00:01
Hello students, it is given that t of t equal to minus t square plus 2t plus 38 and t equal to 9rs.
00:10
First we need to find the average temperature.
00:15
So average temperature equal to integral 0 to 9 t of t into t divided by t minus 0.
00:28
So which is equal to integral 0 to 9 minus t square plus 2t plus 38 into dt divided by 9 which is equal to minus t cube divided by 3 plus 2t square divided by 2 plus 38 t into 0 to 9 divided by 9.
00:58
Now substitute upper limit and lower limit.
01:01
So we get 1 divided by 9 into minus 243 plus 81 plus 342 which is equal to 180 divided by 9.
01:18
So the average temperature is temperature equal to 20 degree celsius.
01:39
For maximum and minimum differentiate t of t with respect to t.
01:45
T of t with respect to t we get t dash of t which is equal to minus 2t plus 2.
02:00
Consider this is equation 1.
02:03
Critical point t dash of t equal to 0.
02:14
So minus 2t plus 2 minus 2t plus 2 equal to 0 minus 2t equal to minus 2...