Mathematics - Definite Integration Question with Solution | TestHub

Since, $f(x)={\int }_a^x\frac{1}{f(x)}dx$.

Differentiating both sides with respect to $x$, then

$f^{\prime }(x)=\frac{1}{f(x)}\Rightarrow 2f(x)f^{\prime }(x)=2$.

Integrating both sides, then

$(f(x){)}^2=2x+c$.

$f(x)=\sqrt{(2x+c)}$.

But ${\int }_0^1\frac{1}{f(x)}dx=\sqrt{2}$.

And $f(1)={\int }_a^1\frac{1}{f(x)}dx=\sqrt{2}$.

$\sqrt{(2+c)}=\sqrt{2}$.

$c=0$.

Then, $f(x)=\sqrt{2x}$.