Proving that $\mu_{X+Y}=\mu_X+\mu_Y$ Prove that $\mu_{X+Y}=\mu_X+\mu_Y$, where $X$ and $Y$ are independent variables and $\mu_X=\int_{-\infty}^\infty xf(x)dx$, $f$ is the pdf of $X$.  So does that mean that $\mu_{X+Y} = \int_{-\infty}^\infty(x+y)f(x+y)dx$?
 A: Are you sure you are trying to "prove" this?
An alternate approach is
\begin{align*}
\mu_X +\mu_Y &= E[X]+E[Y]\\
&= E[X+Y]\\
&=\iint(x+y)f_{X,Y}(x,y)dxdy\\
&=\mu_{X+Y}.
\end{align*}
Obviously, this is not a valid answer if they really wanted you to do all the integration. Maybe they just wanted you to see that connection.
A: Wait!!!
If you have two independent random variables, $X$ and $Y$, with distribution $f_X(x)$ and $f_Y(y)$, then the joint distribution is:
$$f_{X,Y}(x,y) = f_X(x)f_Y(y)$$
Then:
$$\mu_{X+Y} = \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}(x+y)f_{X,Y}(x,y)dxdy = \\
= \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}(x+y)f_{X}(x)f_{Y}(y)dxdy = \\
= \int_{-\infty}^{+\infty}f_{Y}(y)\int_{-\infty}^{+\infty}(x+y)f_{X}(x)dxdy = \\
= \int_{-\infty}^{+\infty}f_{Y}(y)\left[\int_{-\infty}^{+\infty}xf_{X}(x)dx+\int_{-\infty}^{+\infty}yf_{X}(x)dx\right]dy = \\
= \int_{-\infty}^{+\infty}f_{Y}(y)\left[\mu_X+y\cdot 1\right]dy = \\
= \int_{-\infty}^{+\infty}f_{Y}(y)\mu_Xdy + \int_{-\infty}^{+\infty}yf_{Y}(y)dy = \mu_X+\mu_Y.\\$$
At the end of the story, you also need the distribution of $Y$. If $X$ and $Y$ were not independent, then you also need their joint distribution. 
What you wrote is wrong. Indeed, notice that $f(x+y)$ does not make any sense since $f$ is the distribution of $X$ only.
A: This is the_candyman's proof rewritten for not-necessarily-independent random variables. If you have two random variables, $X$ and $Y$, with joint density $f_{X,Y}(x,y)$ and marginal densities $\displaystyle f_X(x)=\int_{-\infty}^{+\infty} f_{X,Y}(x,y)\, dy$ and $\displaystyle  f_Y(y)=\int_{-\infty}^{+\infty} f_{X,Y}(x,y)\, dx$
then
$$\mu_{X+Y} = \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}(x+y)f_{X,Y}(x,y)\, dx\, dy  \\
= \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}x f_{X,Y}(x,y)\,dy \, dx +\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} yf_{X,Y}(x,y) \, dx\,dy  \\
= \int_{-\infty}^{+\infty}x \left(\int_{-\infty}^{+\infty} f_{X,Y}(x,y)\,dy\right) \, dx +\int_{-\infty}^{+\infty}y \left(\int_{-\infty}^{+\infty} f_{X,Y}(x,y) \, dx\right)\,dy  \\
= \int_{-\infty}^{+\infty}x f_{X}(x) \, dx +\int_{-\infty}^{+\infty}y f_{Y}(y) \,dy  \\
= \mu_X+\mu_Y.\\$$
A: No, that is not exactly what is meant. The random variable $X+Y$ will have its own probability density function, say $h,$ and its mean is
$$\mu_{X+Y}=\int_{-\infty}^\infty xh(x)dx$$
