Let f: [ 1 . ∞ ) → ℝ be a differentiable function such that f(1) =...

Given,

$3{∫}_{1}^{x}f\left(t\right)dt = xf\left(x\right) - \dfrac{{x}^{3}}{3}$

Now differentiating both side we get,

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

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

$⇒ {f}^{'}\left(x\right) - \dfrac{2}{x}f\left(x\right) = x$

Which is a linear differential equation,

$⇒ I . F . = {e}^{- \dfrac{2}{x}dx} = \dfrac{1}{{x}^{2}}$

Now solution is given by,

$⇒ y\left(\dfrac{1}{{x}^{2}}\right) = ∫ x \times \dfrac{1}{{x}^{2}}dx = ln x + C$

$⇒ y = {x}^{2} ( ln x + C )$

$⇒ f\left(x\right) = {x}^{2} ( \ln x + C )$

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

$⇒ 0 + C = \dfrac{1}{3} ⇒ C = \dfrac{1}{3}$

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

$⇒ f\left(e\right) = {e}^{2}\left(lne + \dfrac{1}{3}\right)$

$⇒ f\left(e\right) = \dfrac{4{e}^{2}}{3}$