Question

Let $f [1 , \infty) \rightarrow [ 2 , \infty) $ be a differentiable function such that $f (1 ) = \frac{1}{3} .$ If $6 \displaystyle \int _{1}^{\text{x}} f \left(t\right) dt = 3 \text{x} f \left(\text{x}\right) - \left(\text{x}\right)^{3}$ for all $\text{x} \geq 1,$ then the value of $3f \left(2\right)$ is

Answer 1

Given, $ f \left(1\right) = \frac{1}{3}$ and $6 \displaystyle \int _{1}^{\text{x}} f \left(\text{t}\right) \text{dt} = 3 \text{x} f ⁡ \left(\text{x}\right) - \left(\text{x}\right)^{3}$ for all $\text{x} \geq 1$
Using Newton- Leibnitz formula.
Differentiating both sides,
$\Rightarrow 6 f \left(\text{x}\right) \cdot 1 - 0 = 3 f \left(\text{x}\right) + 3 \text{x} \left( f\right)^{'} \left(\text{x}\right) - 3 \left(\text{x}\right)^{2}$
$\Rightarrow 3 \text{x} \left( f\right)^{'} \left(\text{x}\right) - 3 f \left(\text{x}\right) = 3 \left(\text{x}\right)^{2}$
$\Rightarrow \left( f\right)^{'} \left(\text{x}\right) - \frac{1}{\text{x}} f \left(\text{x}\right) = \text{x}$
$\Rightarrow \frac{\text{x} \left( f\right)^{'} \left(\text{x}\right) - f \left(\text{x}\right)}{\left(\text{x}\right)^{2}} = 1$
$\Rightarrow \frac{\text{d}}{\text{dx}} \left\{\frac{ f \left(\text{x}\right)}{\text{x}}\right\} = 1$
Integrating both sides,
$\Rightarrow \frac{ f \left(\text{x}\right)}{\text{x}} = \text{x} + \text{C}$ $\left[\because f \left(1\right) = \frac{1}{3}\right]$
$\frac{1}{3} = 1 + \text{C}$
$\Rightarrow \text{C} = - \frac{2}{3}$
$ f \left(\text{x}\right) = \left(\text{x}\right)^{2} - \frac{2}{3} \text{x}$
$\Rightarrow f \left(2\right) = 4 - \frac{4}{3} = \frac{8}{3}$
$\therefore \, 3 f \left(2\right) = 8$