I was usually given two terms in an exponential distribution which I combined and performed an integral to find the moment generating function.
What should I do here?
Let $X$~$Exponential(\lambda)$ for some $\lambda > 0$.
(a) Compute the moment generating function of $X$.
(b) Compute $E(X^n)$, $n\geq 1$.
(c) Let $Y=\lambda X$. Compute the moment generating function of $Y$.
On
(a) The moment generating function of $X \sim Exp(\lambda)$ is $M_X(t) = E[e^{t X}] = \int_{x \geq 0} \lambda e^{- \lambda x} e^{t x} dx = \frac{\lambda}{\lambda-t}$ for $t<\lambda$.
(b) To get the moments, note that $\frac{\lambda}{\lambda-t} = \lambda \frac{1}{1-\frac{t}{\lambda}} = \lambda \sum_{i=0}^\infty \frac{t^i}{\lambda^i}$ in a neighborhood of zero ($\sum_{i=0}^\infty x^i = \frac{1}{1-x}$). Compare this to $M_X(t) = \sum_{i=0}^\infty \frac{t^n E[X^i]}{i!}$ to read off the moments.
As for (c), note that $M_Y(t) = E[e^{t Y}] = E[e^{t \lambda X}] = E[e^{(t \lambda) X}] = M_X(t \lambda) = \frac{\lambda}{\lambda-t \lambda} = \frac{1}{1-t}$. Match this to the form of $M_X(t)$ for a particular value of $\lambda$ to see what distribution it is.
You know that $M_X(s) = \mathsf E(\boxed ?)$ is the moment generating function.
As $X\sim\mathcal {Exp}(\lambda) \implies f_X(x) = \boxed ?\mathbf 1_{x\in\boxed ?}$ is the density function for an exponential distribution then:
$$\begin{align} M_X(s) = & ~ \int\limits_{\Bbb R} \boxed ?~f_X(x)\operatorname d x \\[1ex] = & ~ \int\limits_{\boxed ?}\boxed ?~\boxed ?\operatorname d x\end{align}$$
Fill in the missing details.
Do the same for the others.