Average kinetic energy of a damped oscillator

[MuIotaTau]

For a damped mechanical oscillator, the energy of the system is given by $$E = \frac{1}{2}m \dot{x}^2 + \frac{1}{2}k x^2$$ where $k$ is the spring constant. From there, I've seen it dictated that the average kinetic energy $\langle T \rangle$ is half of the total energy of the system. This makes sense, since the energy sort of "sloshes" back and forth between kinetic and potential energy, but is there a more formal way to show this is true?

 

[Simon Bridge]

if you have $y=f(x)$ then the average value of $y$ in $a<x<b$ is the height of a rectangle with the same area as the graph of $y$ vs $x$ in that region. $$y_{avg}=\frac{1}{b-a}\int_a^b f(x)\;\text{d}x$$

 

[MuIotaTau]

Oh, that's simple! Thank you!