Let the area of the region (x, y) : 2y leq x2 + 3 , y + |x| leq 3 , y geq |x

Year	2026	2025	2024	Total
Questions	9	11	7	27

Q19 jee_main_2026_24_january_morning Area Under the Curve

Let

A₁

be the bounded area enclosed by the curves

y = x² + 2

,

x + y = 8

and y-axis that lies in the first quadrant. Let

A₂

be the bounded area enclosed by the curves

y = x² + 2

,

y² = x

,

x = 2

, and y-axis that lies in the first quadrant. Then

A₁ - A₂

is equal to

A. $(2)/(3)(2\sqrt(2)+1)$ $\frac{2}{3}(2\sqrt{2}+1)$
B. $(2)/(3)(4\sqrt(2)+1)$ $\frac{2}{3}(4\sqrt{2}+1)$
C. $(2)/(3)(\sqrt(2)+1)$ $\frac{2}{3}(\sqrt{2}+1)$
D. $(2)/(3)(3\sqrt(2)+1)$ $\frac{2}{3}(3\sqrt{2}+1)$

Solution

Related Formula

Area = \intₐb (yupper - ylower) dx

Core Logic

For $A₁$ $A_1$: Intersection of $y = x² + 2$ $y = x^2 + 2$ and $x + y = 8$ $x + y = 8$. $x + (x² + 2) = 8 \Rightarrow x² + x - 6 = 0 \Rightarrow (x+3)(x-2) = 0$ . In first quadrant, $x = 2$ $x = 2$. Intersection point is $(2, 6)$ $(2, 6)$.

Step 1: Calculate A1

A₁ = \int₀² ((8 - x) - (x² + 2)) dx = \int₀² (6 - x - x²) dx

A₁ = [ 6x - (x²)/(2) - (x³)/(3) ]₀² = 12 - 2 - (8)/(3) = 10 - (8)/(3) = (22)/(3)

Step 2: Area A2 Definition

For $A₂$ $A_2$: Enclosed by $y = x² + 2$ $y = x^2 + 2$, $y = \sqrt(x)$ $y = \sqrt{x}$, $x=2$ $x=2$ and y-axis. Upper curve: $y = x²+2$ $y = x^2+2$, Lower curve: $y = \sqrt(x)$ $y = \sqrt{x}$.

Step 3: Calculate A2

A₂ = \int₀² (x² + 2 - \sqrt(x)) dx

A₂ = [ (x³)/(3) + 2x - x3/23/2 ]₀²

A₂ = (8)/(3) + 4 - (2)/(3)(23/2) = (20)/(3) - 4\sqrt(2)3

Step 4: Difference of Areas

A₁ - A₂ = (22)/(3) - ( (20)/(3) - 4\sqrt(2)3 )

= (2)/(3) + 4\sqrt(2)3 = (2)/(3) (1 + 2\sqrt(2))

Pattern Recognition

Carefully identify the bounding regions. $A₁$ $A_1$ uses a line as the upper bound, $A₂$ $A_2$ strips away a root curve as its lower bound. Integrating separately before subtracting avoids algebraic alignment errors.

Chapter Mix

Class 12 Maths: Application of Integrals

Q12 jee_main_2026_24_january_evening Area bounded by Modulus and Parabola

Let

f(α)

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

x = 0

,

x = 1

,

y² = x

and

y = |α x - 5| - |1 - α x| + α x²

. Then

(f(0) + f(1))

is equal to

A. $9$ $9$
B. $14$ $14$
C. $7$ $7$
D. $12$ $12$

Solution

Related Formula

Area = \intx₁x₂ (yupper - ylower) dx

Core Logic

Evaluate $f(0)$ $f(0)$ by setting $α = 0$ $\alpha = 0$ in the function for $y$ $y$.

At $α = 0 y = |0\cdot x - 5| - |1 - 0\cdot x| + 0\cdot x²$

y = |-5| - |1| = 5 - 1 = 4

So for $f(0)$ $f(0)$, the bounded area $A₁$ $A_1$ is between $y = 4$ $y = 4$, $y = \sqrt(x)$ $y = \sqrt{x}$, from $x = 0$ $x = 0$ to $x = 1$ $x = 1$.

Step 1: Calculating f(0)

A₁ = \int₀¹ (4 - \sqrt(x)) dx

= [ 4x - x3/23/2 ]₀¹

= 4(1) - (2)/(3)(1) = (10)/(3)

So, $f(0) = (10)/(3)$ $f(0) = \frac{10}{3}$.

Step 2: Determining curve for f(1)

Evaluate $f(1)$ $f(1)$ by setting $α = 1$ $\alpha = 1$ in the function for $y$ $y$.

At $α = 1 y = |x - 5| - |1 - x| + x²$ . Since the area is restricted to $x in [0, 1]$ $x \in [0, 1]$: $x - 5$ $x - 5$ is negative, so $|x - 5| = 5 - x$ $|x - 5| = 5 - x$. $1 - x$ $1 - x$ is positive, so $|1 - x| = 1 - x$ $|1 - x| = 1 - x$.

y = (5 - x) - (1 - x) + x²

y = 5 - x - 1 + x + x² = 4 + x²

Step 3: Calculating f(1)

For $f(1)$ $f(1)$, the bounded area $A₂$ $A_2$ is between $y = 4 + x²$ $y = 4 + x^2$ and $y = \sqrt(x)$ $y = \sqrt{x}$, from $x = 0$ $x = 0$ to $x = 1$ $x = 1$.

A₂ = \int₀¹ ((4 + x²) - \sqrt(x)) dx

= [ 4x + (x³)/(3) - x3/23/2 ]₀¹

= ( 4 + (1)/(3) - (2)/(3) ) = 4 - (1)/(3) = (11)/(3)

So, $f(1) = (11)/(3)$ $f(1) = \frac{11}{3}$.

Step 4: Final Sum

|f(0) + f(1)| = | A₁ + A₂ | = | (10)/(3) + (11)/(3) |

= (21)/(3) = 7

Pattern Recognition

Modulus functions combined with parametric limits should be instantly resolved by plugging in the bounds of $x$ $x$ (here $[0,1]$ $[0,1]$) to drop the absolute value signs before attempting any integral.

Chapter Mix

Class 12 Maths: Area Under the Curve Class 11 Maths: Modulus Function

Q13 jee_main_2026_28_january_morning Area under Curves

The area of the region

R=(x,y):xy\leq 8, 1\leq y\leq x², x\geq 0

is

A. $(1)/(3)(49 ₑ(2)-15)$
B. $(2)/(3)(20 ₑ(2)+9)$
C. $(2)/(3)(24 ₑ(2)-7)$
D. $(1)/(3)(40 ₑ(2)+27)$

Solution

Core Logic

The region is bounded by three primary curves in the first quadrant (

x\geq0

):

$y \geq 1$ $y \geq 1$ (horizontal line)
$y \leq x²$ $y \leq x^2$ (upward-opening parabola)
$xy \leq 8 y \leq (8)/(x)$ $xy \leq 8 \implies y \leq \frac{8}{x}$ (rectangular hyperbola)

Find points of intersection: Intersection of $y = x²$ $y = x^2$ and $y = (8)/(x)$ $y = \frac{8}{x}$:

x² = (8)/(x) x³ = 8 x = 2

At $x = 2$ $x = 2$, $y = 4$ $y = 4$. So they meet at $(2, 4)$ $(2, 4)$.

Intersection of $y = x²$ $y = x^2$ and $y = 1$ $y = 1$:

x² = 1 x = 1

Intersection of $y = (8)/(x)$ $y = \frac{8}{x}$ and $y = 1$ $y = 1$:

1 = (8)/(x) x = 8

The region spans from $x = 1$ $x = 1$ to $x = 8$ $x = 8$. The upper boundary changes at $x = 2$ $x = 2$.

Step 1: Set up Integrals

For $x in [1, 2]$ $x \in [1, 2]$, the upper curve is $y = x²$ $y = x^2$ and the lower is $y = 1$ $y = 1$. For $x in [2, 8]$ $x \in [2, 8]$, the upper curve is $y = (8)/(x)$ $y = \frac{8}{x}$ and the lower is $y = 1$ $y = 1$.

A = \int₁² (x² - 1) dx + \int₂⁸ ((8)/(x) - 1) dx

Step 2: Integration

Calculate the first integral:

I₁ = [ (x³)/(3) - x ]₁² = ( (8)/(3) - 2 ) - ( (1)/(3) - 1 ) = (2)/(3) - (-(2)/(3)) = (4)/(3)

Calculate the second integral:

I₂ = [ 8 ln x - x ]₂⁸ = (8 ln 8 - 8) - (8 ln 2 - 2)

Since $ln 8 = 3 ln 2$ $\ln 8 = 3 \ln 2$:

I₂ = 24 ln 2 - 8 - 8 ln 2 + 2 = 16 ln 2 - 6

Total Area $A$ $A$:

A = (4)/(3) + 16 ln 2 - 6 = 16 ln 2 - (14)/(3)

A = (48 ln 2 - 14)/(3) = (2)/(3)(24 ln 2 - 7)

Chapter Mix

Class 12 Mathematics: Application of Integrals

Q11 jee_main_2026_28_january_evening Area Bounded by Parabolas and Lines

Let

P₁: y = 4x²

and

P₂: y = x² + 27

be two parabolas. If the area of the bounded region enclosed between

P₁

and

P₂

is six times the area of the bounded region enclosed between the line

y = α x, α > 0

and

P₁

, then

α

is equal to:

A. $8$ $8$
B. $15$ $15$
C. $12$ $12$
D. $6$ $6$

Solution

Related Formula

Area enclosed by $y = f(x)$ $y = f(x)$ and $y = g(x)$ $y = g(x)$ from $x=a$ $x=a$ to $x=b$ $x=b$:

A = \intₐ^b |f(x) - g(x)| dx

Area between $x² = 4ay$ $x^2 = 4ay$ and line $x = my$ $x = my$ is $(8a²)/(3m³)$ $\frac{8a^2}{3m^3}$.

Core Logic

Find Points of Intersection for $P₁$ $P_1$ and $P₂$ $P_2$:

4x² = x² + 27 \Rightarrow 3x² = 27 \Rightarrow x = \pm 3

Area bounded between $P₁$ $P_1$ and $P₂$ $P_2$:

A₁ = \int₋₃³ ((x² + 27) - 4x²) dx = 2 \int₀³ (27 - 3x²) dx

A₁ = 2 [27x - x³]₀³ = 2(81 - 27) = 108 sq. units

Execution

Area between $P₁$ $P_1$ and $y = α x$ $y = \alpha x$ is given as $108 / 6 = 18$ $108 / 6 = 18$ sq. units. For $x² = 4ay$ $x^2 = 4ay$ and $x = my$ $x = my$, area is $(8a²)/(3m³)$ $\frac{8a^2}{3m^3}$. Here, the parabola is $x² = (y)/(4)$ $x^2 = \frac{y}{4}$ and line is $x = (y)/(α)$ $x = \frac{y}{\alpha}$. Thus, $4a = (1)/(4) \Rightarrow a = (1)/(16)$ , and $m = (1)/(α)$ $m = \frac{1}{\alpha}$.

Applying the formula:

(8(1/16)²)/(3(1/α)³) = 18

(8 / 256)/(3 / α³) = 18 \Rightarrow (α³)/(96) = 18

α³ = 18 \times 96 = 1728

$α = 12$ $\alpha = 12$

Pattern Recognition

Using the standard result $(8a²)/(3m³)$ $\frac{8a^2}{3m^3}$ for the area bounded between $x²=4ay$ $x^2=4ay$ and $x=my$ $x=my$ avoids tedious integration for straight lines intersecting parabolas at the origin.

Chapter Mix

Class 12 Maths: Application of Integrals

Q73 jee_main_2025_02_april_morning Area Between Curves

If the area of the region

(x, y): | 4 - x ^ 2 | \leq y \leq x ^ 2, y \leq 4, x \geq 0

is

( 80\sqrt(2)α - β), α, β in N

, then

α + β

is equal to ________.

Numerical Answer. Answer: 22 to 22

Solution

Related Formula

Area bounded by functions integrated with respect to $y$ $y$ axis:

Area = \intcd (xright - xleft) dy

Core Logic

Identify the bounding graphs and intersection coordinates in the first quadrant, then construct standard definite integrals along the vertical axis.

Area Between Curves diagram for Q73 - JEE Main 2025 Morning

Step 1: Unpack Bounding Curves

The condition $|4-x²| \leq y$ $|4-x^2| \le y$ splits into two sections at $x=2$ $x=2$:

For $0 \leq x \leq 2 4 - x² \leq y x² \geq 4 - y x = \sqrt(4-y)$
For $x \geq 2 x² - 4 \leq y x² \leq 4 + y x = \sqrt(4+y)$

Also bounded by $y \leq x² x \geq \sqrt(y)$ $y \le x^2 \implies x \ge \sqrt{y}$, and the outer cap constraint $y \leq 4$ $y \le 4$.

Step 2: Construct the Integral Area Formula

Integrating with respect to $y$ $y$ covers the region bounded on the left by $\sqrt(y)$ $\sqrt{y}$ and $\sqrt(4-y)$ $\sqrt{4-y}$, and on the right by $\sqrt(4+y)$ $\sqrt{4+y}$:

A = \int₀⁴ \sqrt(4+y) dy - \int₀² \sqrt(4-y) dy - \int₂⁴ \sqrt(y) dy

Step 3: Evaluate the Definite Integrals

A = [ (4+y)3/23/2 ]₀⁴ + [ (4-y)3/23/2 ]₀² - [ y3/23/2 ]₂⁴

Evaluating these values precisely:

A = (2)/(3)(83/2 - 43/2) + (2)/(3)(23/2 - 43/2) - (2)/(3)(43/2 - 23/2)

A = 80\sqrt(2)3 - 16

Step 4: Solve for Constants

Compare the final value expression to $( 80\sqrt(2)α - β)$ :

α = 3, β = 16 α + β = 3 + 16 = 22

Pattern Recognition

Integrating along the vertical axis ( $y$ $y$-direction) is significantly faster here because it avoids splitting the domain across multiple vertical segments on the horizontal $x$ $x$-axis.

Chapter Mix

Class 12 Mathematics: Application of Integrals

Let the area of the region (x, y) : 2y ≤ x² + 3$\{(x, y) : 2y \leq x^2 + 3$ , y + |x| ≤ 3$y + |x| \leq 3$ , y ≥ |x - 1|$y \geq |x - 1|\}$ be A. Then 6A is equal to:

Solution & Explanation

Related Formula

Core Logic

Step 1: Set up the integral pieces

Step 2: Evaluate the Integral

Step 3: Calculate 6A

Pattern Recognition

Chapter Mix

Reference Study Guides

More Application of Integrals Previous-Year Questions — Page 2

Solution

Related Formula

Core Logic

Step 1: Calculate A1

Step 2: Area A2 Definition

Step 3: Calculate A2

Step 4: Difference of Areas

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculating f(0)

Step 2: Determining curve for f(1)

Step 3: Calculating f(1)

Step 4: Final Sum

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Set up Integrals

Step 2: Integration

Chapter Mix

Solution

Related Formula

Core Logic

Execution

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Unpack Bounding Curves

Step 2: Construct the Integral Area Formula

Step 3: Evaluate the Definite Integrals

Step 4: Solve for Constants

Pattern Recognition

Chapter Mix

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Let the area of the region $(x, y) : 2y \leq x² + 3$ $\{(x, y) : 2y \leq x^2 + 3$ , $y + |x| \leq 3$ $y + |x| \leq 3$ , $y \geq |x - 1|$ $y \geq |x - 1|\}$ be A. Then 6A is equal to: