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

Q71 jee_main_2025_02_april_morning Definite Integration with GIF

Let

[\cdot]

denote the greatest integer function. If

\int₀^e³[ 1ex - 1]dx = α - ₑ2

, then

α³

is equal to ________.

Numerical Answer. Answer: 8 to 8

Solution

Related Formula

Greatest Integer Function boundaries:

[f(x)] = k for k \leq f(x) < k+1, k in Z

Core Logic

Analyze the value variations of $f(x) = e1-x$ $f(x) = e^{1-x}$ across the integration limits $[0, e³]$ $[0, e^3]$ to break down the integral into distinct piecewise continuous intervals.

Step 1: Determine Step Function Transition Points

Let $y = e1-x$ $y = e^{1-x}$.

At $x = 0 y = e¹ \approx 2.718$ $x = 0 \implies y = e^1 \approx 2.718$
As $x$ $x$ increases, $e1-x$ $e^{1-x}$ decreases monotonically.
Find $x$ $x$ where $y = 2 e1-x = 2 1-x = ln 2 x = 1 - ln 2$ .
Find $x$ $x$ where $y = 1 e1-x = 1 1-x = 0 x = 1$ .
At the final boundary $x = e³ y = e1-e³$ $x = e^3 \implies y = e^{1-e^3}$, which is a very small positive decimal strictly inside $(0,1)$ $(0,1)$.

Step 2: Split the Definite Integral

Rewrite the integral based on the isolated interval blocks:

I = \int₀1-ln 2 2 dx + \int1-ln 2¹ 1 dx + \int₁e³ 0 dx

Step 3: Perform Integrations

I = 2[x]₀1-ln 2 + 1[x]1-ln 2¹ + 0

I = 2(1 - ln 2 - 0) + 1(1 - (1 - ln 2)) = 2 - 2ln 2 + ln 2 = 2 - ln 2

Step 4: Solve for Alpha Cubed

Compare the integrated value to $α - ln 2$ $\alpha - \ln 2$:

α - ln 2 = 2 - ln 2 α = 2

α³ = 2³ = 8

Pattern Recognition

Always map the function values at the extreme boundary points first. Tracking the downward path from $2.71 arrow 2 arrow 1 arrow 0$ reveals exactly where the integer thresholds are crossed.

Chapter Mix

Class 12 Mathematics: Integrals

Q59 jee_main_2025_03_april_evening Definite Integrals

The integral

\int₀^π (8x dx)/(4 ² x + ² x)

is equal to

A. $2π²$ $2\pi^2$
B. $4π²$ $4\pi^2$
C. $π²$ $\pi^2$
D. $(3π²)/(2)$ $\frac{3\pi^2}{2}$

Solution

Related Formula

Using the properties of definite integrals:

\intₐ^b f(x) dx = \intₐ^b f(a+b-x) dx

Also, if $f(2a-x) = f(x)$ $f(2a-x) = f(x)$, then:

\int₀2a f(x) dx = 2 \int₀^a f(x) dx

Core Logic

Let:

I = \int₀^π (8x dx)/(4 ² x + ² x) --- (1)

Applying $x \to π-x$ $x \to \pi-x$:

I = \int₀^π (8(π - x) dx)/(4 ² x + ² x) --- (2)

Step 1: Eliminating the $x$ $x$ term in numerator

Adding (1) and (2):

2I = 8π \int₀^π (dx)/(4 ² x + ² x)

I = 4π \int₀^π (dx)/(4 ² x + ² x)

Since the integrand is symmetric about $x = π/2$ $x = \pi/2$:

I = 8π \int₀π/2 (dx)/(4 ² x + ² x)

Step 2: Integration using $² x$ $\sec^2 x$ substitution

Divide numerator and denominator by $² x$ $\cos^2 x$:

I = 8π \int₀π/2 ( ² x dx)/(4 + ² x)

Let $t = x dt = ² x dx$ At $x = 0$ $x = 0$, $t = 0$ $t = 0$; at $x = π/2$ $x = \pi/2$, $t \to \infty$ $t \to \infty$.

I = 8π \int₀^\infty (dt)/(t² + 2²)

I = 8π [ (1)/(2) ⁻¹((t)/(2)) ]₀^\infty = 4π ( (pi)/(2) - 0 ) = 2π²

Pattern Recognition

The presence of a linear $x$ $x$ factor in the numerator of a definite integral with symmetric trigonometric bounds is almost always eliminated using the $a+b-x$ $a+b-x$ property. This reduces the integral to a standard substitution problem.

Chapter Mix

Class 12 Mathematics: Integrals

Q60 jee_main_2025_07_april_morning Properties of Definite Integrals

The integral

\int₀^π ((x + 3) x)/(1 + 3 ²x) dx

is equal to:

A. $π\sqrt(3) (π + 1)$ $\frac{\pi}{\sqrt{3}} (\pi + 1)$
B. $π\sqrt(3) (π + 2)$ $\frac{\pi}{\sqrt{3}} (\pi + 2)$
C. $π3\sqrt(3) (π + 6)$ $\frac{\pi}{3\sqrt{3}} (\pi + 6)$
D. $π2\sqrt(3) (π + 4)$ $\frac{\pi}{2\sqrt{3}} (\pi + 4)$

Solution

Related Formula

King's Property of Definite Integrals:

\intₐ^b f(x) dx = \intₐ^b f(a + b - x) dx

Core Logic

Let the given integral be:

I = \int₀^π ((x + 3) x)/(1 + 3 ²x) dx (1)

Applying King's property ( $x \to π - x$ $x \to \pi - x$):

I = \int₀^π ((π - x + 3) (π - x))/(1 + 3 ²(π - x)) dx

I = \int₀^π ((π - x + 3) x)/(1 + 3 ²x) dx (2)

Step 1: Eliminate the x Variable

Adding equations (1) and (2):

2I = \int₀^π ([(x + 3) + (π - x + 3)] x)/(1 + 3 ²x) dx

2I = (π + 6)\int₀^π ( x)/(1 + 3 ²x) dx

Using the symmetric property $\int₀2a f(x)dx = 2\int₀^a f(x)dx$ $\int_0^{2a} f(x)dx = 2\int_0^a f(x)dx$ if $f(2a-x)=f(x)$ $f(2a-x)=f(x)$:

2I = 2(π + 6)\int₀π/2 ( x)/(1 + 3 ²x) dx

I = (π + 6)\int₀π/2 ( x)/(1 + 3 ²x) dx

Step 2: Solve Using Substitution

Let $t = \sqrt(3) x$ $t = \sqrt{3}\cos x$. Then $dt = -\sqrt(3) x dx x dx = - dt\sqrt(3)$ . Change in integration boundaries:

When $x = 0 t = \sqrt(3)$ $x = 0 \implies t = \sqrt{3}$
When $x = π/2 t = 0$ $x = \pi/2 \implies t = 0$

Substituting into the integral:

I = (π + 6) \int\sqrt(3)⁰ -dt/\sqrt(3)1 + t² = π + 6\sqrt(3) \int₀\sqrt(3) (dt)/(1 + t²)

I = π + 6\sqrt(3) [ ⁻¹t ]₀\sqrt(3) = π + 6\sqrt(3) ( ⁻¹\sqrt(3) - 0 )

I = π + 6\sqrt(3) \cdot (π)/(3) = π3\sqrt(3)(π + 6)

Pattern Recognition

Whenever you encounter a linear $x$ $x$ factor multiplying trigonometric components in a definite integral with symmetric limits like $0$ $0$ to $π$ $\pi$, executing King's property first is almost guaranteed to cleanly wipe out that variable element.

Chapter Mix

Class 12 Mathematics: Definite Integrals

Q58 jee_main_2025_08_april_evening Properties of Definite Integrals

Let

f(x)

be a a positive function and

I₁ = \int-(1)/(2)¹ 2xf(2x(1 - 2x)) dx

and

I₂ = \int₋₁² f(x(1 - x)) dx

. Then the value of

(I₂)/(I₁)

is equal to

A. $9$ $9$
B. $6$ $6$
C. $12$ $12$
D. $4$ $4$

Solution

Related Formula

\intₐb f(x) dx = \intₐb f(a+b-x) dx

Core Logic

Perform variable substitution to match the arguments and limit bounds across both separate integral functions before invoking King's property.

Step 1: Perform Base Transformation Substitution

In $I₁$ $I_1$, let $2x = t 2dx = dt$ $2x = t \implies 2dx = dt$. Limits mapping: $x = -1/2 t = -1$ $x = -1/2 \implies t = -1$; $x = 1 t = 2$ $x = 1 \implies t = 2$.

I₁ = (1)/(2) \int₋₁² t f(t(1-t)) dt 2I₁ = \int₋₁² t f(t(1-t)) dt

Step 2: Invoke Integral Mirror Properties

Apply the identity using parameters $(a+b-t) = (1-t)$ $(a+b-t) = (1-t)$:

2I₁ = \int₋₁² (1-t) f((1-t)(1-(1-t))) dt

2I₁ = \int₋₁² f(t(1-t)) dt - \int₋₁² t f(t(1-t)) dt

Step 3: Final Matrix Matching Evaluation

Notice component blocks align exactly with $I₂$ $I_2$ definition values:

2I₁ = I₂ - 2I₁ 4I₁ = I₂

(I₂)/(I₁) = 4

Pattern Recognition

Symmetric transformations highlighting factor expressions like $x(1-x)$ $x(1-x)$ coupled with an external linear multiplier term $x$ $x$ naturally simplify to half-weight area forms using reflection rules.

Chapter Mix

Class 12 Mathematics: Definite Integrals

Q61 jee_main_2025_08_april_evening Integration of Absolute Value Functions

The integral

\int₋₁(3)/(2)(|π²x (π x)|) dx

is equal to :

A. $3 + 2π$ $3 + 2\pi$
B. $4 + π$ $4 + \pi$
C. $1 + 3π$ $1 + 3\pi$
D. $2 + 3π$ $2 + 3\pi$

Solution

Related Formula

\int x (π x) dx = -(x)/(π) (π x) + ( (π x))/(π²)

Core Logic

Track sign configurations across the target integration segments to drop absolute modulus walls effectively at clean quadrant intervals.

Step 1: Break Apart the Modulus Domain

For $x in [-1, 1]$ $x \in [-1, 1]$, product value elements $x (π x) \geq 0$ $x \sin(\pi x) \ge 0$. For $x in [1, 3/2]$ $x \in [1, 3/2]$, values drop below zero:

I = π² \int₋₁¹ x (π x) dx - \int₁3/2 x (π x) dx

Step 2: Integrate the Even Function Block

Since $x (π x)$ $x \sin(\pi x)$ is symmetric and even:

\int₋₁¹ x (π x) dx = 2 \int₀¹ x (π x) dx = 2 [ -(x)/(π) (π x) + ( (π x))/(π²) ]₀¹ = (2)/(π)

Step 3: Subtract the Inverse Segment

Evaluating boundary limits across the secondary phase track:

\int₁3/2 x (π x) dx = [ -(x)/(π) (π x) + ( (π x))/(π²) ]₁3/2 = ( 0 - (1)/(π²) ) - ( (1)/(π) ) = -(1)/(π²) - (1)/(π)

I = π² (2)/(π) - (-(1)/(π²) - (1)/(π)) = π² ( (3)/(π) + (1)/(π²) ) = 3π + 1

Pattern Recognition

Products of two odd tracking metrics (like linear variable $x$ $x$ matched with sinusoidal waves) yield overall even systems, enabling rapid evaluation over center-aligned domains.

Chapter Mix

Class 12 Mathematics: Definite Integrals

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 5

Solution

Related Formula

Core Logic

Step 1: Determine Step Function Transition Points

Step 2: Split the Definite Integral

Step 3: Perform Integrations

Step 4: Solve for Alpha Cubed

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Eliminating the x$x$ term in numerator

Step 2: Integration using ² x$\sec^2 x$ substitution

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Eliminate the x Variable

Step 2: Solve Using Substitution

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Perform Base Transformation Substitution

Step 2: Invoke Integral Mirror Properties

Step 3: Final Matrix Matching Evaluation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Break Apart the Modulus Domain

Step 2: Integrate the Even Function Block

Step 3: Subtract the Inverse Segment

Pattern Recognition

Chapter Mix

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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

Step 1: Eliminating the $x$ $x$ term in numerator

Step 2: Integration using $² x$ $\sec^2 x$ substitution