The user wants me to solve for the value of f(2) given an integral equation and an initial condition.

Step-by-step Solution:

1. Start with the given integral equation: We are given the equation for all $x \ge 1$: $6\int\limits_1^x {f(t)dt = 3xf(x) - {x^3} - 5}$

2. Differentiate both sides with respect to x: To eliminate the integral, we can differentiate both sides of the equation with respect to $x$. We will use the Leibniz rule (Fundamental Theorem of Calculus, Part 1) for the left side and the product rule for the right side.

   • Left Hand Side (LHS): $\frac{d}{{dx}}\left( {6\int\limits_1^x {f(t)dt} } \right) = 6f(x)$

   • Right Hand Side (RHS): $\frac{d}{{dx}}\left( {3xf(x) - {x^3} - 5} \right)$ Using the product rule for the term $3xf(x)$: $\frac{d}{{dx}}(3xf(x)) = \left( {\frac{d}{{dx}}(3x)} \right)f(x) + 3x\left( {\frac{d}{{dx}}f(x)} \right) = 3f(x) + 3xf'(x)$ The derivative of the entire RHS is: $3f(x) + 3xf'(x) - 3{x^2} - 0$

3. Form the differential equation: Equating the derivatives of the LHS and RHS: $6f(x) = 3f(x) + 3xf'(x) - 3{x^2}$ Rearranging the terms to simplify: $6f(x) - 3f(x) = 3xf'(x) - 3{x^2}$ $3f(x) = 3xf'(x) - 3{x^2}$ Dividing the entire equation by 3: $f(x) = xf'(x) - {x^2}$ Let $y = f(x)$, so $f'(x) = \frac{{dy}}{{dx}}$. The equation becomes: $y = x\frac{{dy}}{{dx}} - {x^2}$ Rearranging this into the standard form of a linear first-order differential equation, $\frac{{dy}}{{dx}} + P(x)y = Q(x)$: $x\frac{{dy}}{{dx}} - y = {x^2}$ $\frac{{dy}}{{dx}} - \frac{1}{x}y = x$

4. Solve the linear differential equation: This is a linear differential equation with $P(x) = -\frac{1}{x}$ and $Q(x) = x$. First, find the integrating factor (I.F.): $I.F. = {e^{\int {P(x)dx} }} = {e^{\int { - \frac{1}{x}dx} }} = {e^{ - \ln |x|}} = {e^{\ln ({|x|^{ - 1}})}} = \frac{1}{{|x|}}$ Since we are given that $x \ge 1$, we have $|x| = x$. So, the integrating factor is: $I.F. = \frac{1}{x}$ The solution to the differential equation is given by: $y \cdot (I.F.) = \int {Q(x) \cdot (I.F.)dx} + C$ $y \cdot \frac{1}{x} = \int {x \cdot \frac{1}{x}dx} + C$ $\frac{y}{x} = \int {1dx} + C$ $\frac{y}{x} = x + C$ $y = {x^2} + Cx$ So, the function is $f(x) = {x^2} + Cx$.

5. Use the initial condition to find C: We are given the condition $f(1) = 2$. Substitute $x = 1$ into the expression for $f(x)$: $f(1) = {1^2} + C(1)$ $2 = 1 + C$ $C = 1$

6. Determine the specific function f(x): Substituting $C=1$ back into the general solution, we get: $f(x) = {x^2} + x$

7. Calculate the value of f(2): Now, we can find the value of $f(2)$ by substituting $x=2$ into the function: $f(2) = {2^2} + 2$ $f(2) = 4 + 2$ $f(2) = 6$

Final Answer:

The value of f(2) is 6.