$X\sim N(67,4)$
What's the expected value of the portion of the curve $(X>72)$?
I tried to use the definition of expected value ($\int xf(x) \mathrm{d}x$), but my integral was far too complicated and when I tried to evaluate using WolframAlpha the answer was nonsensical...
How do I set up this problem?
Notations: Let $F$ and $f$ be the cdf and pdf of $X$ respectively. Let $\Phi$ and $\phi$ be the cdf and pdf of a standard normal variable.
Part 1: Justification
First of all can you calculate $P(a<X<b|X>72)$?
Yes you can! It is $\displaystyle \frac{P(a<X<b \cap X>72)}{P(X>72)}$.
Now you can treat $Y:=X|X>72$ as a random variable whose pdf (if exists) you would like to compute. First let us calculate the cdf (say $G(a)$).
$$G(a)=P(X\le a|X>72) =\frac{P(X\le a \cap X>72)}{P(X>72)}$$ Now make two cases. If $a\le 72$, then $X\le a \cap X>72 =\emptyset$. And if $a>72$, $$P(X\le a \cap X>72)=P(72<X\le a)=P(X\le a)-P(X\le 72)=F(a)-F(72)$$ Conclusion $$G(a)=\begin{cases}\frac{F(a)-F(72)}{P(X>72)} & a>72\\ 0 & \mbox{otherwise}\end{cases}$$ Hence this random variable $Y$ admits a pdf $g$ given by $$g(a)=\begin{cases}\frac{f(a)}{P(X>72)} & a>72\\ 0 & \mbox{otherwise}\end{cases}$$
Part 2: Calculation
$$\begin{aligned} & E(X|X>72) =\int_{-\infty}^{\infty}xg(x)\,dx \\ & = \frac1{P(X>72)}\int_{72}^{\infty} x\frac1{2\sqrt{2\pi}}\exp\left(-\frac12\left(\frac{x-67}{2}\right)^2\right)\,dx \\ & = \frac{\int\limits_{72}^{\infty} \frac{x-67}{2}\frac{1}{\sqrt{2\pi}}\exp\left(-\frac12\left(\frac{x-67}{2}\right)^2\right)\,dx+67\int\limits_{72}^{\infty}\frac1{2\sqrt{2\pi}}\exp\left(-\frac12\left(\frac{x-67}{2}\right)^2\right)\,dx}{P(X>72)} \\ & = \frac1{P(X>72)}\int\limits_{\frac{(2.5)^2}2}^{\infty} 2\exp(-t)\,dt +67 \\ & =67+\frac{2e^{-\frac{(2.5)^2}2}}{1-\Phi(2.5)} \end{aligned}$$ In the third line $\frac12\left(\frac{x-67}{2}\right)^2=t$ substitution was used. There may computational mistakes, but the trick is essentially present above.
Part 3: Approximation
You can use the following fact $$\lim_{x\to\infty} \frac{1-\Phi(x)}{\phi(x)/x}=1$$ Hence for large enough $x$, $1-\Phi(x) \approx \frac{\phi(x)}{x}$. Replacing this in the dinominator of the last answer, gives a good approximation.