Suppose a 1000 degree liquid is cooling, at the rate of $ C = 22 + 900 e^{-0.3t}$. So after 50 minutes I'd expect $ C = 22 + 900e^{-15}$ = 22. But what is the average value of the temperature over that 50 minutes? I came up with
$ 1/50 ∫_0^{50}\:22\:+\:900e^{-0.3t}dx = 82$
Is this correct?
You are very much correct:
$$T_{avg} = {\sum T(t)\over \triangle t}$$ $$=\displaystyle\int_0^{50} 22+e^{-0.3t}dt\over\displaystyle\int_0^{50} dt$$
$$=22 - {(900)(e^{-15}-e^0)\over50\cdot 0.3}$$
Similarly taking $e^{-15}$ to be negligible
$$= 22 + 60$$
$$\boxed{T_{avg} = 82}$$
However as some commenters have mentioned, the units of $\tau = 0.3$ and $t$ are of importance.