Consider the following function, which describes the fraction of evaporated substance, over time $$ f(t)=1-e^{-a t^2} $$ where $a\in\mathbb{R}^+$. I am asked to deduce the expected time of evaporation, at any point in the substance.
First, I interpreted this as the time it takes for half of the substance to evaporate, which corresponds to the median time $t_{1/2}$, given by solving $$ f(t_{1/2})=0.5\Leftrightarrow t_{1/2}=\sqrt{\frac{\log(2)}{a}}. $$ However, I am told the answer should be $$ t_E=\frac12\sqrt{\frac{\pi}{a}} $$ but I do not really see how I can calculate this from the expression of $f$, given it's a sigmoid approaching $1$. Any ideas?
Edit: I have noticed that, with $f'(t)=2at e^{-a t^2}$, we have $$ t_E=\int_0^\infty t f'(t)\,dt=2a\int_0^\infty t^2 e^{-a t^2}\,dt=\frac12\sqrt{\frac{\pi}{a}}. $$ Is this what they meant?