Question

Let $f[1,\infty )\to [2,\infty )$ be a differentiable function such that $f(1)=\frac{1}{3}.$ If $6{\int }_1^{\text{x}}f\left(t\right)dt=3\text{x}f\left(\text{x}\right)-{\left(\text{x}\right)}^3$ for all $\text{x}\ge 1,$ then the value of $3f\left(2\right)$ is

Answer 1

Answer: 8

Given, $f\left(1\right)=\frac{1}{3}$ and $6{\int }_1^{\text{x}}f\left(\text{t}\right)\text{dt}=3\text{x}f\left(\text{x}\right)-{\left(\text{x}\right)}^3$ for all $\text{x}\ge 1$
Using Newton- Leibnitz formula.
Differentiating both sides,
$\Rightarrow 6f\left(\text{x}\right)\cdot 1-0=3f\left(\text{x}\right)+3\text{x}{\left(f\right)}^{{}^{\prime }}\left(\text{x}\right)-3{\left(\text{x}\right)}^2$
$\Rightarrow 3\text{x}{\left(f\right)}^{{}^{\prime }}\left(\text{x}\right)-3f\left(\text{x}\right)=3{\left(\text{x}\right)}^2$
$\Rightarrow {\left(f\right)}^{{}^{\prime }}\left(\text{x}\right)-\frac{1}{\text{x}}f\left(\text{x}\right)=\text{x}$
$\Rightarrow \frac{\text{x}{\left(f\right)}^{{}^{\prime }}\left(\text{x}\right)-f\left(\text{x}\right)}{{\left(\text{x}\right)}^2}=1$
$\Rightarrow \frac{\text{d}}{\text{dx}}\left\{\frac{f\left(\text{x}\right)}{\text{x}}\right\}=1$
Integrating both sides,
$\Rightarrow \frac{f\left(\text{x}\right)}{\text{x}}=\text{x}+\text{C}$ $\left[\because f\left(1\right)=\frac{1}{3}\right]$
$\frac{1}{3}=1+\text{C}$
$\Rightarrow \text{C}=-\frac{2}{3}$
$f\left(\text{x}\right)={\left(\text{x}\right)}^2-\frac{2}{3}\text{x}$
$\Rightarrow f\left(2\right)=4-\frac{4}{3}=\frac{8}{3}$
$\therefore 3f\left(2\right)=8$