Question

If \(6\int_{1}^{x} f(t)dt = 3x f(x) + x^3 - 4, x \geq 1\) then value of (f(2)-f(3)) is :

Hint:

Equations involving integrals can often be solved by differentiation using Leibniz's rule. Remember to find an initial condition from the original integral equation to determine the constant of integration.

Answer 1

Correct Answer: 3

Step 1: Convert the Integral Equation to a Differential Equation:
We are given an equation involving an integral of an unknown function \(f(x)\). We can convert this to a differential equation by differentiating both sides with respect to x, using the Fundamental Theorem of Calculus (Leibniz's rule). Given: \(6\int_{1}^{x} f(t)dt = 3x f(x) + x^3 - 4\) Differentiating both sides w.r.t x: \[ \frac{d}{dx}\left(6\int_{1}^{x} f(t)dt\right) = \frac{d}{dx}(3x f(x) + x^3 - 4) \] \[ 6f(x) = [3 \cdot f(x) + 3x \cdot f'(x)] + 3x^2 \] \[ 6f(x) = 3f(x) + 3xf'(x) + 3x^2 \]
Step 2: Solve the Differential Equation:
Rearrange the equation to form a standard linear differential equation. \[ 3f(x) - 3xf'(x) = 3x^2 \] \[ f(x) - xf'(x) = x^2 \Rightarrow xf'(x) - f(x) = -x^2 \] Let \(y = f(x)\). The equation is \(x\frac{dy}{dx} - y = -x^2\). Divide by x (since \(x \geq 1\)): \[ \frac{dy}{dx} - \frac{1}{x}y = -x \] This is a linear first-order DE. The integrating factor (I.F.) is \(e^{\int -1/x dx} = e^{-\ln x} = 1/x\). The solution is \(y \cdot (\text{I.F.}) = \int Q(x) \cdot (\text{I.F.}) dx + C\). \[ y \cdot \frac{1}{x} = \int (-x) \cdot \frac{1}{x} dx + C = \int -1 dx + C \] \[ \frac{y}{x} = -x + C \Rightarrow y = -x^2 + Cx \] So, \(f(x) = -x^2 + Cx\).
Step 3: Find the Constant of Integration C:
We can find C by using the original equation. Substitute \(x=1\) into the original equation: \[ 6\int_{1}^{1} f(t)dt = 3(1)f(1) + 1^3 - 4 \] \[ 6(0) = 3f(1) - 3 \Rightarrow 3f(1) = 3 \Rightarrow f(1) = 1 \] Now, use this condition in our solution for \(f(x)\): \[ f(1) = -1^2 + C(1) = 1 \Rightarrow -1 + C = 1 \Rightarrow C = 2 \] Therefore, the specific solution is \(f(x) = -x^2 + 2x\).
Step 4: Calculate the Final Value:
We need to find \(f(2) - f(3)\). \[ f(2) = -(2)^2 + 2(2) = -4 + 4 = 0 \] \[ f(3) = -(3)^2 + 2(3) = -9 + 6 = -3 \] \[ f(2) - f(3) = 0 - (-3) = 3 \]