Probability Question finding variance from marginal Y... 
Hi I was trying to find the variance of marginal $Y$..
And I was trying to find $E(Y^2)$ first but.. it gets nasty...
Am I solving this right way???
 A: We must solve $\int_0^{\infty} 4y^3 e^{-2y}dy=4\int_0^{\infty}y^3e^{-2y}dy$
Let $u=y^3, v=-\frac{1}{2}e^{-2y}, du=3y^2,dv=e^{-2y}$
Then $\int_0^{\infty}y^3e^{-2y}dy=-\frac{1}{2}y^3e^{-2y}+\frac{3}{2}\int y^2e^{-2y}dy$
Let $u=y^2, v=-\frac{1}{2}e^{-2y}, du=2y,dv=e^{-2y}$
Then $\int y^2e^{-2y}dy=-\frac{1}{2}y^2e^{-2y}+\int ye^{-2y}dy$
Let $u=y, v=-\frac{1}{2}e^{-2y}, du=1,dv=e^{-2y}$
Then $\int ye^{-2y}dy=-\frac{1}{2}ye^{-2y}-\int -\frac{1}{2}e^{-2y}dy$
which is finally an integral we can compute without integration by parts.
All in all we have,
$$4\left(-\frac{1}{2}y^3 e^{-2y}+\frac{3}{2}\left(-\frac{1}{2}y^2e^{-2y}-\frac{1}{2}ye^{-2y}-\frac{1}{4}e^{-2y}\right)\right)$$
This gets simplified to
$$-2y^3 e^{-2y} -3y^2e^{-2y}-3ye^{-2y}-\frac{3}{2}e^{-2y}$$
Then we compute the boundaries. Clearly, when $y$ approaches $\infty$, this equals $0$. 
When $y$ approaches $0$, everything is $0$ except for $-\frac{3}{2}e^{-2y}$ and when $y$ approaches $0$, the limit is $-\frac{3}{2}$. Thus,
$$\int_0^{\infty} 4y^3 e^{-2y}dy=\frac{3}{2}$$
Next, we must calculate $E(Y)$. We have
$$E(Y)=\int_0^{\infty} 4y^2e^{-2y}dy=4\int_0^{\infty} y^2e^{-2y}dy$$
Let $u=y^2, v=-\frac{1}{2}e^{-2y}, du=2y,dv=e^{-2y}$
Then $\int_0^{\infty} y^2e^{-2y}dy=-\frac{1}{2}y^2e^{-2y}+\int ye^{-2y}dy$
Let $u=y, v=-\frac{1}{2}e^{-2y}, du=1,dv=e^{-2y}$
Then $\int ye^{-2y}dy=-\frac{1}{2}ye^{-2y}-\int -\frac{1}{2}e^{-2y}dy$
which is finally an integral we can compute without integration by parts.
All in all we have,
$$4\left(-\frac{1}{2}y^2e^{-2y}-\frac{1}{2}ye^{-2y}-\frac{1}{4}e^{-2y}\right)$$
Then we compute the boundaries. Clearly, when $y$ approaches $\infty$, this equals $0$. 
When $y$ approaches $0$, the limit is $-1$. Thus, 
$$\int_0^{\infty} 4y^2 e^{-2y}dy=1$$
Finally,
$$Var(Y)=E(Y^2)-E(Y)^2=\frac{3}{2}-1^2=\frac{1}{2}$$
A: It seems that you want to evaluate
\begin{align}
\int_0^\infty 4y^3\exp(-2y) \, dy &= 2\int_0^\infty y^3[2\exp(-2y)] \, dy \\
&= 2 \frac{3!}{2^3}\\
&=\frac{3!}{2^2}\\
&=\frac32
 \end{align}
where I have used the property that the $n$-th moment of the exponential distribution is $\frac{n!}{\lambda^n}$.
Remark:
Your working is fine though it's a bit long but it's a good practice of integration by parts. You might like to evaluate the limit for those terms that are outside integral earlier to set them to zero and you have less things to worry about. Your mistake at the end seems to be you have forgotten about the multiple of $4$ in the integral.
Now that you have found $E[Y^2]$, give it a try to find $E[Y]$ and find $V(Y)$.
