Let $f:[1.\infty )\to \mathrm{ℝ}$ be a differentiable function such that $f\left(1\right)=\frac{1}{3}$ and $3{\int }_1^{x}f\left(t\right)dt=xf\left(x\right)-\frac{x^3}{3},x\in [1,\infty )$. Let $e$ denote the base of the natural logarithm. Then the value of $f\left(e\right)$ is

Given,

$3{\int }_1^{x}f\left(t\right)dt=xf\left(x\right)-\frac{x^3}{3}$

Now differentiating both side we get,

$\Rightarrow 3f\left(x\right)=f\left(x\right)+x{f}^{\text{'}}\left(x\right)-x^2$

$\Rightarrow x{f}^{\text{'}}\left(x\right)-2f\left(x\right)=x^2$

$\Rightarrow f^{\text{'}}\left(x\right)-\frac{2}{x}f\left(x\right)=x$

Which is a linear differential equation,

$\Rightarrow I.F.=e^{-\frac{2}{x}dx}=\frac{1}{x^2}$

Now solution is given by,

$\Rightarrow y\left(\frac{1}{x^2}\right)=\int x\times \frac{1}{x^2}dx=\ln x+C$

$\Rightarrow y=x^2(\ln x+C)$

$\Rightarrow f\left(x\right)=x^2(\ln x+C)$

Now using given value, $f\left(1\right)=\frac{1}{3}$ we get,

$\Rightarrow 0+C=\frac{1}{3}\Rightarrow C=\frac{1}{3}$

So, $f\left(x\right)=x^2\left(\ln x+C\right)$

$\Rightarrow f\left(e\right)=e^2\left(\ln e+\frac{1}{3}\right)$

$\Rightarrow f\left(e\right)=\frac{4e^2}{3}$