Let f(x) be a a positive function and I1 = -(1/2)1 2xf(2x(1 - 2x)) , dx and I2 = -12 f(x(1

More Definite Integrals Previous-Year Questions — Page 5

Q28 jee_main_2024_01_february_morning Properties of Definite Integrals

\int-π/2π/2 8\sqrt(2) x dx(1+ex)(1+ ⁴x)=απ+β ₑ(3+2\sqrt(2)),

where

α, β

are integers, then

α²+β²

equals

Numerical Answer. Answer: 8 to 8

Solution

Related Formula

King's Property for symmetric integration limits:

\int₋ₐa f(x) dx = \int₋ₐa f(-x) dx

Core Logic

Let the given definite integral be $I$ $I$:

I = \int-π/2π/2 8\sqrt(2) x dx(1+ex)(1+ ⁴x) (1)

Apply King's property by replacing $x$ $x$ with $-x$ $-x$ (since $-π/2 + π/2 = 0$ $-\pi/2 + \pi/2 = 0$):

I = \int-π/2π/2 8\sqrt(2) (-x) dx(1+e(-x))(1+ ⁴(-x))

I = \int-π/2π/2 8\sqrt(2) x dx(1+e- x)(1+ ⁴x) = \int-π/2π/2 8\sqrt(2) x \cdot ex dx(ex+1)(1+ ⁴x) (2)

Step 1: Simplify by Adding Expressions

Adding equations (1) and (2):

2I = \int-π/2π/2 8\sqrt(2) x(1 + ex) dx(1+ex)(1+ ⁴x)

2I = \int-π/2π/2 8\sqrt(2) x dx1+ ⁴x

Since the integrand is even, we can change the limits from $0$ $0$ to $π/2$ $\pi/2$:

2I = 2 \int₀π/2 8\sqrt(2) x dx1+ ⁴x I = \int₀π/2 8\sqrt(2) x dx1+ ⁴x

Step 2: Substitution and Algebraic Deconstruction

Let $x = t x dx = dt$ $\sin x = t \implies \cos x \, dx = dt$. Limits change from $0$ $0$ to $1$ $1$:

I = \int₀¹ 8\sqrt(2) dt1+t⁴ = 4\sqrt(2) \int₀¹ (2 dt)/(1+t⁴)

Dividing the numerator and denominator by $t²$ $t^2$, we write it as two distinct expressions:

I = 4\sqrt(2) [ \int₀¹ (1+(1)/(t²))/(t²+(1)/(t²)) dt - \int₀¹ (1-(1)/(t²))/(t²+(1)/(t²)) dt ]

I = 4\sqrt(2) [ \int-\infty⁰ (dz)/(z²+2) - \int\infty² (dk)/(k²-2) ]

where $z = t - (1)/(t)$ $z = t - \frac{1}{t}$ and $k = t + (1)/(t)$ $k = t + \frac{1}{t}$.

Step 3: Integrate and Evaluate Parameters

Evaluating standard anti-derivatives:

I = 4\sqrt(2) [ 1\sqrt(2) ⁻¹( z\sqrt(2)) ]-\infty⁰ - 4\sqrt(2) [ 12\sqrt(2)ln| k-\sqrt(2)k+\sqrt(2)| ]\infty²

I = 4\sqrt(2)( π2\sqrt(2)) - 2 ln| 2-\sqrt(2)2+\sqrt(2)| = 2π - 2ln(\sqrt(2)-1)²

I = 2π + 2ln(3+2\sqrt(2))

Matching parameters yields $α = 2$ $\alpha = 2$ and $β = 2$ $\beta = 2$. Thus:

α² + β² = 2² + 2² = 8

Pattern Recognition

Sees: Exponential variables causing asymmetry in symmetric integration bounds. Shortcut: Using King's rule completely eliminates the confusing $ex$ $e^{\sin x}$ factor, leaving behind a straightforward rational trigonometric configuration.

Chapter Mix

Class 12 Mathematics: Definite Integrals

Q jee_main_2024_27_jan_morning Properties of Definite Integrals

(a, b)

be the orthocentre of the triangle whose vertices are

(1, 2)

(2, 3)

and

(3, 1)

, and

I₁=\intₐbx~sin(4x-x²)dx

I₂=\intₐbsin(4x-x²)dx

, then

36 I₁I₂

is equal to:

A. $72$ $72$
B. $88$ $88$
C. $80$ $80$
D. $66$ $66$

Solution

Related Formula

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

Core Logic

First, find the orthocentre $(a,b)$ $(a,b)$ of $Δ ABC$ $\Delta ABC$ with vertices $A(1, 2)$ $A(1, 2)$, $B(2, 3)$ $B(2, 3)$, and $C(3, 1)$ $C(3, 1)$. Slope of $AB = (3-2)/(2-1) = 1$ $AB = \frac{3-2}{2-1} = 1$. The altitude from $C$ $C$ onto $AB$ $AB$ must be perpendicular to $AB$ $AB$, so its slope is $-1$ $-1$. Equation of altitude from $C(3,1)$ $C(3,1)$:

y - 1 = -1(x - 3) \Rightarrow x + y = 4

The orthocentre $(a, b)$ $(a, b)$ lies on all altitudes, including this one. Thus, it satisfies $a + b = 4$ $a + b = 4$.

Step 1: Applying Definite Integral Properties

Given $I₁ = \intₐ^b x (4x-x²) dx$ $I_1 = \int_a^b x \sin(4x-x^2) dx$, let's rewrite the argument of sine:

4x - x² = x(4-x)

Apply King's Rule replacing $x$ $x$ with $(a+b-x)$ $(a+b-x)$. Since we proved $a+b = 4$ $a+b = 4$, substitute $x$ $x$ with $(4-x)$ $(4-x)$:

I₁ = \intₐ^b (4-x) ((4-x)(4 - (4-x))) dx

I₁ = \intₐ^b (4-x) ((4-x)x) dx

I₁ = \intₐ^b (4-x) (4x-x²) dx

Step 2: Evaluating the Integral Ratio

Expand the newly formed integral:

I₁ = 4 \intₐ^b (4x-x²) dx - \intₐ^b x (4x-x²) dx

Notice that the second term is $I₁$ $I_1$ and the first integral is $I₂$ $I_2$:

I₁ = 4I₂ - I₁ \Rightarrow 2I₁ = 4I₂ \Rightarrow (I₁)/(I₂) = 2

Step 3: Final Output Evaluation

We need the value of $36 (I₁)/(I₂)$ $36 \frac{I_1}{I_2}$:

36 \times 2 = 72

Pattern Recognition

Whenever you see $\intₐ^b x \cdot f(x(a+b-x)) dx$ $\int_a^b x \cdot f(x(a+b-x)) dx$, immediately apply King's Rule to factor out $x$ $x$. You rarely need the individual values of the integration bounds, only their sum.

Chapter Mix

Class 11 Maths: Straight Lines Class 12 Maths: Definite Integration

Q9 jee_main_2024_27_jan_morning Integration of Irrational Functions

\int₀¹ 1\sqrt(3+x)+\sqrt(1+x)dx=a+b\sqrt(2)+c\sqrt(3)

, where

a

b

c

are rational numbers, then

2a+3b-4c

is equal to:

A. $4$ $4$
B. $10$ $10$
C. $7$ $7$
D. $8$ $8$

Solution

Related Formula

\int xⁿ dx = xⁿ⁺¹n+1

Core Logic

To evaluate integrals with sum of square roots in the denominator, multiply and divide by the conjugate to rationalize it.

I = \int₀¹ \sqrt(3+x)-\sqrt(1+x)(\sqrt(3+x)+\sqrt(1+x))(\sqrt(3+x)-\sqrt(1+x))dx

I = \int₀¹ \sqrt(3+x)-\sqrt(1+x)(3+x) - (1+x)dx

I = (1)/(2) \int₀¹ (\sqrt(3+x) - \sqrt(1+x)) dx

Step 1: Integration and Bounds Setup

Integrate the resulting expression:

I = (1)/(2) [ (3+x)3/23/2 - (1+x)3/23/2 ]₀¹

I = (1)/(2) \cdot (2)/(3) [ (3+x)3/2 - (1+x)3/2 ]₀¹

I = (1)/(3) [ ((4)3/2 - (2)3/2) - ((3)3/2 - (1)3/2) ]

Step 2: Term Simplification

Evaluate the boundary powers: $43/2 = 8$ $4^{3/2} = 8$ $23/2 = 2\sqrt(2)$ $2^{3/2} = 2\sqrt{2}$ $33/2 = 3\sqrt(3)$ $3^{3/2} = 3\sqrt{3}$ $13/2 = 1$ $1^{3/2} = 1$

Substitute back into the expression:

I = (1)/(3) [ 8 - 2\sqrt(2) - 3\sqrt(3) + 1 ] = (1)/(3) [ 9 - 2\sqrt(2) - 3\sqrt(3) ]

I = 3 - (2)/(3)\sqrt(2) - \sqrt(3)

Step 3: Finding Co-efficients

Comparing with $a+b\sqrt(2)+c\sqrt(3)$ $a+b\sqrt{2}+c\sqrt{3}$ yields: $a = 3$ $a = 3$, $b = -(2)/(3)$ $b = -\frac{2}{3}$, $c = -1$ $c = -1$

Compute $2a+3b-4c$ $2a+3b-4c$:

2(3) + 3(-(2)/(3)) - 4(-1)

$6 - 2 + 4 = 8$ $6 - 2 + 4 = 8$

Pattern Recognition

Whenever you see a sum of square roots in the denominator of an integrand, the immediate algorithmic next step is rationalization.

Chapter Mix

Class 12 Maths: Definite Integration

Let f(x)$f(x)$ be a a positive function and I₁ = ∫-(1)/(2)¹ 2xf(2x(1 - 2x)) dx$I_1 = \int_{-\frac{1}{2}}^{1} 2xf(2x(1 - 2x)) \, dx$ and I₂ = ∫₋₁² f(x(1 - x)) dx$I_2 = \int_{-1}^{2} f(x(1 - x)) \, dx$. Then the value of (I₂)/(I₁)$\frac{I_2}{I_1}$ is equal to

Solution & Explanation

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

Reference Study Guides

More Definite Integrals Previous-Year Questions — Page 5

Solution

Related Formula

Core Logic

Step 1: Simplify by Adding Expressions

Step 2: Substitution and Algebraic Deconstruction

Step 3: Integrate and Evaluate Parameters

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Applying Definite Integral Properties

Step 2: Evaluating the Integral Ratio

Step 3: Final Output Evaluation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Integration and Bounds Setup

Step 2: Term Simplification

Step 3: Finding Co-efficients

Pattern Recognition

Chapter Mix

More Definite Integrals Questions — jee_main_2025_08_april_evening

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Let $f(x)$ $f(x)$ be a a positive function and $I₁ = \int-(1)/(2)¹ 2xf(2x(1 - 2x)) dx$ and $I₂ = \int₋₁² f(x(1 - x)) dx$ $I_2 = \int_{-1}^{2} f(x(1 - x)) \, dx$. Then the value of $(I₂)/(I₁)$ $\frac{I_2}{I_1}$ is equal to