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Area under Curves — Standard Curves & Bounded Regions

Master Area under Curves for JEE Main and Advanced. Covers area as definite integral, signed vs total area, standard curves parabola circle ellipse sine sqrt, area using horizontal integration wrt y, area of circle using integration, parabola y2=4ax region, ellipse area pi*ab, sine curve with sign change, with 6 fully verified practice questions.

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The definite integral $\int_{a}^{b} f (x) d x$ computes exactly the signed area between the curve $y = f (x)$ and the $x$ -axis from $x = a$ to $x = b$ . This geometric interpretation transforms integration from an abstract algebraic tool into a powerful technique for computing areas of curved regions — regions that cannot be handled by simple length × width formulas. For JEE, the application of integrals is one of the most visual topics in the syllabus: drawing the correct sketch, identifying the correct formula, finding intersection points, and handling regions below the $x$ -axis are the four skills that separate full marks from partial marks. Every JEE paper (Main and Advanced) includes at least one area question.

1. Area as a Definite Integral — The Core Idea

The area of the region bounded by $y = f (x)$ , the $x$ -axis, and the vertical lines $x = a$ and $x = b$ is:

$Area = \int_{a}^{b} f (x) d x$ (when $f (x) \geq 0$ throughout $[a, b]$ )

When $f (x)$ changes sign in $[a, b]$ , the definite integral gives the net signed area (positive above $x$ -axis, negative below). To find the total area (always positive), use:

$Total Area = \int_{a}^{b} | f (x) | d x$

Key Formula — Three Common Configurations

Region	Formula	Condition
Between $y = f (x)$ and $x$ -axis, $x \in [a, b]$ , $f \geq 0$	$\int_{a}^{b} f (x) d x$	Curve above $x$ -axis
Between $y = f (x)$ and $x$ -axis, $f \leq 0$	$- \int_{a}^{b} f (x) d x = \int_{a}^{b} \| f (x) \| d x$	Curve below $x$ -axis
Between $x = g (y)$ and $y$ -axis, $y \in [c, d]$ , $g \geq 0$	$\int_{c}^{d} g (y) d y$	Curve right of $y$ -axis

2. Standard Curves — Shapes to Know JEE Main & Advanced

Before computing any area, always sketch the curve. These are the standard curves tested in JEE:

Curve	Equation	Shape / Key Features	Area formula
Parabola (upward)	$y = x^{2}$	Vertex at origin, opens upward	$\int_{a}^{b} x^{2} d x = {[\frac{x^{3}}{3}]}_{a}^{b}$
Parabola (right)	$y^{2} = 4 a x$	Vertex at origin, opens right; symmetric about $x$ -axis	$2 \int_{0}^{h} 2 \sqrt{a x} d x$ (for $x \in [0, h]$ )
Circle	$x^{2} + y^{2} = r^{2}$	Centre origin, radius $r$	$π r^{2}$ (standard result); quarter = $\frac{π r^{2}}{4}$
Ellipse	$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$	Semi-axes $a$ (horizontal), $b$ (vertical)	$π a b$ (standard result)
Sine curve	$y = \sin x$	Period $2 π$ ; positive on $(0, π)$ , negative on $(π, 2 π)$	Area $[0, π]$ = 2; total area $[0, 2 π]$ = 4
Square root	$y = \sqrt{x}$	Upper half of parabola $y^{2} = x$ ; passes through $(0, 0), (1, 1)$	$\int_{0}^{a} \sqrt{x} d x = \frac{2}{3} a^{3 / 2}$

3. Area Under Curves — Step-by-Step Method JEE Main & Advanced

Method — 4 Steps

Sketch the curve and shade the required region.
Identify limits of integration from the problem ( $x$ limits or $y$ limits).
Check sign: Is $f (x) \geq 0$ throughout? If not, split at zeros.
Integrate and evaluate.

Worked Example 1 — Area under a parabola

Find the area bounded by $y = x^{2} + 1$ , the $x$ -axis, $x = 0$ and $x = 2$ .

Since $y = x^{2} + 1 \geq 1 > 0$ throughout $[0, 2]$ , the curve is always above the $x$ -axis.

$A = \int_{0}^{2} (x^{2} + 1) d x = {[\frac{x^{3}}{3} + x]}_{0}^{2} = \frac{8}{3} + 2 = \frac{14}{3}$

Worked Example 2 — Area of circle using integration

Find the area of the circle $x^{2} + y^{2} = 16$ .

Upper semicircle: $y = \sqrt{16 - x^{2}}$ , $x \in [- 4, 4]$ .

$A = 2 \int_{- 4}^{4} \sqrt{16 - x^{2}} d x$

Using the standard result $\int_{- r}^{r} \sqrt{r^{2} - x^{2}} d x = \frac{π r^{2}}{2}$ :

$A = 2 \cdot \frac{π \cdot 16}{2} = 16 π$

Worked Example 3 — Area with sign change

Find the total area enclosed between $y = \sin x$ and the $x$ -axis over $[0, 2 π]$ .

$\sin x \geq 0$ on $[0, π]$ and $\sin x \leq 0$ on $[π, 2 π]$ . Split:

$A = \int_{0}^{π} \sin x d x + \int_{π}^{2 π} (- \sin x) d x$

$= [- \cos x]_{0}^{π} + [\cos x]_{π}^{2 π} = (1 + 1) + (1 + 1) = 2 + 2 = 4$

Note: $\int_{0}^{2 π} \sin x d x = 0$ (signed area), but total area $= 4$ (unsigned). These are different!

Worked Example 4 — Horizontal integration (parabola $y^{2} = 4 x$ )

Find the area bounded by the parabola $y^{2} = 4 x$ and the line $x = 1$ .

The parabola opens right, symmetric about $x$ -axis. At $x = 1$ : $y = \pm 2$ .

$A = 2 \int_{0}^{1} 2 \sqrt{x} d x = 4 {[\frac{x^{3 / 2}}{3 / 2}]}_{0}^{1} = 4 \cdot \frac{2}{3} = \frac{8}{3}$

4. Areas of Standard Regions JEE Main & Advanced

Area using parametric curves

If the curve is given parametrically as $x = f (t)$ , $y = g (t)$ :

$A = \int_{t_{1}}^{t_{2}} y \frac{d x}{d t} d t = \int_{t_{1}}^{t_{2}} g (t) \cdot f^{'} (t) d t$

Substitute limits $t_{1}, t_{2}$ corresponding to $x = a, b$ .

Area of Ellipse via integration

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ : upper half gives $y = \frac{b}{a} \sqrt{a^{2} - x^{2}}$ .

$A = 4 \int_{0}^{a} \frac{b}{a} \sqrt{a^{2} - x^{2}} d x = 4 \cdot \frac{b}{a} \cdot \frac{π a^{2}}{4} = π a b$

Area Problems — Decision Framework

Situation	Action
Curve entirely above $x$ -axis on $[a, b]$	$\int_{a}^{b} f (x) d x$ directly
Curve crosses $x$ -axis in $[a, b]$	Split at zeros; add $\int \| f \| d x$ for each sub-interval
Curve given as $x = g (y)$ (horizontal)	Integrate w.r.t. $y$ : $\int_{c}^{d} g (y) d y$
Circle $x^{2} + y^{2} = r^{2}$ — whole circle area	Use $π r^{2}$ directly (or compute quarter and multiply by 4)
Ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$	$π a b$ directly
Parabola $y^{2} = 4 a x$ , region up to $x = h$	$2 \int_{0}^{h} 2 \sqrt{a x} d x = \frac{8}{3} \sqrt{a} \cdot h^{3 / 2}$ ; use symmetry about $x$ -axis

JEE Power Tips — Avoid These Common Mistakes

Signed area vs total area — the most common JEE trap. $\int_{0}^{2 π} \sin x d x = 0$ (positive and negative regions cancel). The question "find the area bounded by $y = \sin x$ and the $x$ -axis from $0$ to $2 π$ " asks for $4$ (total unsigned area). Always read whether "signed" or "total/enclosed" area is asked.
For $y^{2} = 4 a x$ (parabola opening right): use horizontal symmetry. The region is symmetric about the $x$ -axis. Compute area above $x$ -axis and double it: $A = 2 \int_{0}^{h} y d x = 2 \int_{0}^{h} 2 \sqrt{a x} d x$ .
Area of circle and ellipse: just use the formula. $π r^{2}$ for circle, $π a b$ for ellipse. You don't need to integrate from scratch — these are standard results. Show the formula in your answer.
When finding area by integrating w.r.t. $y$ : the integral is $\int_{c}^{d} x d y$ where $x = g (y)$ . This is better when the curve is more naturally expressed as $x = f (y)$ — e.g., $x = y^{2}$ (parabola opening right), $x = \sqrt{4 - y^{2}}$ (semicircle on right).
Always find intersection points correctly before setting up the area. If limits are not given, you must solve simultaneously. For $y = x^{2}$ and $y = x$ : set equal $x^{2} = x$ , so $x = 0$ or $x = 1$ — these are the limits of integration.

Practice Questions (JEE Level)

Q1: Find the area bounded by $y = x^{2} + 1$ , the x-axis, $x = 0$ , and $x = 2$ .

Answer: $\frac{14}{3}$

Explanation:

The required area $A$ is the definite integral of the curve from $x = 0$ to $x = 2$ :
$A = \int_{0}^{2} (x^{2} + 1) d x$

Evaluate the integral:
$A = {[\frac{x^{3}}{3} + x]}_{0}^{2}$
$A = (\frac{8}{3} + 2) - 0$
$A = \frac{8}{3} + \frac{6}{3} = \frac{14}{3}$

Q2: Find the total area enclosed between $y = \sin x$ and the x-axis over the interval $[0, 2 π]$ .

Answer: 4

Explanation:

Because $\sin x$ dips below the x-axis from $π$ to $2 π$ , a standard definite integral over $[0, 2 π]$ would yield 0. To find the total area, we must split the integral at the zero-crossing ( $x = π$ ) and take the absolute value of the negative region:

$A = \int_{0}^{π} \sin x d x + | \int_{π}^{2 π} \sin x d x |$

$A = [- \cos x]_{0}^{π} + | [- \cos x]_{π}^{2 π} |$
$A = (- \cos π + \cos 0) + | (- \cos 2 π + \cos π) |$
$A = (1 + 1) + | (- 1 - 1) |$
$A = 2 + | - 2 | = 2 + 2 = 4$

Q3: Find the area of the circle $x^{2} + y^{2} = 16$ .

Answer: $16 π$

Explanation:

This is the equation of a circle centered at the origin with a radius $r = 4$ . Using the standard geometric formula: $A = π r^{2} = π (4)^{2} = 16 π$ .

Via Integration:
Using the symmetry of the circle, find the area in the first quadrant and multiply by 4:
$A = 4 \int_{0}^{4} \sqrt{16 - x^{2}} d x$

Using the standard integral formula for $\sqrt{a^{2} - x^{2}}$ :
$A = 4 {[\frac{x}{2} \sqrt{16 - x^{2}} + \frac{16}{2} \arcsin (\frac{x}{4})]}_{0}^{4}$
$A = 4 [0 + 8 \arcsin (1) - (0 + 0)]$
$A = 4 [8 (\frac{π}{2})] = 16 π$

Q4: Find the area bounded by the parabola $y^{2} = 4 x$ and the line $x = 1$ .

Answer: $\frac{8}{3}$

Explanation:

The parabola $y^{2} = 4 x ⟹ y = \pm 2 \sqrt{x}$ is symmetric about the x-axis. To find the total enclosed area, we calculate the area in the first quadrant (from $x = 0$ to $x = 1$ ) and multiply by 2.

$A = 2 \int_{0}^{1} 2 \sqrt{x} d x$
$A = 4 \int_{0}^{1} x^{1 / 2} d x$
$A = 4 {[\frac{x^{3 / 2}}{3 / 2}]}_{0}^{1}$
$A = 4 \times \frac{2}{3} (1 - 0) = \frac{8}{3}$

Q5 (JEE-type): Find the area enclosed by the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ .

Answer: $6 π$

Explanation:

The standard formula for the area of an ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ is $A = π a b$ . Here, $a^{2} = 9 ⟹ a = 3$ , and $b^{2} = 4 ⟹ b = 2$ .
Area = $π (3) (2) = 6 π$ .

Via Integration:
$y = \pm \frac{2}{3} \sqrt{9 - x^{2}}$ . By symmetry, calculate the first quadrant area and multiply by 4:
$A = 4 \int_{0}^{3} \frac{2}{3} \sqrt{9 - x^{2}} d x$
$A = \frac{8}{3} {[\frac{x}{2} \sqrt{9 - x^{2}} + \frac{9}{2} \arcsin (\frac{x}{3})]}_{0}^{3}$
$A = \frac{8}{3} [0 + \frac{9}{2} \arcsin (1)] = \frac{8}{3} (\frac{9}{2} \cdot \frac{π}{2}) = 6 π$

Q6 (MCQ): The area bounded by $y = | x |$ , $x = - 2$ , $x = 2$ and the x-axis is:

A) 0
B) 2
C) 4
D) 8

Answer: C) 4.

Explanation:

The function $y = | x |$ is an even function (symmetric about the y-axis). Therefore, the area from $x = - 2$ to $2$ is twice the area from $x = 0$ to $2$ .

$A = \int_{- 2}^{2} | x | d x = 2 \int_{0}^{2} x d x$
$A = 2 {[\frac{x^{2}}{2}]}_{0}^{2}$
$A = 2 \times \frac{4}{2} = 4$

(Geometrically, this forms two identical right-angled triangles with base 2 and height 2. Area = $2 \times (\frac{1}{2} \times 2 \times 2) = 4$ .)

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Area under Curves — Standard Curves & Bounded Regions

1. Area as a Definite Integral — The Core Idea

Key Formula — Three Common Configurations

2. Standard Curves — Shapes to Know JEE Main & Advanced

3. Area Under Curves — Step-by-Step Method JEE Main & Advanced

Method — 4 Steps

Worked Example 1 — Area under a parabola

Worked Example 2 — Area of circle using integration

Worked Example 3 — Area with sign change

Worked Example 4 — Horizontal integration (parabola $y^{2} = 4 x$ )

4. Areas of Standard Regions JEE Main & Advanced

Area using parametric curves

Area of Ellipse via integration

Area Problems — Decision Framework

Practice Questions (JEE Level)

Related Study Material

Practice Questions

Topic Information

Quick Actions

Track Your Learning

Study Tips

Edvaya Target

We Value Your Privacy

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Edvaya Aspire

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Target Subjects

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Subject Mastery

Area under Curves — Standard Curves & Bounded Regions

1. Area as a Definite Integral — The Core Idea

Key Formula — Three Common Configurations

2. Standard Curves — Shapes to Know JEE Main & Advanced

3. Area Under Curves — Step-by-Step Method JEE Main & Advanced

Method — 4 Steps

Worked Example 1 — Area under a parabola

Worked Example 2 — Area of circle using integration

Worked Example 3 — Area with sign change

Worked Example 4 — Horizontal integration (parabola y2=4x)

4. Areas of Standard Regions JEE Main & Advanced

Area using parametric curves

Area of Ellipse via integration

Area Problems — Decision Framework

Practice Questions (JEE Level)

Related Study Material

Practice Questions

Topic Information

Quick Actions

Track Your Learning

Study Tips

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We Value Your Privacy

Worked Example 4 — Horizontal integration (parabola $y^{2} = 4 x$ )