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

Q71 jee_main_2025_24_jan_morning Differentiating Under the Integral Sign

Let

f

be a differentiable function such that

2(x+2)²f(x) - 3(x+2)² = 10\int₀x(t+2)f(t)dt

for

x \geq 0

. Then

f(2)

is equal to ________.

Numerical Answer. Answer: 19

Solution

Related Formula

The Leibniz Integral Rule template allows direct differentiation of an integral with variable limits:

(d)/(dx)( \int₀x g(t) dt ) = g(x)

Core Logic

Differentiate both sides of the given functional equation with respect to $x$ $x$ using the product rule:

(d)/(dx)[ 2(x+2)² f(x) - 3(x+2)² ] = (d)/(dx)[ 10\int₀x(t+2)f(t)dt ]

4(x+2)f(x) + 2(x+2)² f'(x) - 6(x+2) = 10(x+2)f(x)

Since $x \geq 0$ $x \geq 0$, the factor $(x+2)$ $(x+2)$ is strictly non-zero. Divide the entire equation by $2(x+2)$ $2(x+2)$:

2f(x) + (x+2)f'(x) - 3 = 5f(x)

(x+2)f'(x) - 3f(x) = 3

Step 1: Solve the First-Order Differential Equation

Rearrange the expression into standard linear differential equation form where $y = f(x)$ $y = f(x)$:

(dy)/(dx) - (3)/(x+2)y = (3)/(x+2)

Compute the Integrating Factor (I.F.):

I.F. = e\int -(3)/(x+2) dx = e-3ln(x+2) = (x+2)⁻³

Multiply through by the I.F. and integrate:

y \cdot (x+2)⁻³ = \int (3)/(x+2) \cdot (x+2)⁻³ dx = \int 3(x+2)⁻⁴ dx

(f(x))/((x+2)³) = 3 \cdot (x+2)⁻³-3 + C = -(x+2)⁻³ + C

f(x) = -1 + C(x+2)³

Step 2: Apply the Boundary Condition

Find the boundary condition by substituting $x = 0$ $x = 0$ into the original integral equation equation:

2(0+2)² f(0) - 3(0+2)² = 10 \int₀⁰ (t+2)f(t) dt

8f(0) - 12 = 0 f(0) = (12)/(8) = (3)/(2)

Substitute $x = 0$ $x = 0$ into our general solution formula:

f(0) = -1 + C(0+2)³ (3)/(2) = -1 + 8C

(5)/(2) = 8C C = (5)/(16)

Thus, the explicit function is:

f(x) = -1 + (5)/(16)(x+2)³

Step 3: Evaluate at target point x = 2

Substitute $x = 2$ $x = 2$ into the final function equation:

f(2) = -1 + (5)/(16)(2+2)³ = -1 + (5)/(16)(64)

f(2) = -1 + 5(4) = -1 + 20 = 19

Pattern Recognition

When an equation contains a variable integral limit $\int₀^x$ $\int_0^x$, differentiating both sides using the Leibniz rule converts it into a standard differential equation. The initial value is found by setting $x = 0$ $x = 0$ directly in the original expression.

Chapter Mix

Class 12 Mathematics: Definite Integrals Class 12 Mathematics: Differential Equations

Q jee_main_2025_28_jan_evening Integration by Parts

Let

f: Rarrow R

be a twice differentiable function such that

f(2)=1

. If

F(x)=xf(x)

for all

xin R

,

\int₀²xF(x)dx=6

and

\int₀²x²F(x)dx=40

, then

F(2)+\int₀²F(x)dx

is equal to:

A. $11$ $11$
B. $15$ $15$
C. $6$ $6$
D. $13$ $13$

Solution

Related Formula

Integration by Parts formula:

\int u \cdot v dx = u \int v dx - \int ( u' \int v dx ) dx

Core Logic

Given $F(x) = xf(x)$ $F(x) = xf(x)$ and $f(2) = 1 F(2) = 2f(2) = 2$ $f(2) = 1 \implies F(2) = 2f(2) = 2$.

Let's apply Integration by Parts to the first given integral:

\int₀² x F'(x) dx = 6

[ xF(x) ]₀² - \int₀² F(x) dx = 6

2F(2) - 0 - \int₀² F(x) dx = 6

2(2) - \int₀² F(x) dx = 6 \int₀² F(x) dx = 4 - 6 = -2

Step 1: Simplify Second Integral via Integration by Parts

Now look at the second integral:

\int₀² x² F''(x) dx = 40

Applying Integration by parts (taking $u = x²$ $u = x^2$ and $v = F''(x)$ $v = F''(x)$):

[ x² F'(x) ]₀² - \int₀² 2x F'(x) dx = 40

4 F'(2) - 0 - 2 \int₀² x F'(x) dx = 40

We already know $\int₀² x F'(x) dx = 6$ $\int_{0}^{2} x F'(x) dx = 6$:

4 F'(2) - 2(6) = 40

4 F'(2) - 12 = 40 4 F'(2) = 52 F'(2) = 13

Step 2: Sum the Values

We need to find $F'(2) + \int₀² F(x) dx$ $F'(2) + \int_{0}^{2} F(x) dx$: $13 + (-2) = 11$ $13 + (-2) = 11$

Pattern Recognition

Notice how the definition of $f(x)$ $f(x)$ is mostly a distraction to find $F(2)=2$ $F(2)=2$. The problem is fundamentally testing consecutive applications of integration by parts to reduction structures.

Chapter Mix

Class 12 Mathematics: Definite Integration

Q66 jee_main_2025_29_jan_morning Integration by Substitution

The integral

80\int₀(π)/(4)(( θ + θ)/(9 + 16 2θ))dθ

is equal to:

A. $3 ₑ4$ $3\log_{e}4$
B. $6 ₑ₄$ $6\log_{\mathrm{e}4}$
C. $4 ₑ₃$ $4\log_{\mathrm{e}3}$
D. $2 ₑ3$ $2\log_{e}3$

Solution

Related Formula

\int (dt)/(a² - b² t²) = (1)/(2a) ln | (a+bt)/(a-bt) |

2θ = 1 - ( θ - θ)²

Core Logic

Let $θ - θ = t$ $\sin\theta - \cos\theta = t$. Then $( θ + θ)dθ = dt$ $(\cos\theta + \sin\theta)d\theta = dt$. Transform limits: When $θ = 0 t = 0 - 1 = -1$ $\theta = 0 \implies t = 0 - 1 = -1$ When $θ = (π)/(4) t = 1\sqrt(2) - 1\sqrt(2) = 0$

Step 1: Perform the algebraic substitution

Express the denominator base:

9 + 16 2θ = 9 + 16[1 - t²] = 25 - 16t²

The integral transforms to:

I = 80 \int₋₁⁰ (dt)/(25 - 16t²) = (80)/(16) \int₋₁⁰ (dt)/(((5)/(4))² - t²)

Step 2: Execute Integral Calculation

I = 5 [ (1)/(2((5)/(4))) ln | ((5)/(4) + t)/((5)/(4) - t) | ]₋₁⁰

I = 2 [ ln(1) - ln( (1/4)/(9/4) ) ] = 2 [ 0 - ln((1)/(9)) ] = 2ln(9) = 4ln(3)

Pattern Recognition

Whenever $( θ + θ)$ $(\sin\theta + \cos\theta)$ sits inside the numerator, instantly choose $t = θ - θ$ $t = \sin\theta - \cos\theta$ as your core linear substitution engine to clean denominators.

Chapter Mix

Class 12 Mathematics: Definite Integrals

Q71 jee_main_2025_29_jan_morning Functional Equations with Integrals

Let

f: (0, \infty) \to R

be a twice differentiable function. If for some

a \neq 0

,

\int₀¹ f(λ x) dλ = a f(x)

,

f(1) = 1

and

f(16) = (1)/(8)

, then

16 - f'((1)/(16))

is equal to

Numerical Answer. Answer: 112

Solution

Related Formula

Leibniz Integral Rule for differentiation: (d)/(dx)\int₀x f(t) dt = f(x)

Core Logic

Perform variable substitution inside the integral: let $λ x = t dλ = (1)/(x) dt$ . When $λ = 0 t = 0$ $\lambda = 0 \implies t = 0$; when $λ = 1 t = x$ $\lambda = 1 \implies t = x$. The equation transforms to:

(1)/(x) \int₀x f(t) dt = a f(x) \int₀x f(t) dt = a x f(x)

Step 1: Differentiate with respect to x

Using Leibniz rule and product rule:

f(x) = a [x f'(x) + f(x)]

(1 - a)f(x) = a x f'(x) (f'(x))/(f(x)) = (1-a)/(a) (1)/(x)

Integrating both sides yields:

ln f(x) = ((1-a)/(a))ln x + c f(x) = C x(1-a)/(a)

Step 2: Calculate Constants using boundaries

Given $f(1) = 1 C = 1$ $f(1) = 1 \implies C = 1$. Given $f(16) = (1)/(8) (1)/(8) = (16)(1-a)/(a) 2⁻³ = (2⁴)(1-a)/(a)$

-3 = (4(1-a))/(a) -3a = 4 - 4a a = 4

Therefore, power exponent $= (1-4)/(4) = -(3)/(4) f(x) = x-(3)/(4)$ .

Step 3: Evaluate target derivative value

Find the derivative:

f'(x) = -(3)/(4) x-(7)/(4)

Substitute $x = (1)/(16)$ $x = \frac{1}{16}$:

f'((1)/(16)) = -(3)/(4) (2⁻⁴)-(7)/(4) = -(3)/(4) (2⁷) = -(3)/(4) \times 128 = -96

Final requested computation calculation:

16 - f'((1)/(16)) = 16 - (-96) = 112

Pattern Recognition

Scaling inputs inside functional definite integrals tracks closely to homogenous Euler equation properties. Converting integrations quickly to local power functions reduces processing parameters.

Chapter Mix

Class 12 Mathematics: Definite Integrals Class 12 Mathematics: Differential Equations

Q2 jee_main_2024_01_february_morning Properties of Definite Integrals

The value of the integral

\int₀(π)/(4) x dx ⁴(2x)+ ⁴(2x)

equals:

A. $\sqrt(2)π²8$ $\frac{\sqrt{2}\pi^{2}}{8}$
B. $\sqrt(2)π²16$ $\frac{\sqrt{2}\pi^{2}}{16}$
C. $\sqrt(2)π²32$ $\frac{\sqrt{2}\pi^{2}}{32}$
D. $\sqrt(2)π²64$ $\frac{\sqrt{2}\pi^{2}}{64}$

Solution

Related Formula

King's Property of Definite Integrals:

\intₐb f(x) dx = \intₐb f(a+b-x) dx

Core Logic

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

I = \int₀(π)/(4) x dx ⁴(2x)+ ⁴(2x)

Substitute $2x = t 2dx = dt dx = (1)/(2)dt$ . When $x = 0 t = 0$ $x = 0 \implies t = 0$, and when $x = (π)/(4) t = (π)/(2)$ .

I = (1)/(4) \int₀(π)/(2) t dt ⁴t + ⁴t (1)

Step 1: Apply King's Property

Applying the property $\int₀a f(t) dt = \int₀a f(a-t) dt$ :

I = (1)/(4) \int₀(π)/(2) ((π)/(2) - t) dt ⁴((π)/(2)-t) + ⁴((π)/(2)-t)

I = (1)/(4) \int₀(π)/(2) ((π)/(2) - t) dt ⁴t + ⁴t (2)

Adding equations (1) and (2):

2I = (1)/(4) \int₀(π)/(2) (π)/(2) dt ⁴t + ⁴t

2I = (π)/(8) \int₀(π)/(2) dt ⁴t + ⁴t

2I = (π)/(8) \int₀(π)/(2) ⁴t dt ⁴t + 1

2I = (π)/(8) \int₀(π)/(2) (1 + ²t) ²t dt ⁴t + 1

Step 2: Substitution and Algebraic Limits

Let $t = y ²t dt = dy$ . When $t = 0 y = 0$ $t = 0 \implies y = 0$, and when $t = (π)/(2) y = \infty$ $t = \frac{\pi}{2} \implies y = \infty$.

2I = (π)/(8) \int₀\infty (1 + y²) dy1 + y⁴

I = (π)/(16) \int₀\infty 1 + 1y²y² + 1y² dy

Now put $y - (1)/(y) = p (1 + 1y²) dy = dp$ . When $y \to 0^+ p \to -\infty$ $y \to 0^+ \implies p \to -\infty$, and when $y \to \infty p \to \infty$ $y \to \infty \implies p \to \infty$. Also, $y² + 1y² = p² + 2 = p² + (\sqrt(2))²$ .

I = (π)/(16) \int-\infty\infty dpp² + (\sqrt(2))²

I = π16\sqrt(2) [ ⁻¹( p\sqrt(2)) ]-\infty\infty

$ $I = π16\sqrt(2) ( (π)/(2) - (-(π)/(2)) ) = π²16\sqrt(2) = \sqrt(2)π²32$

Pattern Recognition

Sees: Integrand containing $x$ $x$ in the numerator and symmetric trigonometric functions in the denominator. Shortcut: The elimination of $x$ $x$ using King's property is standard. For integrals containing $(1+y²)/(1+y⁴)$ $\frac{1+y^2}{1+y^4}$, dividing by $y²$ $y^2$ transforms the denominator into a perfect square form $(y-(1)/(y))²+2$ $\left(y-\frac{1}{y}\right)^2+2$, making substitution trivial.

Chapter Mix

Class 12 Mathematics: Definite Integrals Class 11 Mathematics: Trigonometric Identities

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 4

Solution

Related Formula

Core Logic

Step 1: Solve the First-Order Differential Equation

Step 2: Apply the Boundary Condition

Step 3: Evaluate at target point x = 2

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Simplify Second Integral via Integration by Parts

Step 2: Sum the Values

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Perform the algebraic substitution

Step 2: Execute Integral Calculation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Differentiate with respect to x

Step 2: Calculate Constants using boundaries

Step 3: Evaluate target derivative value

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Apply King's Property

Step 2: Substitution and Algebraic Limits

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