Exponential distribution moment generating function to find the mean With mean = 2 with exponential distribution
Calculate 
$ E(200 + 5Y^2 + 4Y^3) = 432 $
$E(200) = 200 $
$E(5Y^2) = 5E(Y^2) = 5(8) = 40 $
$E(4Y^3) = 4E(Y^3) = 4(48) = 192 $
$E(Y^2) = V(Y) + [E(Y)]^2 = 2^2+2^2$
$E(Y^3) = m_Y^3(0) = 48(1-2(0))^{-4} = 48$
is this right?
 A: Your work is correct.  I'm guessing you got your computation for the third moment by differentiating the moment generating function; it might be worth making that explicit if that's what you did.
For the exponential distributed random variable $Y$, one can show that the moments $E(Y^n)$ are $\displaystyle E(Y^n) = \frac{n!}{\lambda^n}$ where $\displaystyle E(Y) = \frac{1}{\lambda}$. If $E(Y) = 2$, then $\lambda = \frac{1}{2}$ and \begin{eqnarray} E(200+5Y^2+4Y^3) &=& 200 + 5E(Y^2) + 4E(Y^3)\\ &=& 200 + 5\cdot\frac{2!}{(\frac{1}{2})^2} + 4\cdot\frac{3!}{(\frac{1}{2})^3}\\ &=& 200 + 5\cdot 2\cdot 4 + 4\cdot 6 \cdot 8 = 432.\end{eqnarray} 
A: Let $Y\sim\operatorname{Exp}(\lambda)$ that is with pdf
$$
f(y)=
\begin{cases}
\lambda \operatorname{e}^{-\lambda y} & y\ge 0 \\
0 & y<0
\end{cases}
$$
with parameter $\lambda>0$.
The moments can be evaluated directly or using the moment generating function.
Direct Evaluation
$$
\begin{align}
\Bbb{E}(Y^n)&=\int_{-\infty}^{\infty}y^n f(y)\operatorname{d}y=\int_{0}^{\infty}y^n \lambda \operatorname{e}^{-\lambda y}\operatorname{d}y\\
&=\frac{1}{\lambda^n}\int_{0}^{\infty}t^n  \operatorname{e}^{-t}\operatorname{d}t\qquad\qquad\qquad\qquad\qquad(\text{putting }\lambda y=t)\\
&=\frac{1}{\lambda^n}\Gamma(n+1)=\frac{n!}{\lambda^n}\tag 1
\end{align}
$$
using the Gamma Function $\Gamma(x) = \int_0^\infty t^{x-1} {\rm e}^{-t}\,{\rm d}t$ and that $\Gamma(n+1)=n!$ for any positive integer $n$.
Using the Moment Generating Function
The moment generating function (mgf) is
$$
\begin{align}
M_Y(t)=\Bbb{E}\left(\operatorname{e}^{tY}\right)&=\int_{-\infty}^{\infty} \operatorname{e}^{ty}f(y)\operatorname{d}y=\int_{0}^{\infty} \operatorname{e}^{ty}\lambda \operatorname{e}^{-\lambda y}\operatorname{d}y\\
&=\lambda\int_{0}^{\infty} \operatorname{e}^{(t-\lambda) y}\operatorname{d}y=\frac{\lambda}{t-\lambda}\left[\operatorname{e}^{(t-\lambda) y}\right]_{0}^{\infty}=\frac{\lambda}{\lambda-t}\hspace{1cm}\text{for } t<\lambda
\end{align}
$$
Recalling the geometric series
$$
\frac{1}{1-z}=\sum_{n=0}^{\infty}z^n \hspace{1cm}\text{for } |z|<1
$$
we can expand the mgf for $t<\lambda$ as
$$
M_Y(t)=\frac{\lambda}{\lambda-t}=\frac{1}{1-\frac{t}{\lambda}}=\sum_{n=0}^{\infty}\frac{1}{\lambda^n}t^n\tag 2
$$
Observing that 
$$
M_Y(t)=\sum_{n=0}^{\infty}\frac{M^{(n)}_Y(0)}{n!}t^n=\sum_{n=0}^{\infty}\frac{\Bbb{E}(Y^n)}{n!}t^n\tag 3
$$
equating the coefficients of (2) and (3) we find
$$
M^{(n)}_Y(0)=\Bbb{E}(Y^n)=\frac{n!}{\lambda^n}.\tag 4
$$
Solution
Let $Z=200+5Y^2+4Y^3$ with $Y\sim\operatorname{Exp}(\lambda)$. If the mean of $Y$ is $2$, from (1) or (4) we have
$$
\Bbb{E}(Y)=\frac{1}{\lambda}=2
$$
so that $\lambda=\frac{1}{2}$ and $Y\sim\operatorname{Exp}\left(\frac{1}{2}\right)$.
From (1) or (4) we can evaluate the moments $\Bbb{E}(Y^n)=\left(\frac{1}{\lambda}\right)^n n!$ for $n=2$ and $n=3$
$$
\begin{align}
\Bbb{E}(Y^2)&=\left(\frac{1}{\lambda}\right)^2 2!=2^2\cdot 2=8\\
\Bbb{E}(Y^3)&=\left(\frac{1}{\lambda}\right)^3 3!=2^3\cdot 6=48\\
\end{align}
$$
By the linearity of the expectation, the mean of $Z$ is
$$
\begin{align}
\Bbb{E}(Z)&=\Bbb{E}(200+5Y^2+4Y^3)\\
&=200+5\Bbb{E}(Y^2)+4\Bbb{E}(Y^3)\\
&=200+5\cdot 8+ 4\cdot 48\\
&=432
\end{align}
$$
