Solution & Explanation

### Related Formula Area of an ellipse

fracx^2a^2 + fracy^2b^2 = 1

is given by:

textArea = pi a b

Area of a rhombus bounded by

|x| + |y| = a

2a^2

. ### Core Logic The given curves form a bounded geometric area. Ellipse:

x^2 + 4y^2 = 4 Rightarrow fracx^24 + fracy^21 = 1

. Here,

a = 2

b = 1

. The region to be excluded is bounded by

y = |x| - 1

and

y = 1 - |x|

, which rearranges to

|x| + |y| = 1

. This forms a square/rhombus centered at the origin with vertices at

(1, 0), (0, 1), (-1, 0), (0, -1)

. ### Step 1: Calculate Total and Excluded Areas Total Area of the Ellipse:

textArea = pi (2)(1) = 2pi

Excluded Area (Rhombus

|x| + |y| = 1

): The rhombus consists of 4 identical right-angled triangles in each quadrant. Area of one triangle =

frac12 times textbase times textheight = frac12 times 1 times 1 = frac12

. Total excluded area =

4 times frac12 = 2

. ### Step 2: Calculate Required Area

Ellipse and Modulus area diagram for Q2 - JEE Main 2026 Morning

Required Area = Area of ellipse

-

Shaded Area

= 2pi - 2

= 2(pi - 1)

### Pattern Recognition Transform absolute value equations

y = pm(|x| - a)

into

|x| + |y| = a

to instantly recognize a standard rhombus, allowing direct geometry formulas instead of integration. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Area Under Curves Class 11 Maths: Conic Sections

More Area Under Curves Previous-Year Questions — Page 3

Q59 jee_main_2025_28_jan_evening Area Bounded by Algebraic and Rational Curves

The area of the region bounded by the curves

x(1+y^2)=1

and

y^2=2x

is :

A. $2left(fracpi2-frac13right)$
B. $fracpi4-frac13$ $\frac{\pi}{4}-\frac{1}{3}$
C. $fracpi2-frac13$ $\frac{\pi}{2}-\frac{1}{3}$
D. $frac12left(fracpi2-frac13right)$

Solution

### Related Formula Area integrating with respect to

y

A = int_y_1^y_2 (x_textright - x_textleft) dy

Standard integral:

int frac11+y^2 dy = tan^-1(y)

### Core Logic The boundary curves are: 1)

x = frac11+y^2

x = fracy^22

Find intersection points by setting

x

equal:

frac11+y^2 = fracy^22 implies 2 = y^2(1+y^2) implies y^4 + y^2 - 2 = 0

(y^2 + 2)(y^2 - 1) = 0

Since

y

is real,

y^2 = 1 implies y = pm 1

. When

y = pm 1

x = frac12

. Intersection points are

left(frac12, 1right)

and

left(frac12, -1right)

. ### Step 1: Set up and Compute Area Integral Between

y = -1

and

y = 1

frac11+y^2 ge fracy^22

textArea = int-1^1 left( frac11+y^2 - fracy^22 right) dy

Since the integrand is an even function of

y

textArea = 2 int_0^1 left( frac11+y^2 - fracy^22 right) dy

textArea = 2 left[ tan^-1(y) - fracy^36 right]_0^1

textArea = 2 left[ tan^-1(1) - frac16 - (0) right] = 2 left( fracpi4 - frac16 right) = fracpi2 - frac13

### Pattern Recognition Whenever curves are functions of

y^2

, integrating along the

y

-axis avoids dealing with messy radical functions (square roots) and naturally accounts for symmetry across the

x

-axis. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Application of Integrals

Q57 jee_main_2025_29_jan_morning Area Under Curves

Let the area of the region

\(x, y) : 2y leq x^2 + 3

y + |x| leq 3

y geq |x - 1|\

be A. Then 6A is equal to:

A. 16
B. 12
C. 18
D. 14

Solution

### Related Formula

textArea = int_a^b (y_textupper - y_textlower) \, dx

### Core Logic Plotting the boundary lines and tracking intersection points yields a composite geometric region bounding a central area.

Area Under Curves diagram for Q57 - JEE Main 2025 Morning

### Step 1: Set up the integral pieces The bounded region

A

can be conceptualized as a total bounding box/rectangle minus specific external integrals:

A = 4 - 2 int_0^1 left[ (3 - x) - left( fracx^2 + 32 right) right] \, dx

### Step 2: Evaluate the Integral

A = 4 - 2 left[ 3x - fracx^22 - fracx^36 - frac32x right]_0^1

A = 4 - 2 left[ 3 - frac12 - frac16 - frac32 right] = 4 - 2 left[ frac56 right] = 4 - frac53 = frac73

### Step 3: Calculate 6A

6A = 6 times frac73 = 14

### Pattern Recognition When dealing with multiple absolute functions (

|x|

|x-1|

), check for coordinate mirror symmetry across vertical axes to slash total required calculus computations in half. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Application of Integrals

Q10 jee_main_2024_01_february_morning Area Between Curves

The area enclosed by the curves

xy+4y=16

and

x+y=6

is equal to:

A. $28-30 log_e2$ $28-30 \log_{e}2$
B. $30-28 log_e2$ $30-28 \log_{e}2$
C. $30-32 log_e2$ $30-32 \log_{e}2$
D. $32-30 log_e2$ $32-30 \log_{e}2$

Solution

### Related Formula Area enclosed between two intersecting curves

y_1 = f(x)

and

y_2 = g(x)

from boundary limits

x = a

x = b

textArea = int_a^b (y_textupper - y_textlower) \, dx

### Core Logic Given the two boundary curves: 1.

xy + 4y = 16 implies y(x+4) = 16 implies y = frac16x+4

x + y = 6 implies y = 6 - x

Find the intersection points by equating the two expressions for

y

frac16x+4 = 6 - x implies 16 = (6-x)(x+4)

16 = 6x + 24 - x^2 - 4x implies x^2 - 2x - 8 = 0

(x-4)(x+2) = 0 implies x = 4, \, x = -2

### Step 1: Integral Formulation Between

x = -2

and

x = 4

, the line

y = 6 - x

lies above the curve

y = frac16x+4

. Therefore, the required enclosed area is:

textArea = int_-2^4 left( (6-x) - frac16x+4 right) dx

textArea = left[ 6x - fracx^22 - 16 ln|x+4| right]_-2^4

Area Between Curves diagram for Q10 - JEE Main 2024 01 February Morning — The diagram displays the shaded region enclosed between the straight line and the hyperbola between limits minus two and four.

### Step 2: Apply Limits and Simplify Substitute the upper limit

x = 4

U = 6(4) - frac4^22 - 16 ln|4+4| = 24 - 8 - 16 ln 8 = 16 - 16 ln 8

Substitute the lower limit

x = -2

L = 6(-2) - frac(-2)^22 - 16 ln|-2+4| = -12 - 2 - 16 ln 2 = -14 - 16 ln 2

Subtract the lower limit value from the upper limit value:

textArea = U - L = (16 - 16 ln 8) - (-14 - 16 ln 2)

textArea = 30 - 16 ln(2^3) + 16 ln 2 = 30 - 48 ln 2 + 16 ln 2

textArea = 30 - 32 ln 2 = 30 - 32 log_e2

### Pattern Recognition Sees: Area bounded by a straight line and a shifting rectangular hyperbola. Shortcut: The roots of the difference equation

x^2 - 2x - 8 = 0

directly give the limits. Always check graph orientations to place the upper linear equation before the curve equation inside the integral bracket to preserve absolute area values. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Application of Integrals Class 11 Mathematics: Conic Sections (Hyperbola)

Q25 jee_main_2024_29_january_evening Area Under Curves

Let the area of the region

\(x,y):0leq xleq 3,0leq yleq min \x^2 +2,2x + 2\ \

A

. Then

12A

is equal to

Numerical Answer. Answer: 164 to 164

Solution

### Related Formula

textArea A = int min\f(x), g(x)\\, dx

### Core Logic Let us find the point of intersection of the curves

y = x^2 + 2

and

y = 2x + 2

x^2 + 2 = 2x + 2 implies x^2 - 2x = 0 implies x = 0 text or x = 2

Evaluating behaviors over boundaries: * For

0 leq x leq 2

x^2 + 2 leq 2x + 2 implies min = x^2 + 2

* For

2 leq x leq 3

2x + 2 leq x^2 + 2 implies min = 2x + 2

### Step 1: Integration Resolution Setting up continuous area integral steps:

A = int_0^2 (x^2 + 2)\, dx + int_2^3 (2x + 2)\, dx

A = left[ fracx^33 + 2x right]_0^2 + left[ x^2 + 2x right]_2^3

A = left( frac83 + 4 right) + left( (9 + 6) - (4 + 4) right)

A = frac203 + (15 - 8) = frac203 + 7 = frac413

Area Under Curves diagram for Q25 - JEE Main 2024 Evening

### Step 2: Scaling the Output Value We need to compute

12A

12A = 12 times frac413 = 4 times 41 = 164

### Pattern Recognition For min/max boundary sets, always compute intersections first to accurately split integration domains into separate regions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Application of Integrals

Q24 jee_main_2024_27_jan_morning Area Under Curve

Let the area of the region

\(x, y): x^2y + 4 ge 0, x+2y^2ge0, x+4y^2le8, yge0\

fracmn

where

m

and

n

are coprime numbers. Then

m+n

is equal to:

Numerical Answer. Answer: 119 to 119

Solution

### Related Formula

textArea = int_y_1^y_2 (x_textright - x_textleft) dy

### Core Logic We need to find the area bounded by the curves in the first quadrant (since

y ge 0

). Note that the first inequality

x^2y + 4 ge 0

is trivially satisfied for all

x, y ge 0

. So we focus on bounding

x

using: Right curve:

x = 8 - 4y^2

Left curve:

x = -2y^2

However, there might be constraints where these cross or hit the axes. We must inspect intersections. ### Step 1: Finding Intersection Points Where do the left and right parabolas intersect?

8 - 4y^2 = -2y^2 Rightarrow 2y^2 = 8 Rightarrow y^2 = 4 Rightarrow y = 2

(Since

y ge 0

). Wait, does

x

have boundaries? Since

x

can't drop arbitrarily into negative territory if it's restricted by other axes. Let's check

x^2y + 4 ge 0

. If

x = -2y^2

, then

y

must be bounded, but wait - the question limits are actually split into two regions depending on

x^2y+4 ge 0

? Wait, the first inequality

x^2y+4 ge 0

might not be trivial if

x

is negative. If

x = -2y^2

, then

(-2y^2)^2 y + 4 ge 0 Rightarrow 4y^5 + 4 ge 0

, which is true for all

y ge 0

. Wait, there seems to be a misinterpretation of the first inequality. Let's look closer at the PDF solution boundaries. The integration is broken at

y=1

. Why?

x+2y^2 ge 0 Rightarrow x ge -2y^2

. Wait, the solution states another left curve:

2y-4

. Is

x^2y+4 ge 0

actually

x + 2y - 4 ge 0

? Yes, OCR shows `x^2y+4` but the solution integrates `(2y-4)`. Thus the original condition is likely

x - 2y + 4 ge 0 Rightarrow x ge 2y - 4

! Let's assume the left boundary splits between

x = -2y^2

and

x = 2y - 4

. ### Step 2: Region Bounds Intersection of

x = -2y^2

and

x = 2y - 4

-2y^2 = 2y - 4 Rightarrow y^2 + y - 2 = 0 Rightarrow (y+2)(y-1) = 0 Rightarrow y = 1

. Intersection of

x = 2y - 4

and

x = 8 - 4y^2

2y - 4 = 8 - 4y^2 Rightarrow 4y^2 + 2y - 12 = 0 Rightarrow 2y^2 + y - 6 = 0 Rightarrow (2y-3)(y+2) = 0 Rightarrow y = 3/2

. So the region shifts left boundary at

y=1

and closes entirely at

y=3/2

. ### Step 3: Setting up the Integration Region 1 (from

y=0

y=1

A_1 = int_0^1 ((8 - 4y^2) - (-2y^2)) dy = int_0^1 (8 - 2y^2) dy

A_1 = left[ 8y - frac2y^33 right]_0^1 = 8 - frac23 = frac223

Region 2 (from

y=1

y=3/2

A_2 = int_1^3/2 ((8 - 4y^2) - (2y - 4)) dy = int_1^3/2 (12 - 2y - 4y^2) dy

A_2 = left[ 12y - y^2 - frac4y^33 right]_1^3/2

A_2 = left( 12left(frac32right) - frac94 - frac43left(frac278right) right) - left( 12 - 1 - frac43 right)

A_2 = left( 18 - frac94 - frac92 right) - left( 11 - frac43 right) = left( 18 - frac274 right) - frac293 = frac454 - frac293 = frac135 - 11612 = frac1912

### Step 4: Final Output Total Area

A = A_1 + A_2

A = frac223 + frac1912 = frac88 + 1912 = frac10712

This implies

m = 107

and

n = 12

. Since 107 and 12 are coprime,

m + n = 107 + 12 = 119

. ### Pattern Recognition When dealing with multiple inequalities bounded by

y ge 0

, always project horizontally (integrate wrt

y

) as the bounds natively trace left-to-right distances. Find intersection nodes to partition the integral correctly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Application of Integrals

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

Solution & Explanation

Reference Study Guides

More Area Under Curves Previous-Year Questions — Page 3

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Solution

Solution

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The area of the region, inside the ellipse x^2 + 4y^2 = 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

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