\[\begin{align*} E(X) &= \int_0^{\infty} x \lambda e^{-\lambda x} dx\\ \text{let: } t &= \lambda x \\ \therefore x &= \frac{t}{\lambda} \\ \therefore dx &= \frac{dt}{\lambda} \\ \therefore E(X) & = \int_0^\infty \frac{te^t}{\lambda}dt \\ &= \lambda\int_0^{\infty} te^{-t}dt \\ &= \frac{1}{\lambda}\Gamma(2) \\ &= \frac{1}{\lambda} \end{align*}\]
\[\begin{align*} E(X^2) &= \int_0^{\infty} x^2 \lambda e^{-\lambda x} dx\\ \text{let: } t &= \lambda x \\ \therefore x &= \frac{t}{\lambda} \\ \therefore dx &= \frac{dt}{\lambda} \\ \therefore E(X^2) & = \int_o^\infty x (\lambda x)e^{-\lambda x} dx\\ &= \int_0^\infty \frac{t}{\lambda} \frac{te^t}{\lambda}dt \\ &= \frac{1}{\lambda^2} \int_0^\infty t^2e^tdt \\ &= \frac{\Gamma(3)}{\lambda^2} \\ &= \frac{2}{\lambda^2} \\ \therefore Var(X) &= \frac{2}{\lambda^2} - \left(\frac{1}{\lambda}\right)^2 \\ &= \frac{2}{\lambda^2} - \frac{1}{\lambda^2} \\ &= \frac{1}{\lambda^2} \end{align*}\]
\[\begin{align*} M_X(t) &= \int_0^{\infty} e^{tx} \lambda e^{-\lambda x} dx\\ &= \lambda \int_0^\infty e^{x(t-\lambda)}dx \\ &= \lambda \left[ \frac{e^{x(t-\lambda)}}{(t-\lambda)} \right]_0^\infty \\ &= \lim_{x\to\infty} \left[\frac{\lambda}{t-\lambda} e^{x(t-\lambda)} \right] - \frac{\lambda}{t - \lambda}e^{0(t-\lambda)} \\ &= \frac{\lambda}{t-\lambda}\left[\lim_{x\to\infty} e^{x(t-\lambda)} -1 \right] \\ \end{align*}\] Note that \(t \ne \lambda\), or else \(M_x(t)\) becomes undefined. If \(t>\lambda\), then \(\lim_{x\to\infty} \exp\left(x(t-\lambda)\right) = \infty\).
This implies the \(t<\lambda\): \[\begin{align*} &= \frac{\lambda}{t-\lambda}\left[0-1 \right] \\ &= \frac{-\lambda}{t-\lambda} \\ &= \frac{\lambda}{\lambda - t}, t< \lambda \end{align*}\]