First, we consider the graph of the situation and estimate that the average should be around 14 to 16 degrees.

612-6-12510152025tTOpen image in a new page

The graph of temperature `T` at time `t`.

`T_["ave"]=(int_a^bf(x) text[d]x)/(b-a)`

`=(int_-12^12(0.001t^4-0.28t^2+25)\ dt)/(12-(-12))`

`=1/24[(0.001t^5)/(5)-(0.28t^3)/(3)+25t]_-12^12`

`=2/24[(0.001t^5)/(5)-(0.28t^3)/(3)+25t]_0^12`

`=1/12[49.7664-161.28+300]`

`=15.7 ^@ text[C]`

Our earlier estimate was quite reasonable.