Let $f:[1, \infty) \rightarrow[2, \infty)$ be a differentiable function. If $10 \int_1^x f(\mathrm{t}) \mathrm{dt}=5 x f(x)-x^5-9$ for all $x \geqslant 1$, then the value of $f(3)$ is :

<p>To solve the problem, we start with the given equation:</p> <p>$ 10 \int_1^x f(t) \, dt = 5x f(x) - x^5 - 9 $</p> <p>By differentiating both sides with respect to $ x $, we have:</p> <p>$ \frac{d}{dx}\left(10 \int_1^x f(t) \, dt\right) = \frac{d}{dx}(5x f(x) - x^5 - 9) $</p> <p>The left side simplifies to $ 10 f(x) $. For the right side, using the product rule and the power rule, we get:</p> <p>$ 10 f(x) = 5f(x) + 5x \frac{d}{dx} f(x) - 5x^4 $</p> <p>Rearranging terms, we obtain:</p> <p>$ 5 f(x) = 5x \frac{d}{dx} f(x) - 5x^4 $</p> <p>Let $ y = f(x) $. Thus:</p> <p>$ 5 y = 5x \frac{dy}{dx} - 5x^4 $</p> <p>Dividing by 5, we have:</p> <p>$ y = x \frac{dy}{dx} - x^4 $</p> <p>Rewriting, we get:</p> <p>$ \frac{dy}{dx} - \frac{y}{x} = x^3 $</p> <p>This is a linear differential equation. The integrating factor (I.F.) is calculated as:</p> <p>$ \text{I.F.} = e^{\int \frac{-1}{x} \, dx} = e^{-\ln x} = \frac{1}{x} $</p> <p>Multiplying through by the integrating factor, we have:</p> <p>$ y \cdot \frac{1}{x} = \int x^3 \cdot \frac{1}{x} \, dx = \int x^2 \, dx = \frac{x^3}{3} + C $</p> <p>Thus, we solve for $ y $:</p> <p>$ y = \frac{x^4}{3} + Cx $</p> <p>Substituting back, we need to find $ C $ using the condition at $ x = 1 $:</p> <p>Since $ \int_1^1 f(t) \, dt = 0 $, we substitute:</p> <p>$ 10 \cdot 0 = 5 \cdot 1 \cdot f(1) - 1^5 - 9 \quad \Rightarrow \quad 0 = 5f(1) - 1 - 9 $</p> <p>$ 5f(1) = 10 \quad \Rightarrow \quad f(1) = 2 $</p> <p>Now use $ f(1) = 2 $ in the function:</p> <p>$ 2 = \frac{1}{3} + C \cdot 1 \quad \Rightarrow \quad C = \frac{5}{3} $</p> <p>The function $ f(x) $ is given by:</p> <p>$ f(x) = \frac{x^4}{3} + \frac{5}{3}x $</p> <p>To find $ f(3) $:</p> <p>$ f(3) = \frac{3^4}{3} + \frac{5}{3} \cdot 3 = 27 + 5 = 32 $</p> <p>Therefore, the value of $ f(3) $ is 32.</p>