Question

Let $f:R\to R$ be a continuous function which satisfies $f\left(x\right)={\int }_0^{x}f\left(t\right)dt$ Then the value of $f\left(ln5\right)$ is

Answer 1

$f\left(x\right)={\int }_0^{x}f\left(t\right)dt$
Substitute $x=0$, then integral of $f\left(t\right)$ from $0$ to $0$ will become $0$.
$\Rightarrow f\left(0\right)=0$
Now, differentiate both side of given integral
$f^{{}^{\prime }}\left(x\right)=f\left(x\right),x>0$
$\Rightarrow f\left(x\right)=k{e}^x,x>0$
$\because f\left(0\right)=0$ and $f\left(x\right)$ is continuous
$\Rightarrow f\left(x\right)=0$ $x>0$
$\therefore f\left(ln5\right)=0$