Discussion Forum

Let  $f: [1,\infty) \rightarrow (2, \infty)$  be a differentiable function such that /(1) = 2. If $6\int_{1}^{x} f(t)dt=3x f(x)-x^{3}$  for all $x  \geq 1$  , then the value of f(2) is 

Answer:

Explanation:

 Given, $f(1)= \frac{1}{3}$ and  $6\int_{1}^{x} f(t) dt$

  = $3x f(x)-x^{3}$ for all x $\geq$ 1

 Using (Newton-Leibnitz formula),

On Differentiating both sides,

 $6 f(x).1-0-3f(x)+3x f'(x)-3x^{2}$

$\Rightarrow$    $3 x f'(x)-3f(x)=3x^{2}$

$\Rightarrow$   $f'(x)-\frac{1}{x} f(x)=x$

 $\Rightarrow$   $\frac{x f'(x)-f(x)}{x^{2}}=1 \Rightarrow  \frac{d}{dx}\left\{\frac{f(x)}{x}\right\}=1$

 On integreating  both sides,

 $\frac{f(x)}{x}=x+C$                $[ \because f(1)=\frac{1}{3}]$

 $\Rightarrow$  $\frac{1}{3}=1+C \Rightarrow C=-\frac{2}{3}$

 Now,  $f(x)=x^{2}-\frac{2}{3} x$

 $\Rightarrow$    $f(2)=4-\frac{4}{3}=\frac{8}{3}$

Note      Here, f(1) =2  does not satisfy given function

 $\therefore$    $f(1)=\frac{1}{3}$

 For that  , $f(x)=x^{2}-\frac{2}{3} x$

 and $f(2)=4-\frac{4}{3}= \frac{8}{3}$