Let f$f$ be a differentiable function such that 2(x+2)²f(x) - 3(x+2)² = 10∫₀x(t+2)f(t)dt$$2(x+2)^{2}f(x) - 3(x+2)^{2} = 10\int_{0}^{x}(t+2)f(t)dt$$ for x ≥ 0$x \geq 0$. Then f(2)$f(2)$ is equal to ________.

Related Formula

Greatest Integer Function boundaries:

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 ≈ 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$y = 2 \implies e^{1-x} = 2 \implies 1-x = \ln 2 \implies x = 1 - \ln 2$.
• Find x$x$ where y = 1 e1-x = 1 1-x = 0 x = 1$y = 1 \implies e^{1-x} = 1 \implies 1-x = 0 \implies x = 1$.
• 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:

Step 3: Perform Integrations

Step 4: Solve for Alpha Cubed

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

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$2.71 \rightarrow 2 \rightarrow 1 \rightarrow 0$ reveals exactly where the integer thresholds are crossed.

Chapter Mix

Class 12 Mathematics: Integrals

Related Formula

Using the properties of definite integrals:

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

Core Logic

Let:

Applying x → π-x$x \to \pi-x$:

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

Adding (1) and (2):

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

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

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

Let t = x dt = ² x dx$t = \tan x \implies dt = \sec^2 x \, dx$ At x = 0$x = 0$, t = 0$t = 0$; at x = π/2$x = \pi/2$, t → ∞$t \to \infty$.

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

Related Formula

King's Property of Definite Integrals:

Core Logic

Let the given integral be:

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

Step 1: Eliminate the x Variable

Adding equations (1) and (2):

Using the symmetric property ∫₀2a f(x)dx = 2∫₀^a f(x)dx$\int_0^{2a} f(x)dx = 2\int_0^a f(x)dx$ if f(2a-x)=f(x)$f(2a-x)=f(x)$:

Step 2: Solve Using Substitution

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

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

Substituting into the integral:

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

Related Formula

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$x = 1 \implies t = 2$.

Step 2: Invoke Integral Mirror Properties

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

Step 3: Final Matrix Matching Evaluation

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

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

Related Formula

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) ≥ 0$x \sin(\pi x) \ge 0$. For x in [1, 3/2]$x \in [1, 3/2]$, values drop below zero:

Step 2: Integrate the Even Function Block

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

Step 3: Subtract the Inverse Segment

Evaluating boundary limits across the secondary phase track:

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