The problem asks for the value of $f(3)$ for a differentiable function $f: [1, \infty) \to [2, \infty)$ that satisfies the given integral equation.

Integral Equation Differentiation

The given integral equation is:

$10 \int_{1}^{x} f(t)dt = 5xf(x) - x^5 - 9$

Differentiate both sides of the equation with respect to $x$. Using the Fundamental Theorem of Calculus on the left side and the product rule on the $5xf(x)$ term on the right side:

$ \frac{d}{dx} \left( 10 \int_{1}^{x} f(t)dt \right) = \frac{d}{dx} (5xf(x) - x^5 - 9) $

$ 10 f(x) = \left( 5 \cdot f(x) + 5x \cdot \frac{d}{dx}f(x) \right) - 5x^4 - 0 $

$ 10 f(x) = 5f(x) + 5xf'(x) - 5x^4 $

Differential Equation Formulation

Rearrange the terms to simplify and identify the differential equation:

$ 10 f(x) - 5f(x) = 5xf'(x) - 5x^4 $

$ 5 f(x) = 5xf'(x) - 5x^4 $

Divide the entire equation by 5:

$ f(x) = xf'(x) - x^4 $

Rearrange to the standard form $x f'(x) - f(x) = x^4$. Divide by $x^2$ (since $x \ge 1$):

$ \frac{xf'(x) - f(x)}{x^2} = \frac{x^4}{x^2} $

Recognize the left side as the derivative of the quotient $\frac{f(x)}{x}$:

$ \frac{d}{dx} \left( \frac{f(x)}{x} \right) = x^2 $

Differential Equation Solution

Integrate both sides with respect to $x$:

$ \int \frac{d}{dx} \left( \frac{f(x)}{x} \right) dx = \int x^2 dx $

$ \frac{f(x)}{x} = \frac{x^3}{3} + C $

where $C$ is the constant of integration.

Solve for $f(x)$:

$ f(x) = \frac{x^4}{3} + Cx $

Function Constant Determination

To find the constant $C$, evaluate the original integral equation at $x=1$:

$ 10 \int_{1}^{1} f(t)dt = 5(1)f(1) - 1^5 - 9 $

Since $\int_{1}^{1} f(t)dt = 0$:

$ 10 \cdot 0 = 5f(1) - 1 - 9 $

$ 0 = 5f(1) - 10 $

$ 5f(1) = 10 \implies f(1) = 2 $

Now, substitute $x=1$ into the derived expression for $f(x)$:

$ f(1) = \frac{1^4}{3} + C(1) $

$ 2 = \frac{1}{3} + C $

$ C = 2 - \frac{1}{3} = \frac{6}{3} - \frac{1}{3} = \frac{5}{3} $

The function is therefore:

$ f(x) = \frac{x^4}{3} + \frac{5}{3}x $

f(3) Value Calculation

Substitute $x=3$ into the determined function $f(x)$:

$ f(3) = \frac{3^4}{3} + \frac{5}{3}(3) $

$ f(3) = \frac{81}{3} + 5 $

$ f(3) = 27 + 5 $

$ f(3) = 32 $