Let the line x = -1 divide the area of the region (x,y):1+x2leq yleq3-x in the ratio m:n, gcd (m,n)=1. Then m+n is…

Q71 jee_main_2025_04_april_morning Area Bounded by Multiple Curves

If the area of the region

(x,y) : |x - 5| \leq y \leq 4\sqrt(x)

is A, then 3A is equal to

Numerical Answer. Answer: 368 to 368

Solution

Related Formula

Area tracking equation via horizontal slices or split verticals:

Area = \intx₁x₂ (yupper - ylower) dx

Core Logic

Find the intersections of $y = |x - 5|$ $y = |x - 5|$ and $y = 4\sqrt(x)$ $y = 4\sqrt{x}$: Branch 1: $5 - x = 4\sqrt(x) x + 4\sqrt(x) - 5 = 0 (\sqrt(x) + 5)(\sqrt(x) - 1) = 0 x = 1, y = 4$ . Branch 2: $x - 5 = 4\sqrt(x) x - 4\sqrt(x) - 5 = 0 (\sqrt(x) - 5)(\sqrt(x) + 1) = 0 x = 25, y = 20$ .

Area Bounded by Multiple Curves diagram for Q71 - JEE Main 2025 Morning

Step 1: Set Up Area Definite Integral

Split integration intervals around vertex $x = 5$ $x = 5$ or compute via simple boundary differences:

A = \int₁²⁵ 4\sqrt(x) dx - Area of Left Triangle - Area of Right Triangle

Area of Left Triangle = (1)/(2) \times (5 - 1) \times 4 = 8

Area of Right Triangle = (1)/(2) \times (25 - 5) \times 20 = 200

Step 2: Complete Computations

\int₁²⁵ 4\sqrt(x) dx = [ (8)/(3)x3/2 ]₁²⁵ = (8)/(3)(125 - 1) = (8 \times 124)/(3) = (992)/(3)

A = (992)/(3) - 208 = (992 - 624)/(3) = (368)/(3)

$3A = 368$ $3A = 368$

Pattern Recognition

Subtracting standard geometric triangles beneath linear configurations from total absolute root curves saves significant time compared to managing multiple separate analytical integral pieces.

Chapter Mix

Class 12 Mathematics: Area Under Curves

Q67 jee_main_2025_24_jan_evening Area Under Curves

The area of the region enclosed by the curves

y=ex,

y=|ex-1|

and y-axis is:

A. $1+ ₑ2$ $1+\log_{e}2$
B. $ₑ2$ $\log_{e}2$
C. $2 ₑ2-1$ $2\log_{e}2-1$
D. $1- ₑ2$ $1-\log_{e}2$

Solution

Related Formula

The area between two curves $y₁(x)$ $y_1(x)$ and $y₂(x)$ $y_2(x)$ from $x = a$ $x = a$ to $x = b$ $x = b$ is given by:

Area = \intₐb |y₁(x) - y₂(x)| dx

Core Logic

Analyze the curves to locate intersection points :

Curve 1: $y = e^x$ $y = e^x$
Curve 2: $y = |e^x - 1|$ $y = |e^x - 1|$
Boundary: y-axis ( $x = 0$ $x = 0$)

For $x < 0$ $x < 0$, $e^x < 1 \Rightarrow |e^x - 1| = 1 - e^x$ .

Find intersection: $e^x = 1 - e^x \Rightarrow 2e^x = 1 \Rightarrow e^x = (1)/(2) \Rightarrow x = -ln 2$ .

Area bounded by exponential curves diagram for Q67 - JEE Main 2025 Evening

Step 1: Set up the Definite Integral

The integration spans from $x = -ln 2$ $x = -\ln 2$ to $x = 0$ $x = 0$. In this interval, $e^x \geq 1 - e^x$ $e^x \ge 1 - e^x$:

Area = \int-ln 2⁰ [ e^x - (1 - e^x) ] dx

Area = \int-ln 2⁰ (2e^x - 1) dx

Step 2: Integration Evaluation

Integrate term-by-term :

Area = [ 2e^x - x ]-ln 2⁰

= (2e⁰ - 0) - (2e-ln 2 - (-ln 2))

= 2 - (2((1)/(2)) + ln 2) = 2 - (1 + ln 2) = 1 - ln 2

Pattern Recognition

Modulus graphs split at their critical point ( $e^x = 1 \Rightarrow x = 0$ $e^x = 1 \Rightarrow x = 0$). Recognizing that the required segment lies strictly in the negative quadrant simplifies the definition of the upper and lower functions immediately.

Chapter Mix

Class 12 Mathematics: Application of Integrals

Q63 jee_main_2025_24_jan_morning Area Bounded by Multiple Curves

The area of the region

(x,y) x² + 4x + 2 \leq y \leq |x + 2|

is equal to :

A. $7$ $7$
B. $24/5$ $24/5$
C. $20/3$ $20/3$
D. $5$ $5$

Solution

Related Formula

The net area between two intersecting functions $y₂(x)$ $y_2(x)$ and $y₁(x)$ $y_1(x)$ from lower limit $a$ $a$ to upper limit $b$ $b$ is evaluated using the definite integral:

Area = \intₐb [y₂(x) - y₁(x)] dx

Core Logic

Let's perform a horizontal shift transformation by defining $X = x + 2$ $X = x + 2$ to simplify the expressions:

Parabola: $y = x² + 4x + 2 = (x+2)² - 2 y = X² - 2$
Absolute curve: $y = |x + 2| y = |X|$ $y = |x + 2| \implies y = |X|$

This coordinate translation preserves area completely while shifting the axis to the origin.

Step 1: Compute Points of Intersection

Find where the transformed curves intersect by equating the functions: $|X|² - 2 = |X|$ $|X|^2 - 2 = |X|$

|X|² - |X| - 2 = 0

(|X| - 2)(|X| + 1) = 0

Since $|X| \geq 0$ $|X| \geq 0$, we discard $|X| = -1$ $|X| = -1$. This gives $|X| = 2 X = \pm 2$ $|X| = 2 \implies X = \pm 2$.

Step 2: Set up and Evaluate the Transformed Integral

Since both curves are symmetric about the vertical line $X = 0$ $X = 0$, we can compute the area for the positive half and double it:

Area = 2 \int₀² [ X - (X² - 2) ] dX = 2 \int₀² ( 2 + X - X² ) dX

Integrate the polynomial row-by-row:

= 2 [ 2X + (X²)/(2) - (X³)/(3) ]₀²

= 2 [ 2(2) + (4)/(2) - (8)/(3) ] = 2 [ 4 + 2 - (8)/(3) ]

= 2 [ 6 - (8)/(3) ] = 2 \cdot (10)/(3) = (20)/(3)

Pattern Recognition

Applying a horizontal variable substitution matching $X = x + \pm c$ $X = x + ± c$ shifts complex quadratic forms to centered formats, which avoids tedious arithmetic across asymmetrical integration limits.

Chapter Mix

Class 12 Mathematics: Application of Integrals

Q jee_main_2025_28_jan_evening Area Bounded by Algebraic and Rational Curves

The area of the region bounded by the curves

x(1+y²)=1

and

y²=2x

is :

A. $2((π)/(2)-(1)/(3))$
B. $(π)/(4)-(1)/(3)$ $\frac{\pi}{4}-\frac{1}{3}$
C. $(π)/(2)-(1)/(3)$ $\frac{\pi}{2}-\frac{1}{3}$
D. $(1)/(2)((π)/(2)-(1)/(3))$

Solution

Related Formula

Area integrating with respect to $y$ $y$:

A = \inty₁y₂ (xright - xleft) dy

Standard integral:

\int (1)/(1+y²) dy = ⁻¹(y)

Core Logic

The boundary curves are:

$x = (1)/(1+y²)$ $x = \frac{1}{1+y^2}$
$x = (y²)/(2)$ $x = \frac{y^2}{2}$

Find intersection points by setting $x$ $x$ equal:

(1)/(1+y²) = (y²)/(2) 2 = y²(1+y²) y⁴ + y² - 2 = 0

(y² + 2)(y² - 1) = 0

Since $y$ $y$ is real, $y² = 1 y = \pm 1$ $y^2 = 1 \implies y = \pm 1$. When $y = \pm 1$ $y = \pm 1$, $x = (1)/(2)$ $x = \frac{1}{2}$. Intersection points are $((1)/(2), 1)$ $\left(\frac{1}{2}, 1\right)$ and $((1)/(2), -1)$ $\left(\frac{1}{2}, -1\right)$.

Step 1: Set up and Compute Area Integral

Between $y = -1$ $y = -1$ and $y = 1$ $y = 1$, $(1)/(1+y²) \geq (y²)/(2)$ $\frac{1}{1+y^2} \ge \frac{y^2}{2}$.

Area = \int₋₁¹ ( (1)/(1+y²) - (y²)/(2) ) dy

Since the integrand is an even function of $y$ $y$:

Area = 2 \int₀¹ ( (1)/(1+y²) - (y²)/(2) ) dy

Area = 2 [ ⁻¹(y) - (y³)/(6) ]₀¹

Area = 2 [ ⁻¹(1) - (1)/(6) - (0) ] = 2 ( (π)/(4) - (1)/(6) ) = (π)/(2) - (1)/(3)

Pattern Recognition

Whenever curves are functions of $y²$ $y^2$, integrating along the $y$ $y$-axis avoids dealing with messy radical functions (square roots) and naturally accounts for symmetry across the $x$ $x$-axis.

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 x² + 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

Area = \intₐb (yupper - ylower) 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 $A$ $A$ can be conceptualized as a total bounding box/rectangle minus specific external integrals:

A = 4 - 2 \int₀¹ [ (3 - x) - ( (x² + 3)/(2) ) ] dx

Step 2: Evaluate the Integral

A = 4 - 2 [ 3x - (x²)/(2) - (x³)/(6) - (3)/(2)x ]₀¹

A = 4 - 2 [ 3 - (1)/(2) - (1)/(6) - (3)/(2) ] = 4 - 2 [ (5)/(6) ] = 4 - (5)/(3) = (7)/(3)

Step 3: Calculate 6A

$6A = 6 \times (7)/(3) = 14$ $6A = 6 \times \frac{7}{3} = 14$

Pattern Recognition

When dealing with multiple absolute functions ( $|x|$ $|x|$, $|x-1|$ $|x-1|$), check for coordinate mirror symmetry across vertical axes to slash total required calculus computations in half.

Chapter Mix

Class 12 Mathematics: Application of Integrals

Let the line x = -1$x = -1$ divide the area of the region (x,y):1+x²≤ y≤3-x$\left\{(x,y):1+x^{2}\leq y\leq3-x\right\}$ in the ratio m:n$m:n$, gcd (m,n)=1$(m,n)=1$. Then m+n$m+n$ is equal to

Solution & Explanation

Related Formula

Core Logic

Step 1: Setting up the Areas

Step 2: Integration

Step 3: Finding the Ratio

Pattern Recognition

Chapter Mix

Reference Study Guides

More Applications of Integrals Previous-Year Questions — Page 4

Solution

Related Formula

Core Logic

Step 1: Set Up Area Definite Integral

Step 2: Complete Computations

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Set up the Definite Integral

Step 2: Integration Evaluation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Compute Points of Intersection

Step 2: Set up and Evaluate the Transformed Integral

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Set up and Compute Area Integral

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Set up the integral pieces

Step 2: Evaluate the Integral

Step 3: Calculate 6A

Pattern Recognition

Chapter Mix

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Let the line $x = -1$ $x = -1$ divide the area of the region $(x,y):1+x²\leq y\leq3-x$ in the ratio $m:n$ $m:n$, gcd $(m,n)=1$ $(m,n)=1$. Then $m+n$ $m+n$ is equal to