Question

Let $f:R\to R$ be a continuous function satisfying $f\left(x\right)=\underset{0}{\overset{x}{\int }}f\left(t\right)dt.$ Then the value of $f\left(1\right)$ is

• A. $0$
• B. $e$
• C. $1$
• D. $4$

Answer 1

The correct option is A $0$

$f\left(x\right)=\underset{0}{\overset{x}{\int }}f\left(t\right)dt$
$\Rightarrow f^{\prime }\left(x\right)=f\left(x\right)$
$\Rightarrow \int \frac{f^{\prime }\left(x\right)}{f\left(x\right)}dx=\int dx$
$\Rightarrow \ln f\left(x\right)=x+\ln C$ where $C$ is a constant.
$\Rightarrow f\left(x\right)=C{e}^x$

$f\left(0\right)=0\Rightarrow C=0$
$\therefore f\left(x\right)=0\forall x\in R$
So, $f\left(1\right)=0$