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$\large X \sim exp(\lambda)$

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. $\wedge$
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