Intuitive explanation of 'Normal' integral

Question

For $f$ given as (standard normal density) \begin{equation} f(x) = \frac{1}{{\sqrt {2\pi } }}{e^{ - \frac{{{x^2}}}{2}}}, \end{equation} the follwing holds \begin{equation} \int\limits_a^\infty {xf(x)dx} = f(a). \end{equation} This result isn't hard to derive. However, I'm having a difficulty to intuitively understand why 'expected' value of $X\sim N(0,1)$ over a region is equal to the density at the lower bound. I know it isn't a true expected value but still. Why a result of this integral is equal to $f(a)$?

Answer 1

A trick is that $f(x)=\frac1{\sqrt{2\pi}}e^{-x^2/2}$ satisfies the differential equation $f'(x)=-xf(x)$ and its limit as $x\to\infty$ is $0$ (this condition is already a consequence of being a solution of the differential equation, but there's no harm in mentioning it). Therefore $\int_x^\infty tf(t)\,dt=f(x)$.