $y$ when the region is easier to describe horizontally. For the region bounded by $x = y^{2}$ and $x = y + 2$ — integrating w.r.t. $x$ requires splitting and is messy. Integrating w.r.t. $y$ gives a single clean integral from $y = - 1$ to $y = 2$ .
Area between $y = \sin x$ and $y = \cos x$ : they cross at $x = π / 4 + n π$ — always split there. On $[0, π / 4]$ , $\cos > \sin$ . On $[π / 4, π]$ , $\sin > \cos$ . Missing the split and integrating blindly gives zero or a smaller-than-correct answer.
The formula $A = \int_{a}^{b} [f - g] d x$ requires $f \geq g$ throughout — not just at the endpoints. Always check that the curves don't cross in the interior. If they do, you must split.

Practice Questions (JEE Level)

Q1: Find the area enclosed between $y = x$ and $y = x^{2}$ .

Answer: $\frac{1}{6}$

Explanation:

First, find the points of intersection by setting the equations equal:
$x = x^{2} ⟹ x^{2} - x = 0 ⟹ x (x - 1) = 0 ⟹ x = 0, 1$ .

Test a point in the interval $(0, 1)$ to see which curve is on top. Let $x = 0.5$ :
$0.5 > 0.25$ , so the line $y = x$ is above the parabola $y = x^{2}$ .

Set up and evaluate the integral:
$A = \int_{0}^{1} (x - x^{2}) d x$
$A = {[\frac{x^{2}}{2} - \frac{x^{3}}{3}]}_{0}^{1}$
$A = (\frac{1}{2} - \frac{1}{3}) - 0 = \frac{1}{6}$ .

Q2: Find the area bounded by $y = \sqrt{x}$ and $y = x^{2}$ .

Answer: $\frac{1}{3}$

Explanation:

First, find the intersection points:
$\sqrt{x} = x^{2}$
Square both sides: $x = x^{4} ⟹ x^{4} - x = 0 ⟹ x (x^{3} - 1) = 0 ⟹ x = 0, 1$ .

On the interval $(0, 1)$ , $\sqrt{x}$ is greater than $x^{2}$ .

Set up and evaluate the integral:
$A = \int_{0}^{1} (\sqrt{x} - x^{2}) d x$
$A = {[\frac{x^{3 / 2}}{3 / 2} - \frac{x^{3}}{3}]}_{0}^{1} = {[\frac{2 x^{3 / 2}}{3} - \frac{x^{3}}{3}]}_{0}^{1}$
$A = (\frac{2}{3} - \frac{1}{3}) - 0 = \frac{1}{3}$ .

Q3: Find the area enclosed between the parabola $y = x^{2}$ and the line $y = 2 x$ .

Answer: $\frac{4}{3}$

Explanation:

Find the intersection points:
$x^{2} = 2 x ⟹ x^{2} - 2 x = 0 ⟹ x (x - 2) = 0 ⟹ x = 0, 2$ .

Test a point in the interval $(0, 2)$ , such as $x = 1$ :
$2 (1) > 1^{2} ⟹ 2 > 1$ , so the line $y = 2 x$ is above the parabola $y = x^{2}$ .

Set up and evaluate the integral:
$A = \int_{0}^{2} (2 x - x^{2}) d x$
$A = {[x^{2} - \frac{x^{3}}{3}]}_{0}^{2}$
$A = (4 - \frac{8}{3}) - 0$
$A = \frac{12}{3} - \frac{8}{3} = \frac{4}{3}$ .

Q4: Find the area between $y = x^{2} - 4$ and $y = x + 2$ .

Answer: $\frac{125}{6}$

Explanation:

Find the intersection points:
$x^{2} - 4 = x + 2$
$x^{2} - x - 6 = 0$
$(x - 3) (x + 2) = 0 ⟹ x = - 2, 3$ .

Within the interval $(- 2, 3)$ , the line $y = x + 2$ lies above the parabola $y = x^{2} - 4$ .

Set up the integral:
$A = \int_{- 2}^{3} [(x + 2) - (x^{2} - 4)] d x$
$A = \int_{- 2}^{3} (- x^{2} + x + 6) d x$

Evaluate:
$A = {[- \frac{x^{3}}{3} + \frac{x^{2}}{2} + 6 x]}_{- 2}^{3}$
Upper limit ( $x = 3$ ): $- \frac{27}{3} + \frac{9}{2} + 18 = - 9 + 4.5 + 18 = 13.5$
Lower limit ( $x = - 2$ ): $- \frac{- 8}{3} + \frac{4}{2} - 12 = \frac{8}{3} + 2 - 12 = \frac{8}{3} - 10 = - \frac{22}{3}$

$A = 13.5 - (- \frac{22}{3}) = \frac{27}{2} + \frac{22}{3} = \frac{81}{6} + \frac{44}{6} = \frac{125}{6}$ .

Q5 (JEE-type): Find the area between $y = x^{2}$ and the line $y = 4$ .

Answer: $\frac{32}{3}$

Explanation:

Find the intersection points:
$x^{2} = 4 ⟹ x = \pm 2$ .

The line $y = 4$ is above the parabola $y = x^{2}$ on the interval $(- 2, 2)$ . Because both functions are even (symmetric about the y-axis), we can evaluate from 0 to 2 and double the result:

$A = \int_{- 2}^{2} (4 - x^{2}) d x$
$A = 2 \int_{0}^{2} (4 - x^{2}) d x$
$A = 2 {[4 x - \frac{x^{3}}{3}]}_{0}^{2}$
$A = 2 (8 - \frac{8}{3})$
$A = 2 (\frac{24}{3} - \frac{8}{3}) = 2 (\frac{16}{3}) = \frac{32}{3}$ .

Q6 (MCQ — JEE-type): The area bounded by the curves $y = \cos x$ , $y = \sin x$ and $x = 0$ on the interval $[0, π / 4]$ is:

A) $\sqrt{2}$
B) $\sqrt{2} - 1$
C) $2 \sqrt{2} - 1$
D) 1

Answer: B) $\sqrt{2} - 1$ .

Explanation:

On the interval from $0$ to $π / 4$ , the cosine curve starts at 1 and drops, while the sine curve starts at 0 and rises. Therefore, $\cos x \geq \sin x$ throughout this interval.

Set up the integral with the upper curve minus the lower curve:
$A = \int_{0}^{π / 4} (\cos x - \sin x) d x$

Evaluate:
$A = [\sin x + \cos x]_{0}^{π / 4}$
$A = (\sin \frac{π}{4} + \cos \frac{π}{4}) - (\sin 0 + \cos 0)$
$A = (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) - (0 + 1)$
$A = \frac{2}{\sqrt{2}} - 1 = \sqrt{2} - 1$ .

Q7 (JEE-type): Find the area enclosed by $y = | x |$ and $y = 1$ .

Answer: 1

Explanation:

Find the intersection points:
$| x | = 1 ⟹ x = \pm 1$ .

The line $y = 1$ is a horizontal cap above the V-shaped curve $y = | x |$ over the interval $(- 1, 1)$ . Because both functions are symmetric about the y-axis, we can integrate from 0 to 1 and double the result. For $x > 0$ , $| x | = x$ .

$A = \int_{- 1}^{1} (1 - | x |) d x$
$A = 2 \int_{0}^{1} (1 - x) d x$
$A = 2 {[x - \frac{x^{2}}{2}]}_{0}^{1}$
$A = 2 (1 - \frac{1}{2}) = 2 (\frac{1}{2}) = 1$ .

(Geometrically, this forms a triangle with a base of 2 (from $x = - 1$ to $x = 1$ ) and a height of 1. Area = $\frac{1}{2} \times base \times height = \frac{1}{2} \times 2 \times 1 = 1$ .)

Related Study Material

More Mathematics Notes More Calculus Notes All Application Of Integrals Topics

Practice Questions

Mathematics Practice Application Of Integrals Questions

Topic Information

DifficultyIntermediate

Est. Read Time24 minutes

CategoryApplication Of Integrals

Quick Actions

Track Your Learning

Study Tips

Take notes of key points while reading

Practice related questions after studying

Review topics marked as "Learning" regularly

Test your knowledge with practice questions

Edvaya Target

Detailed Notes & Solved Examples

Master complex concepts with our rich, topic-wise study notes. Access hundreds of step-by-step solved examples and interactive flashcards right from your phone.

Edvaya Target

Edvaya Aspire

Take a Free Mock Test

Target Subjects

Aspire Subjects

Subject Mastery

Area between Two Curves

1. The Core Formula JEE Main & Advanced

When Curves Cross — Splitting the Region

2. Standard Examples — Line and Parabola JEE Main & Advanced

Example 1 — Area between $y = x$ and $y = x^{2}$

Example 2 — Area between $y = \sqrt{x}$ and $y = x^{2}$

Example 3 — Area between parabola and line: $y = x^{2} - 4$ and $y = x + 2$

3. Horizontal Slicing — Integrating w.r.t. $y$ JEE Main & Advanced

Example — Area between $y^{2} = x$ and $x = y + 2$

4. Area between Trig Curves JEE Main & Advanced

Area between $y = \sin x$ and $y = \cos x$ on $[0, π / 4]$

Area between $y = \sin x$ and $y = \cos x$ on $[0, π]$

5. Useful Symmetry Tricks JEE Main & Advanced

Area between $y = x^{2}$ and $y = 4$

Area Between Curves — 5-Step Checklist

Practice Questions (JEE Level)

Related Study Material

Practice Questions

Topic Information

Quick Actions

Track Your Learning

Study Tips

Edvaya Target

We Value Your Privacy

Edvaya Target

Edvaya Aspire

Take a Free Mock Test

Target Subjects

Aspire Subjects

Subject Mastery

Area between Two Curves

1. The Core Formula JEE Main & Advanced

When Curves Cross — Splitting the Region

2. Standard Examples — Line and Parabola JEE Main & Advanced

Example 1 — Area between y=x and y=x2

Example 2 — Area between y=x and y=x2

Example 3 — Area between parabola and line: y=x2−4 and y=x+2

3. Horizontal Slicing — Integrating w.r.t. y JEE Main & Advanced

Example — Area between y2=x and x=y+2

4. Area between Trig Curves JEE Main & Advanced

Area between y=sin⁡x and y=cos⁡x on [0,π/4]

Area between y=sin⁡x and y=cos⁡x on [0,π]

5. Useful Symmetry Tricks JEE Main & Advanced

Area between y=x2 and y=4

Area Between Curves — 5-Step Checklist

Practice Questions (JEE Level)

Related Study Material

Practice Questions

Topic Information

Quick Actions

Track Your Learning

Study Tips

Edvaya Target

We Value Your Privacy

Example 1 — Area between $y = x$ and $y = x^{2}$

Example 2 — Area between $y = \sqrt{x}$ and $y = x^{2}$

Example 3 — Area between parabola and line: $y = x^{2} - 4$ and $y = x + 2$

3. Horizontal Slicing — Integrating w.r.t. $y$ JEE Main & Advanced

Example — Area between $y^{2} = x$ and $x = y + 2$

Area between $y = \sin x$ and $y = \cos x$ on $[0, π / 4]$

Area between $y = \sin x$ and $y = \cos x$ on $[0, π]$

Area between $y = x^{2}$ and $y = 4$