Let the area of the region bounded by the curve y = x, x, lines x = 0, x = (3pi/2), and the x-axis be A. Then, A + A2…

Year	2026	2025	2024	Total
Questions	9	11	7	27

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Core Logic

To evaluate the area under $y = x, x$ $y = \max \{\sin x, \cos x\}$, we must identify where one function is greater than the other in $[0, (3π)/(2)]$ $[0, \frac{3\pi}{2}]$.

Area Under the Curve diagram for Q22 - JEE Main 2026 Morning

From

0

(π)/(4)

x > x \Rightarrow y = x

From

(π)/(4)

(5π)/(4)

x > x \Rightarrow y = x

From

(5π)/(4)

(3π)/(2)

x > x \Rightarrow y = x

Step 1: Formulate the Area Integral

Since area is bounded by the x-axis, we must take the absolute value if the function goes below the axis. Wait, the function $x$ $\sin x$ is negative from $π$ $\pi$ to $(5π)/(4)$ $\frac{5\pi}{4}$, and $x$ $\cos x$ is negative from $(5π)/(4)$ $\frac{5\pi}{4}$ to $(3π)/(2)$ $\frac{3\pi}{2}$. Let's integrate carefully:

A = \int₀π/4 x dx + \intπ/4π x dx + \intπ5π/4 | x| dx + \int5π/43π/2 | x| dx

Because area is geometric, absolute values are explicitly integrated:

A = \int₀π/4 x dx + \intπ/4π x dx + \intπ5π/4 (- x) dx + \int5π/43π/2 (- x) dx

Step 2: Evaluate Integrals

\int₀π/4 x dx = [ x]₀π/4 = 1\sqrt(2) - 0 = 1\sqrt(2)

\intπ/4π x dx = [- x]π/4π = -(-1) - (- 1\sqrt(2)) = 1 + 1\sqrt(2)

\intπ5π/4 (- x) dx = [ x]π5π/4 = - 1\sqrt(2) - (-1) = 1 - 1\sqrt(2)

\int5π/43π/2 (- x) dx = [- x]5π/43π/2 = -(-1) - ( -(- 1\sqrt(2)) ) = 1 - 1\sqrt(2)

Step 3: Total Area

Sum the areas together:

A = 1\sqrt(2) + (1 + 1\sqrt(2)) + (1 - 1\sqrt(2)) + (1 - 1\sqrt(2))

A = 1\sqrt(2) + 1 + 1\sqrt(2) + 1 - 1\sqrt(2) + 1 - 1\sqrt(2) = 3

Now calculate $A² + A$ $A^2 + A$:

A² + A = (3)² + 3 = 9 + 3 = 12

Pattern Recognition

The expression $( x, x)$ $\max(\sin x, \cos x)$ always splits segments exactly at $nπ + π/4$ $n\pi + \pi/4$. When bounded by the x-axis, parts below $y=0$ $y=0$ explicitly require negation. Charting the piecewise transitions is non-negotiable.

Chapter Mix

Class 12 Maths: Application of Integrals

More Application of Integrals Previous-Year Questions — Page 6

Q11 jee_main_2024_31_jan_evening Area Under Curves

The area of the region enclosed by the parabola

y = 4x - x²

and

3y = (x - 4)²

is equal to

A. $(32)/(9)$ $\frac{32}{9}$
B. $4$ $4$
C. $6$ $6$
D. $(14)/(3)$ $\frac{14}{3}$

Solution

Related Formula

Area = \intₐb (yupper - ylower) dx

Core Logic

Area Under Curves diagram for Q11 - JEE Main 2024 Evening

Find intersection points of $y = 4x - x²$ $y = 4x - x^2$ and $3y = (x - 4)²$ $3y = (x - 4)^2$:

3(4x - x²) = x² - 8x + 16

12x - 3x² = x² - 8x + 16

4x² - 20x + 16 = 0 x² - 5x + 4 = 0

Roots are $x = 1, 4$ $x = 1, 4$.

Area integral:

Area = \int₁⁴ [ (4x - x²) - ((x - 4)²)/(3) ] dx

= [ (4x²)/(2) - (x³)/(3) - ((x - 4)³)/(9) ]₁⁴

= [ 2(16) - (64)/(3) - 0 ] - [ 2(1) - (1)/(3) - ((-3)³)/(9) ]

= ( 32 - (64)/(3) ) - ( 2 - (1)/(3) + 3 )

= (32)/(3) - ( 5 - (1)/(3) ) = (32)/(3) - (14)/(3) = (18)/(3) = 6

Chapter Mix

Class 12 Maths: Applications of the Integrals

Q5 jee_main_2024_31_jan_morning Area bounded by Parabolas and Inequalities

The area of the region

(x,y): y² \leq 4x, x < 4, (xy(x - 1)(x - 2))/((x - 3)(x - 4)) > 0, x \neq 3

A. $(16)/(3)$ $$\frac{16}{3}$$
B. $(64)/(3)$ $$\frac{64}{3}$$
C. $(8)/(3)$ $$\frac{8}{3}$$
D. $(32)/(3)$ $$\frac{32}{3}$$

Solution

Core Logic

Given $y² \leq 4x$ $y^2 \le 4x$ and $x < 4$ $x < 4$. Analyze the inequality $(xy(x-1)(x-2))/((x-3)(x-4)) > 0$ $\frac{xy(x-1)(x-2)}{(x-3)(x-4)} > 0$ considering $y > 0$ $y > 0$ and $y < 0$ $y < 0$ separately.

Area bounded by Parabolas and Inequalities diagram for Q5 - JEE Main 2024 Morning

Step 1: Case I (y > 0)

If $y > 0$ $y > 0$, the inequality reduces to $(x(x-1)(x-2))/((x-3)(x-4)) > 0$ $\frac{x(x-1)(x-2)}{(x-3)(x-4)} > 0$. Using wavy curve method and given $x in (0, 4)$ $x \in (0, 4)$: $x in (0, 1) (2, 3)$ $x \in (0, 1) \cup (2, 3)$.

Step 2: Case II (y < 0)

If $y < 0$ $y < 0$, the inequality reduces to $(x(x-1)(x-2))/((x-3)(x-4)) < 0$ $\frac{x(x-1)(x-2)}{(x-3)(x-4)} < 0$. Using wavy curve method and given $x in (0, 4)$ $x \in (0, 4)$: $x in (1, 2) (3, 4)$ $x \in (1, 2) \cup (3, 4)$.

Step 3: Area Computation

Because the regions map perfectly without overlap in opposite quadrants relative to the x-axis, they form complete parabolic strips when combined: Area $= 2 \int₀⁴ \sqrt(x) dx = 2 \cdot (2)/(3)[x3/2]₀⁴ = (4)/(3) \cdot 8 = (32)/(3)$ .

Chapter Mix

Class 12 Maths: Area Under Curves Class 11 Maths: Linear Inequalities

Let the area of the region bounded by the curve $y = x, x$ $y = \max \{\sin x, \cos x\}$, lines $x = 0, x = (3π)/(2)$ $x = 0, x = \frac{3\pi}{2}$, and the $x$ $x$-axis be $A$ $A$. Then, $A + A²$ $A + A^2$ is equal to _____.

Solution & Explanation

Core Logic

Step 1: Formulate the Area Integral

Step 2: Evaluate Integrals

Step 3: Total Area

Pattern Recognition

Chapter Mix

Reference Study Guides

More Application of Integrals Previous-Year Questions — Page 6

Solution

Related Formula

Core Logic

Chapter Mix

Solution

Core Logic

Step 1: Case I (y > 0)

Step 2: Case II (y < 0)

Step 3: Area Computation

Chapter Mix

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Let the area of the region bounded by the curve y = x, x$y = \max \{\sin x, \cos x\}$, lines x = 0, x = (3π)/(2)$x = 0, x = \frac{3\pi}{2}$, and the x$x$-axis be A$A$. Then, A + A²$A + A^2$ is equal to _____.

Solution & Explanation

Core Logic

Step 1: Formulate the Area Integral

Step 2: Evaluate Integrals

Step 3: Total Area

Pattern Recognition

Chapter Mix

Reference Study Guides

More Application of Integrals Previous-Year Questions — Page 6

Solution

Related Formula

Core Logic

Chapter Mix

Solution

Core Logic

Step 1: Case I (y > 0)

Step 2: Case II (y < 0)

Step 3: Area Computation

Chapter Mix

More Application of Integrals Questions — jee_main_2026_23_january_morning

Select Your Target Exam

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Exam Questions

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Let the area of the region bounded by the curve $y = x, x$ $y = \max \{\sin x, \cos x\}$, lines $x = 0, x = (3π)/(2)$ $x = 0, x = \frac{3\pi}{2}$, and the $x$ $x$-axis be $A$ $A$. Then, $A + A²$ $A + A^2$ is equal to _____.