Let-f-1-R-be-a-differentiable-function-such-that-f-1-1-3-and-3-1-x-f-t-dt-x-f-x-x-3-3-x-1-find-tbe-value-of-f-e-

Question Number 204645 by cortano12 last updated on 24/Feb/24

Let f : [\bar{1} \infty) \to R be a differentiable

function such that f (1) = \frac{1}{3} and

3 \underset{1}{\int^{x}} f (t) d t = x f (x) - \frac{x^{3}}{3}, x \in [1, \infty)

find tbe value of f (e)

Commented by universe last updated on 24/Feb/24

\frac{4 e^{2}}{3} ?

Answered by mr W last updated on 24/Feb/24

y = f (x)

3 y = y + {x y}^{'} - x^{2}

y^{'} - \frac{2}{x} y = x

- 2 \int \frac{d x}{x} = - 2 \ln x = \ln \frac{1}{x^{2}}

\frac{1}{x^{2}} y = \int (x \times \frac{1}{x^{2}}) d x + C = \ln x + C

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

\frac{1}{3} = 1^{2} (0 + C) \Rightarrow C = \frac{1}{3}

\Rightarrow y = f (x) = x^{2} (\ln x + \frac{1}{3})

\Rightarrow f (e) = e^{2} (\ln e + \frac{1}{3}) = \frac{4 e^{2}}{3}

Commented by cortano12 last updated on 25/Feb/24

yes