Prove that the covariance of $(X_{2}-X_{bar}, X_{bar})$ Let $(X_{1},...,X_{n})$ be a normal distribution $N(\mu, \sigma^2)$.
Firstly, by $X_{bar}$ I mean the average of the random sample so we know that the distribution for that part will change into  $N(\mu, \sigma^2/n)$.
So I know that the formula needed $Cov(X,Y)=E[XY]-E[X]E[Y]$.  I can't seem to understand how to get $X_{2}$ or why its even significant in this regard.  Can someone clarify that for me?  
From there we just take the integral to get EX, EY and EXY, right?  In that regard, is the integral from negative to positive infinity?  And if so, how would you take it?  I ask because it seems like you have the exponential term in the pdf which can prove quite troublesome if you don't restrict it somehow.  Can anyone offer some tips?

By linearity, our covariance is $\text{Cov}(X_2,\bar{X})-\text{Cov}(\bar{X},\bar{X})$.
The second term is known to you, it is the variance of $\bar{X}$.
For the first term, again expand. We get
$$\sum_1^n \text{Cov}(X_2, X_i/n).$$
Almost all the terms are $0$, by independence. The only surviving term is the covariance of $(X_2,X_2/n)$.
