The area of the region, inside the ellipse x2 + 4y2 = 4 and outside the region bounded by the curves y = |x|

More Area Under Curves Previous-Year Questions

Q11 jee_main_2026_21_jan_evening Area Under Curve

(x, y) : 1 - 2x \leq y \leq 4 - x², x \geq 0, y \geq 0

(α)/(β), α, β, in N, (α, β) = 1

, then the value of

(α + β)

is :

A. $73$ $73$
B. $85$ $85$
C. $91$ $91$
D. $67$ $67$

Solution

Related Formula

Area = \intx₁x₂ (f(x) - g(x)) dx

Area of Parabola piece: \int₀² (4-x²) dx

Core Logic

Area under curve diagram for Q11 - JEE Main 2026 Evening

The region is bounded above by

y = 4 - x²

, below by

y = 1 - 2x

, and constrained to

x \geq 0, y \geq 0

. The parabola intersects the x-axis at

x=2

(since

4-x²=0, x\geq 0

). The line intersects the x-axis at

x=(1)/(2)

(since

1-2x=0

) and y-axis at

y=1

. The required area is the area under the parabola in the first quadrant minus the small triangular region bounded by

y = 1-2x, x=0, y=0

Step 1: Calculate the Area

Total area under parabola in 1st quadrant = \int₀² (4 - x²) dx = [ 4x - (x³)/(3) ]₀² = 8 - (8)/(3) = (16)/(3)

Area of the small right triangle formed by the line $y=1-2x$ $y=1-2x$ in the first quadrant: Vertices are $(0,0), (1/2,0), (0,1)$ $(0,0), (1/2,0), (0,1)$.

Area of triangle = (1)/(2) \times base \times height = (1)/(2) \times (1)/(2) \times 1 = (1)/(4)

Step 2: Subtraction and Format Match

Required Area = (16)/(3) - (1)/(4) = (64 - 3)/(12) = (61)/(12)

Here, $α = 61, β = 12$ $\alpha = 61, \beta = 12$. Check $(61, 12) = 1$ $\gcd(61, 12) = 1$. This matches. So, $α + β = 61 + 12 = 73$ $\alpha + \beta = 61 + 12 = 73$.

Pattern Recognition

For areas defined by $y \geq g(x)$ $y \geq g(x)$ when $g(x)$ $g(x)$ forms a simple geometric shape (like a line), subtract the geometric area directly rather than splitting the integral algebraically. It eliminates integration errors.

Chapter Mix

Class 12 Maths: Application of Integrals

Q3 jee_main_2026_22_january_morning Area Between Two Curves

Let the line

x = -1

divide the area of the region

(x,y):1+x²\leq y\leq3-x

in the ratio

m:n

, gcd

(m,n)=1

. Then

m+n

is equal to

A. $25$ $25$
B. $28$ $28$
C. $26$ $26$
D. $27$ $27$

Solution

Related Formula

Area = \intₐb (yupper - ylower) dx

Core Logic

First, find the points of intersection for the curves $y = 1 + x²$ $y = 1 + x^2$ and $y = 3 - x$ $y = 3 - x$:

$1 + x² = 3 - x$ $1 + x^2 = 3 - x$

$x² + x - 2 = 0$ $x^2 + x - 2 = 0$

(x + 2)(x - 1) = 0 x = -2, x = 1

So the total region is bounded between $x = -2$ $x = -2$ and $x = 1$ $x = 1$. The line $x = -1$ $x = -1$ divides this region into two parts.

Area Between Two Curves diagram for Q3 - JEE Main 2026 Morning

Step 1: Setting up the Areas

Let the area to the left of $x = -1$ $x = -1$ be proportional to $m$ $m$ and the area to the right be proportional to $n$ $n$.

Am = \int₋₂⁻¹ [(3 - x) - (1 + x²)] dx

Aₙ = \int₋₁¹ [(3 - x) - (1 + x²)] dx

The integrand simplifies to $2 - x - x²$ $2 - x - x^2$.

Step 2: Integration

\int (2 - x - x²) dx = 2x - (x²)/(2) - (x³)/(3)

Evaluate $Am$ $A_m$:

Am = [2x - (x²)/(2) - (x³)/(3)]₋₂⁻¹

Am = (-2 - (1)/(2) + (1)/(3)) - (-4 - 2 + (8)/(3)) = (-(13)/(6)) - (-(10)/(3)) = (7)/(6)

Evaluate $Aₙ$ $A_n$:

Aₙ = [2x - (x²)/(2) - (x³)/(3)]₋₁¹

Aₙ = (2 - (1)/(2) - (1)/(3)) - (-2 - (1)/(2) + (1)/(3)) = ((7)/(6)) - (-(13)/(6)) = (20)/(6)

Step 3: Finding the Ratio

The ratio of the areas $m:n$ $m:n$ is:

(m)/(n) = (Aₙ)/(Am) or (Am)/(Aₙ)

Wait, the solution designates $(m)/(n) = \int₋₁¹\int₋₂⁻¹ = (20/6)/(7/6) = (20)/(7)$ . (Since $(20,7)=1$ $\gcd(20,7)=1$, $m=20$ $m=20$ and $n=7$ $n=7$).

Therefore, $m+n = 20 + 7 = 27$ $m+n = 20 + 7 = 27$.

Pattern Recognition

When a vertical line divides an area into a ratio, calculate the definite integral on both sides of the splitting line independently. Keep fractions with a common denominator until the final ratio step.

Chapter Mix

Class 12 Maths: Applications of Integrals

Q22 jee_main_2026_23_january_morning Area Under the Curve

Let the area of the region bounded by the curve

y = x, x

, lines

x = 0, x = (3π)/(2)

, and the

x

-axis be

A

. Then,

A + A²

is equal to _____.

Numerical Answer. Answer: 12 to 12

Solution

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

Q12 jee_main_2026_23_january_evening Area Between Two Curves

The area of the region enclosed between the circles

x² + y² = 4

and

x² + (y - 2)² = 4

is:

A. $(2)/(3)(2π-3\sqrt(3))$ $\frac{2}{3}(2\pi-3\sqrt{3})$
B. $(4)/(3)(2π-3\sqrt(3))$ $\frac{4}{3}(2\pi-3\sqrt{3})$
C. $(4)/(3)(2π-\sqrt(3))$ $\frac{4}{3}(2\pi-\sqrt{3})$
D. $(2)/(3)(4π-3\sqrt(3))$ $\frac{2}{3}(4\pi-3\sqrt{3})$

Solution

Related Formula

\int \sqrt(a² - x²) dx = (x)/(2)\sqrt(a² - x²) + (a²)/(2) ⁻¹((x)/(a))

Core Logic

Area Between Two Curves diagram for Q12 - JEE Main 2026 Evening

The intersection points of

x² + y² = 4

and

x² + (y-2)² = 4

: Subtracting the equations gives

y² - (y-2)² = 0 y² - (y² - 4y + 4) = 0 4y = 4 y = 1

. Substitute

y=1

back:

x² + 1 = 4 x = \pm \sqrt(3)

. The points of intersection are

(\sqrt(3), 1)

and

(-\sqrt(3), 1)

The area is symmetric about the y-axis, so we integrate from $x = 0$ $x = 0$ to $x = \sqrt(3)$ $x = \sqrt{3}$ and double the result. Upper curve is the lower arc of the top circle: $y = 2 - \sqrt(4-x²)$ $y = 2 - \sqrt{4-x^2}$ -- wait, the area is bounded by the top arc of the bottom circle and the bottom arc of the top circle. Upper curve for area: $y = \sqrt(4-x²)$ $y = \sqrt{4-x^2}$ (from $x²+y²=4$ $x^2+y^2=4$) Lower curve for area: $y = 2 - \sqrt(4-x²)$ $y = 2 - \sqrt{4-x^2}$ (from $x²+(y-2)²=4$ $x^2+(y-2)^2=4$)

A = 2\int₀\sqrt(3) [\sqrt(4-x²) - (2 - \sqrt(4-x²))] dx

Step 1: Integration

A = 2\int₀\sqrt(3) (2\sqrt(4-x²) - 2) dx = 4\int₀\sqrt(3) (\sqrt(4-x²) - 1) dx

A = 4[ (1)/(2)(x\sqrt(4-x²) + 4 ⁻¹(x)/(2)) - x ]₀\sqrt(3)

Evaluating the limits:

= 4[ (1)/(2)(\sqrt(3)(1) + 4 ⁻¹( \sqrt(3)2)) - \sqrt(3) - (0) ]

= 4[ (1)/(2)(\sqrt(3) + 4((π)/(3))) - \sqrt(3) ]

= 4[ \sqrt(3)2 + (2π)/(3) - \sqrt(3) ]

= 4[ (2π)/(3) - \sqrt(3)2 ]

= (8π)/(3) - 2\sqrt(3) = (2)/(3)(4π - 3\sqrt(3)) (Sq. units)

Pattern Recognition

For intersecting identical circles with centres on an axis, symmetry simplifies the integration drastically. Recognize that integrating the circle arc function handles the bulk of the calculation.

Chapter Mix

Class 12 Maths: Area Under Curves

The area of the region, inside the ellipse x² + 4y² = 4$x^{2} + 4y^{2} = 4$ and outside the region bounded by the curves y = |x| - 1$y = |x| - 1$ and y = 1 - |x|$y = 1 - |x|$ , is:

Solution & Explanation

Related Formula

Core Logic

Step 1: Calculate Total and Excluded Areas

Step 2: Calculate Required Area

Pattern Recognition

Chapter Mix

Reference Study Guides

More Area Under Curves Previous-Year Questions

Solution

Related Formula

Core Logic

Step 1: Calculate the Area

Step 2: Subtraction and Format Match

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Setting up the Areas

Step 2: Integration

Step 3: Finding the Ratio

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Formulate the Area Integral

Step 2: Evaluate Integrals

Step 3: Total Area

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Integration

Pattern Recognition

Chapter Mix

More Area Under Curves Questions — jee_main_2026_21_jan_morning

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Welcome back

The area of the region, inside the ellipse $x² + 4y² = 4$ $x^{2} + 4y^{2} = 4$ and outside the region bounded by the curves $y = |x| - 1$ $y = |x| - 1$ and $y = 1 - |x|$ $y = 1 - |x|$ , is: