Question

If $f:(0,\infty )\to R$ be a continuous and differentiable function, such that $f^3\left(x\right)=\underset{0}{\overset{x}{\int }}t\cdot f^2\left(t\right)dt,f\left(x\right)\ne 0,f\left(1\right)=\frac{1}{6}$ for every $x>0,$ then the value of $f\left(6\right)$ is

Answer 1

Given, $f^3\left(x\right)=\underset{0}{\overset{x}{\int }}t\cdot f^2\left(t\right)dt$
On differentiating, we have: $3f^2\left(x\right)f^{\prime }\left(x\right)=x{f}^2\left(x\right)\because f\left(x\right)\ne 0\Rightarrow f^{\prime }\left(x\right)=\frac{x}{3}\therefore f\left(x\right)=\frac{x^2}{6}+C$
but, $f\left(1\right)=\frac{1}{6}\Rightarrow C=0$
$\Rightarrow f\left(6\right)=6$