The area of the region bounded by the curves x(1+y2)=1 and y2=2x is :

Solution & Explanation

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

More Area Under Curves 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

The area of the region bounded by the curves $x(1+y²)=1$ $x(1+y^{2})=1$ and $y²=2x$ $y^{2}=2x$ is :

Solution & Explanation

Related Formula

Core Logic

Step 1: Set up and Compute Area Integral

Pattern Recognition

Chapter Mix

More Area Under Curves 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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The area of the region bounded by the curves x(1+y²)=1$x(1+y^{2})=1$ and y²=2x$y^{2}=2x$ is :

Solution & Explanation

Related Formula

Core Logic

Step 1: Set up and Compute Area Integral

Pattern Recognition

Chapter Mix

More Area Under Curves 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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The area of the region bounded by the curves $x(1+y²)=1$ $x(1+y^{2})=1$ and $y²=2x$ $y^{2}=2x$ is :