X follows an exponential distribution, calculate Expected value of sqrt(X). Problem: Let X follow an exponential distribution with expected value of 1. Define Y=sqrt(X). Calculate E(Y).
This is my first course in probability theory (5 weeks ≈ about 5*40 hours of workload) so the tools we have learned are not that many.
Tip: if X follows N(0,1) then E(X^2)=1.
Attempt: 
Correct answer it sqrt(Pi/2)
 A: An exponential distribution of a random variable $X$ of expected value $1$ (i.e., mean) has PDF $f_X(x) = e^{-x}$ for $x \gt 0$, and zero otherwise.  You want $E(\sqrt{X})$. 
In general, to compute $E[(g(X)]$ for a distribution $f_X(x)$ is
$$\int_{-\infty}^{\infty} dx \, g(x) f(x) $$
In your case, the integral is
$$\int_0^{\infty} dx \, \sqrt{x} \, e^{-x}$$
To evaluate the integral, sub $x=y^2$ and get
$$2 \int_0^{\infty} dy \, y^2 \, e^{-y^2} = \int_{-\infty}^{\infty} dy \, y^2 e^{-y^2}$$
The result is, in fact $\sqrt{\pi}/2$. 
A: If $X$ follows an exponential distribution with parameter $\lambda$, we know $E(X) = \frac{1}{\lambda}$. Because $E(X) = 1$, it is fairly simple to see $\lambda = 1$ as well.
Thus, the PDF of $X$ is given by
$$f(x;\lambda=1) = \left\{ \begin{array}{rl}
                           e^{-x} & x \geq 0\\
                           0 & \mbox{elsewhere}
                           \end{array} \right.$$
We know $$E(X) = \displaystyle\int_{\mathbb{R}} x\cdot f(x) dx = 0 + \displaystyle \int_0^\infty x\cdot e^{-x}dx.$$ The question is $E(Y)$ though. Hopefully we know that if $h(X)$ is any function of $X$ then,
$$E\left[h(X)\right] = \int_{\mathbb R} h(x) \cdot f(x) dx.$$
$Y = \sqrt X$ is a function of $X$, so what would $E(Y)$ be?
