Integration (9)

Calculus Level 4

Let $f\left( x \right)$ is a continuous function which takes only positive values for

$x\ge 0$ and satisfy $f\left( 1 \right) =0.5$

and $\int _{ 0 }^{ x }{ f\left( t \right) dt} =x\sqrt { f\left( x \right) }$ .

Then the value of $\frac { 1 }{ f\left( \sqrt { 2 } +1 \right) }$

1 solution

Tanishq Varshney
Jan 24, 2016

Using Newton leibnitz theorem

$\large{f(x)=\sqrt{f(x)}+x \frac{f^{\prime} (x)}{2 \sqrt{f(x)}}}$

let $f(x)=y$

$\large{\Rightarrow x \frac{d y}{dx}=2 \sqrt{y} (y- \sqrt{y})}$

Using variable separable

$\large{\int \frac{dy}{2 \sqrt{y}(y-\sqrt{y})}=\int \frac{dx}{x}}$

put $\sqrt{y}=t$

$\large{\ln \left(\frac{\sqrt{y}-1}{\sqrt{y}}\right)=\ln cx}$

where $c$ is constant of integration.

and $f(1)=0.5$ gives value of $c=1-\sqrt{2}$

$\Large{\frac{1}{f(x)}=(1+(\sqrt{2}-1)x)^2}$