To find the average temperature, we will integrate the temperature function over the interval 18 hour period (that is from 0 to 18)
[tex]\begin{gathered} Average\text{ value = }\int_0^{18}(50\text{ + 6t - }\frac{1}{2}t^2)dt \\ =\int_0^{18}50\text{ dt+ 6tdt - }\frac{1}{2}t^2dt \\ =[50t\text{ + }\frac{6t^2}{2}\text{- }\frac{1}{2}\frac{t^3}{3}]\text{ interval 0 to 18} \\ =[50t\text{ + 3t}^2\text{ - }\frac{t^3}{6}]\text{ interval 0 to 18} \end{gathered}[/tex][tex]\begin{gathered} substitute\text{ the intervals:} \\ =\text{ \lbrack50\lparen18\rparen+ 3\lparen18\rparen}^2\text{ - \lparen}\frac{18^3}{6})]\text{ - \lbrack50\lparen0\rparen+ 3\lparen0\rparen}^2\text{ - \lparen}\frac{0^3}{6})]\text{ } \\ =\text{ 900 + 972 - 972 - 0} \\ Average\text{ value }=\text{ 900} \end{gathered}[/tex]The average temperature will be the average value gotten above divided by the time period
[tex]\begin{gathered} Average\text{ temperature = }\frac{900}{(18-0)}\text{ = }\frac{900}{18} \\ \\ Average\text{ temperature = 50\degree} \end{gathered}[/tex]