Evaluate Value of Function from Integral Equation

To find the value of $f(2)$, we examine the given integral equation:
$6 \int_{1}^{x} f(t) \, dt = 3x f(x) - x^3 \quad \text{for all } x \geq 1$

Step 1: Consistency check at the lower limit $x = 1$

Substitute $x = 1$ into both sides of the given equation:

• Left-Hand Side (LHS):
  $\text{LHS} = 6 \int_{1}^{1} f(t) \, dt = 0$

• Right-Hand Side (RHS):
  $\text{RHS} = 3(1) f(1) - (1)^3$
  Given that $f(1) = 2$:
  $\text{RHS} = 3(1)(2) - 1 = 6 - 1 = 5$

This leads to:
$0 = 5$
which is a contradiction. Thus, no such function $f(x)$ exists that simultaneously satisfies the given integral equation and the condition $f(1) = 2$.

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Step 2: Formal analysis via differentiation (ignoring the contradiction)

Differentiating both sides of the integral equation with respect to $x$ using Leibniz's rule:
$\frac{d}{dx} \left( 6 \int_{1}^{x} f(t) \, dt \right) = \frac{d}{dx} \left( 3x f(x) - x^3 \right)$

$6 f(x) = 3 f(x) + 3x f'(x) - 3x^2$

Simplifying:
$3 f(x) = 3x f'(x) - 3x^2$
$f(x) = x f'(x) - x^2$
$x f'(x) - f(x) = x^2$

Dividing both sides by $x^2$ (since $x \geq 1 > 0$):
$\frac{x f'(x) - f(x)}{x^2} = 1$
$\frac{d}{dx} \left( \frac{f(x)}{x} \right) = 1$

Integrating both sides with respect to $x$:
$\frac{f(x)}{x} = x + C \implies f(x) = x^2 + Cx$

If we use the initial condition $f(1) = 2$:
$2 = 1^2 + C(1) \implies C = 1$
which would give:
$f(x) = x^2 + x$

Under this form, evaluating at $x = 2$ yields:
$f(2) = 2^2 + 2 = 6$

However, substituting $f(x) = x^2 + x$ back into the original integral equation:
$\text{LHS} = 6 \int_{1}^{x} (t^2 + t) \, dt = 6 \left[ \frac{t^3}{3} + \frac{t^2}{2} \right]_1^x = 2x^3 + 3x^2 - 5$
$\text{RHS} = 3x(x^2 + x) - x^3 = 2x^3 + 3x^2$
The two sides differ by the constant $5$ for all $x \geq 1$.

Conclusion

Because the integral equation itself is intrinsically inconsistent with the given data (evaluating at $x = 1$ leads to $0 = 5$), no valid function exists. Consequently, this question has no consistent solution, and the official answer is MARKS TO ALL.