The area of the region enclosed by the parabola (y - 2)2 = x - 1 , the line x

Year	2026	2025	2024	Total
Questions	18	26	18	62

Q12 jee_main_2026_22_january_evening Area Under Curves

The area of the region

A = (x,y) : 4x² + y² \leq 8 and y² \leq 4x

is:

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

Solution

Related Formula

Area enclosed between curves is computed by breaking into integration regions at intersection points.

Core Logic

Area bounded by ellipse and parabola for Q12 - JEE Main 2026 Evening

Find intersection of $4x² + y² = 8$ $4x^2 + y^2 = 8$ and $y² = 4x$ $y^2 = 4x$:

4x² + 4x - 8 = 0 x² + x - 2 = 0 x = 1 (x > 0)

For $x in [0, 1]$ $x \in [0, 1]$, region bounded by parabola $y = \pm 2\sqrt(x)$ $y = \pm 2\sqrt{x}$. For $x in [1, \sqrt(2)]$ $x \in [1, \sqrt{2}]$, region bounded by ellipse $y = \pm \sqrt(8 - 4x²)$ $y = \pm \sqrt{8 - 4x^2}$.

Step 1: Definite Integration

Area = 2 \int₀¹ 2\sqrt(x) dx + 2 \int₁\sqrt(2) \sqrt(8 - 4x²) dx

= 4 [ (2)/(3) x3/2 ]₀¹ + 4 \int₁\sqrt(2) \sqrt(2 - x²) dx

= (8)/(3) + 4 \cdot (1)/(2) [ x\sqrt(2-x²) + 2 ⁻¹( x\sqrt(2)) ]₁\sqrt(2)

= (8)/(3) + 2 [ (0 + 2 \cdot (π)/(2)) - (1 + 2 \cdot (π)/(4)) ] = (8)/(3) + 2π - 2 - π = π + (2)/(3)

Area $= π + (2)/(3)$ $= \pi + \frac{2}{3}$ sq. units.

Pattern Recognition

Split area integration at intersection $x=1$ $x=1$ between parabola boundary and ellipse boundary.

Chapter Mix

Class 12 Maths: Integral Calculus

Q22 jee_main_2026_22_january_evening Definite Integral and King's Property

Let

[\cdot]

be the greatest integer function. If

α = \int₀⁶⁴ (x1/3 - [x1/3]) dx

, then

(1)/(π) \int₀απ ( ( ²θ)/( ⁶θ + ⁶θ) ) dθ

is equal to ____.

Numerical Answer. Answer: 36 to 36

Solution

Related Formula

King's property of definite integrals: $\intₐ^b f(x)dx = \intₐ^b f(a+b-x)dx$ $\int_a^b f(x)dx = \int_a^b f(a+b-x)dx$.

Core Logic

Evaluate $α$ $\alpha$:

\int₀⁶⁴ x1/3 dx = [ (3)/(4) x4/3 ]₀⁶⁴ = (3)/(4) (256) = 192

\int₀⁶⁴ [x1/3] dx = \int₀¹ 0 dx + \int₁⁸ 1 dx + \int₈²⁷ 2 dx + \int₂₇⁶⁴ 3 dx

= 0 + 7(1) + 19(2) + 37(3) = 7 + 38 + 111 = 156

Thus, $α = 192 - 156 = 36$ $\alpha = 192 - 156 = 36$.

Step 1: Evaluate Trigonometric Integral

Let $E = (1)/(π) \int₀36π ( ²θ)/( ⁶θ + ⁶θ) dθ = (36)/(π) \int₀π ( ²θ)/( ⁶θ + ⁶θ) dθ$

= (72)/(π) \int₀π/2 ( ²θ)/( ⁶θ + ⁶θ) dθ

Let$ Let $J = \int_{0}^{\pi/2} \frac{\sin^2\theta}{\sin^6\theta + \cos^6\theta} d\theta. By King's property,$. By King's property, $J = \int_{0}^{\pi/2} \frac{\cos^2\theta}{\sin^6\theta + \cos^6\theta} d\theta.

2J = \int₀π/2 (1)/( ⁶θ + ⁶θ) dθ = \int₀\infty (1 + λ²)/(λ⁴ - λ² + 1) dλ = (π)/(2) J = (π)/(4)

Wait, checking addition:$ Wait, checking addition: $2J = \frac{\pi}{2} \implies J = \frac{\pi}{2}when evaluating$ when evaluating $\int_0^{\infty} \frac{1 + 1/\lambda^2}{\lambda^2 - 1 + 1/\lambda^2} d\lambda = \pi. So$. So $J = \frac{\pi}{2}. Step 2: Final Calculation

E = (72)/(π) \times (π)/(2) = 36

$

Pattern Recognition

Break fractional part integral into piecewise constant steps; convert periodic trigonometric integral using symmetry.

Chapter Mix

Class 12 Maths: Integral Calculus

Q2 jee_main_2026_23_january_morning Indefinite Integration

Let

f(x) = \int (2 - x²)e^x(\sqrt(1 + x))(1 - x)(3)/(2) dx

. If

f(0) = 0

, then

f((1)/(2))

is equal to:

A. $\sqrt(3e) - 1$ $\sqrt{3e} - 1$
B. $\sqrt(2e) + 1$ $\sqrt{2e} + 1$
C. $\sqrt(2e) - 1$ $\sqrt{2e} - 1$
D. $\sqrt(3e) + 1$ $\sqrt{3e} + 1$

Solution

Related Formula

\int e^x [g(x) + g'(x)] dx = e^x g(x) + C

Core Logic

Rewrite the numerator $(2 - x²)$ $(2 - x^2)$ as $(1 - x²) + 1$ $(1 - x^2) + 1$ to split the integral into two recognizable parts matching the $e^x [g(x) + g'(x)]$ $e^x [g(x) + g'(x)]$ form.

I = \int e^x ( (1 - x²) + 1\sqrt(1 + x) \cdot (1 - x)3/2 ) dx

= \int e^x ( 1 - x²\sqrt(1 + x) \cdot (1 - x)3/2 + 1\sqrt(1 + x) \cdot (1 - x)3/2 ) dx

Step 1: Simplify Terms

Simplify the first term:

1 - x²\sqrt(1 + x) \cdot (1 - x)3/2 = (1 - x)(1 + x)\sqrt(1 + x) \cdot (1 - x)\sqrt(1 - x) = \sqrt(1 + x)\sqrt(1 - x) = \sqrt((1 + x)/(1 - x))

Now verify if the derivative of $g(x) = \sqrt((1 + x)/(1 - x))$ $g(x) = \sqrt{\frac{1 + x}{1 - x}}$ matches the second term.

g'(x) = 12\sqrt((1+x)/(1-x)) \cdot ((1-x)(1) - (1+x)(-1))/((1-x)²) = \sqrt(1-x)2\sqrt(1+x) \cdot (2)/((1-x)²) = 1\sqrt(1+x) \cdot (1-x)3/2

This perfectly matches the second term. Thus, the integral evaluates to:

f(x) = e^x \sqrt((1 + x)/(1 - x)) + C

Step 2: Apply Boundary Conditions

Given $f(0) = 0$ $f(0) = 0$:

e⁰ \sqrt((1 + 0)/(1 - 0)) + C = 0 \Rightarrow 1 + C = 0 \Rightarrow C = -1

So, $f(x) = e^x \sqrt((1 + x)/(1 - x)) - 1$ .

Step 3: Evaluate at x = 1/2

f((1)/(2)) = e1/2 \sqrt((1 + 1/2)/(1 - 1/2)) - 1 = e1/2 \sqrt((3/2)/(1/2)) - 1

$= \sqrt(3e) - 1$ $= \sqrt{3e} - 1$

Pattern Recognition

Any integral involving $e^x$ $e^x$ multiplied by an algebraic fraction heavily signals the use of $\int e^x [g(x) + g'(x)] dx$ $\int e^x [g(x) + g'(x)] dx$. Breaking the numerator into $(1-x²) + 1$ $(1-x^2) + 1$ is the classic key to unlock this.

Chapter Mix

Class 12 Maths: Integrals

Q10 jee_main_2026_23_january_morning Definite Integration Properties

The value of the integral

\int (π)/(24)(5π)/(24) dx1+3\sqrt( 2x)

is:

A. $(π)/(12)$ $\frac{\pi}{12}$
B. $(π)/(18)$ $\frac{\pi}{18}$
C. $(π)/(6)$ $\frac{\pi}{6}$
D. $(π)/(3)$ $\frac{\pi}{3}$

Solution

Related Formula

\intₐ^b f(x) dx = \intₐ^b f(a + b - x) dx (King's Property)

Core Logic

Let $I = \int(π)/(24)(5π)/(24) dx1+3\sqrt( 2x) (1)$ Apply King's property. Here $a + b = (π)/(24) + (5π)/(24) = (6π)/(24) = (π)/(4)$ . Replace $x$ $x$ with $(π)/(4) - x$ $\frac{\pi}{4} - x$:

I = \int(π)/(24)(5π)/(24) dx1+3\sqrt( 2((π)/(4) - x))

Step 1: Apply Trigonometric Identity

Notice that $2((π)/(4) - x) = ((π)/(2) - 2x) = 2x$ . So, $I = \int(π)/(24)(5π)/(24) dx1+3\sqrt( 2x)$ . Convert $2x$ $\cot 2x$ to $(1)/( 2x)$ $\frac{1}{\tan 2x}$:

I = \int(π)/(24)(5π)/(24) 3\sqrt( 2x)3\sqrt( 2x)+1 dx (2)

Step 2: Add the Integrals

Add equations (1) and (2):

2I = \int(π)/(24)(5π)/(24) 1 + 3\sqrt( 2x)1 + 3\sqrt( 2x) dx = \int(π)/(24)(5π)/(24) (1) dx

2I = [ x ](π)/(24)(5π)/(24) = (5π)/(24) - (π)/(24) = (4π)/(24) = (π)/(6)

I = (1)/(2) ( (π)/(6) ) = (π)/(12)

Pattern Recognition

Limits of $π/24$ $\pi/24$ and $5π/24$ $5\pi/24$ add up to $π/4$ $\pi/4$. When dealing with $(2x)$ $\tan(2x)$, multiplying by $2$ $2$ transforms this upper/lower sum to $π/2$ $\pi/2$, establishing the classic $/$ $\tan/\cot$ symmetry map triggered by King's Property.

Chapter Mix

Class 12 Maths: Integrals

Q14 jee_main_2026_23_january_evening Integration by Substitution

Let

I(x) = \int 3dx(4x + 6)( 4x² + 8x + 3)

and

I(0) = \sqrt(3)4 + 20

. If

I((1)/(2)) = a\sqrt(2)b + c

, where

a, b, c in N

,

(a,b) = 1

, then

a + b + c

is equal to :

A. $29$ $29$
B. $28$ $28$
C. $31$ $31$
D. $30$ $30$

Solution

Related Formula

\int xⁿ dx = xⁿ⁺¹n+1 + c

For integrals containing linear terms multiplied by the root of a quadratic, use the standard substitution $L = (1)/(t)$ $L = \frac{1}{t}$.

Core Logic

Let $4x + 6 = (1)/(t) x = (1/t - 6)/(4)$ . Differentiating, $4 dx = -(1)/(t²) dt dx = -(dt)/(4t²)$ . We also need to map the quadratic part $4x² + 8x + 3 = 4(x² + 2x + 1) - 1 = 4(x+1)² - 1$ .

Substitute $x+1 = (1/t - 6)/(4) + 1 = (1/t - 2)/(4)$ :

4(x+1)² - 1 = 4((1/t - 2)/(4))² - 1 = ((1-2t)²)/(4t²) - 1

The integral becomes:

I(x) = \int 3 (-(dt)/(4t²))((1)/(t))\sqrt(((1-2t)²)/(4t²) - 1)

Step 1: Simplify Integral

I(x) = -(3)/(4) \int dtt \sqrt((1 - 4t + 4t² - 4t²)/(4t²))

I(x) = -(3)/(4) \int dtt \sqrt(1-4t)2t

I(x) = -(3)/(2) \int dt\sqrt(1-4t)

I(x) = -(3)/(2) (1-4t)1/2(1/2)(-4) + C = (3)/(4) \sqrt(1-4t) + C

Substituting back $t = (1)/(4x+6)$ $t = \frac{1}{4x+6}$:

I(x) = (3)/(4) \sqrt(1 - (4)/(4x+6)) + C = (3)/(4) \sqrt((4x+2)/(4x+6)) + C

Step 2: Final Calculation

Given $I(0) = \sqrt(3)4 + 20$ $I(0) = \frac{\sqrt{3}}{4} + 20$:

I(0) = (3)/(4)\sqrt((2)/(6)) + C = (3)/(4) 1\sqrt(3) + C = \sqrt(3)4 + C

Thus, $C = 20$ $C = 20$.

Now, evaluate $I(1/2)$ $I(1/2)$:

I((1)/(2)) = (3)/(4)\sqrt((4(1/2)+2)/(4(1/2)+6)) + 20 = (3)/(4)\sqrt((4)/(8)) + 20 = (3)/(4) 1\sqrt(2) + 20 = 3\sqrt(2)8 + 20

Comparing with $a\sqrt(2)b + c$ $\frac{a\sqrt{2}}{b} + c$, we get $a=3, b=8, c=20$ $a=3, b=8, c=20$. $(3,8) = 1$ $\gcd(3,8) = 1$, condition is met.

a + b + c = 3 + 8 + 20 = 31

Pattern Recognition

The integral format $\int dx(ax+b)\sqrt(px²+qx+r)$ classically demands substituting the linear outer piece as $ax+b = 1/t$ $ax+b = 1/t$.

Chapter Mix

Class 12 Maths: Indefinite Integration

The area of the region enclosed by the parabola (y - 2)² = x - 1$(y - 2)^2 = x - 1$ , the line x - 2y + 4 = 0$x - 2y + 4 = 0$ and the positive coordinate axes is

Solution & Explanation

Related Formula

Core Logic

Step 1: Setting up the Integral w.r.t y

Step 2: Evaluating the Integral

Pattern Recognition

Chapter Mix

Reference Study Guides

More Integral Calculus Previous-Year Questions — Page 2

Solution

Related Formula

Core Logic

Step 1: Definite Integration

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Evaluate Trigonometric Integral

Step 2: Final Calculation

Step 2: Final Calculation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Simplify Terms

Step 2: Apply Boundary Conditions

Step 3: Evaluate at x = 1/2

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Apply Trigonometric Identity

Step 2: Add the Integrals

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Simplify Integral

Step 2: Final Calculation

Pattern Recognition

Chapter Mix

More Integral Calculus Questions — jee_main_2024_30_january_evening

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 enclosed by the parabola $(y - 2)² = x - 1$ $(y - 2)^2 = x - 1$ , the line $x - 2y + 4 = 0$ $x - 2y + 4 = 0$ and the positive coordinate axes is