I'm having trouble getting the answer here.

$\displaystyle g(X)=3X-2$

Density function:

$\displaystyle f(x)=\left\{\begin{array}{cc}\frac{1}{4}e^\frac{-x}{4},&\mbox{ if }x>0 \\0, & \mbox{ elsewhere }\end{array}\right.$

I used:

$\displaystyle var(g(X))=E(g(X)^2)-[E(g(X))]^2$

So I got:

$\displaystyle E(g(X))= \int g(X)f(x)=\int(3X-2)(\frac{1}{4}e^\frac{-x}{4})$

$\displaystyle =\int\frac{3}{4}Xe^\frac{-x}{4}-\int\frac{1}{2}e^\frac{-x}{4}$

$\displaystyle =\frac{3}{4}[-4Xe^\frac{-x}{4}+4\int e^\frac{-x}{4}dx]-\int\frac{1}{2}e^\frac{-x}{4}$

$\displaystyle =\frac{3}{4}[-4Xe^\frac{-x}{4}-16 e^\frac{-x}{4}]+2e^\frac{-x}{4}$

$\displaystyle =-3Xe^\frac{-x}{4}-10e^\frac{-x}{4}\mid \begin{array}{cc}\infty\\0\end{array}$

$\displaystyle =0-(-10)=10$

Nevermind, I think I've got it right but if someone would double check my work that'd be great. Thanks.

However, the answer should be 10, so the signs are switched up somewhere in there but I can't find where.Apparently, this first part of the problem is the only part that's wrong. If someone could help me, that'd be awesome. After typing the second part out and redoing the math, see that it's right.

$\displaystyle E(g(X)^2)= \int(3X-2)^2 f(x) = \int(9X^2 -12X+4)(\frac{1}{4}e^\frac{-x}{4})$

$\displaystyle =\frac{9}{4}\int X^2 e^\frac{-x}{4}-3\int Xe^\frac{-x}{4}+\int e^\frac{-x}{4}$

$\displaystyle =\frac{9}{4}[X^2 e^\frac{-x}{4}+4\int 2Xe^\frac{-x}{4}]-3\int Xe^\frac{-x}{4}+\int e^\frac{-x}{4}$

$\displaystyle =\frac{9}{4}[-4X^2 e^\frac{-x}{4}+8[-4Xe^\frac{-x}{4}+4\int e^\frac{-x}{4}]]-3\int Xe^\frac{-x}{4}+\int e^\frac{-x}{4}$

$\displaystyle =\frac{9}{4}[-4X^2 e^\frac{-x}{4}+8[-4Xe^\frac{-x}{4}-16e^\frac{-x}{4}]]-3\int Xe^\frac{-x}{4}+\int e^\frac{-x}{4}$

$\displaystyle =\frac{9}{4}[-4X^2 e^\frac{-x}{4}+8[-4Xe^\frac{-x}{4}-16e^\frac{-x}{4}]]-3[-4Xe^\frac{-x}{4}+4\int e^\frac{-x}{4}]+\int e^\frac{-x}{4}$

$\displaystyle =\frac{9}{4}[-4X^2 e^\frac{-x}{4}+4[-8Xe^\frac{-x}{4}-16e^\frac{-x}{4}]]-3[-4Xe^\frac{-x}{4}-16e^\frac{-x}{4}]+\int e^\frac{-x}{4}$

$\displaystyle =\frac{9}{4}[-4X^2 e^\frac{-x}{4}+8[-4Xe^\frac{-x}{4}-16e^\frac{-x}{4}]]-3[-4Xe^\frac{-x}{4}-16e^\frac{-x}{4}]-4e^\frac{-x}{4}$

$\displaystyle =-9X^2e^\frac{-x}{4}-72Xe^\frac{-x}{4}-288e^\frac{-x}{4}+12Xe^\frac{-x}{4}+48e^\frac{-x}{4}-4e^\frac{-x}{4}$

$\displaystyle =-9X^2e^\frac{-x}{4}-60Xe^\frac{-x}{4}-244e^\frac{-x}{4}\mid \begin{array}{cc}\infty\\0\end{array}$

$\displaystyle =0-(0-0-244)=244$

This part is correct because the final answer is supposed to be 144

So finally:

$\displaystyle var(g(X))= 244-(-10)^2= 144$

It's the right answer, but for the first part it should be 10, not-10.