题目
一道英文的微积分题目?
If a cup of tea has temperature 980C in a room where the temperature is C,then according to Newton's Law of Cooling the temperature of the tea after t minutes is,T(t) = 18 + 80e^(-t/50)
(a) What is the average temperature of the tea during the first 25 minutes?
(b) At what time does the tea reach the average temperature that you found in (a)?
温度一个是98度，一个是18度
If a cup of tea has temperature 980C in a room where the temperature is C,then according to Newton's Law of Cooling the temperature of the tea after t minutes is,T(t) = 18 + 80e^(-t/50)
(a) What is the average temperature of the tea during the first 25 minutes?
(b) At what time does the tea reach the average temperature that you found in (a)?
温度一个是98度，一个是18度
提问时间：2021-04-04
答案
平均值=曲边梯形面积/底边长
曲边梯形面积=T积分(t) = 18t + 4000(1-e^(-t/50))
(a):
平均值=曲边梯形面积/底边长=T积分(25)/25=(18*25 + 4000(1-e^(-25/50)))/25=……（用计算器算出）
(b):
令已求出的平均值=T(t) = 18 + 80e^(-t/50),反解出t=……即可
曲边梯形面积=T积分(t) = 18t + 4000(1-e^(-t/50))
(a):
平均值=曲边梯形面积/底边长=T积分(25)/25=(18*25 + 4000(1-e^(-25/50)))/25=……（用计算器算出）
(b):
令已求出的平均值=T(t) = 18 + 80e^(-t/50),反解出t=……即可
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