$\large X \sim exp(\lambda)$
Its pdf is
$f(x) = \left\{\begin{matrix} \lambda e^{-\lambda x} , & x \geq 0 \\ 0, & x < 0 \end{matrix}\right.$
Then mean of exponential distribution,
$E[X] = \int_{- \infty}^{\infty} x f(x)dx$
$E[X] = \int_{0}^{\infty} x \ \lambda \ e^{-\lambda x} dx$
$E[X] = $$ \large \int_{0}^{\infty} \frac{\lambda x}{e^{\lambda x}} dx$
$E[X] =$$ \large \left [ - \frac{e^{\lambda\left ( -x \right )}\left ( \lambda x + 1 \right )}{\lambda} \right ]^{\infty} _ 0$
$E[X] =$$ \large \frac{1}{\lambda}$
No, value of pdf and $E[X]$ is not same. $\wedge$