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

水杯的温度为98℃，环境温度是C，根据牛顿定律，杯中的水温度与时间的关系T(t) = 18 + 80e^(-t/50)
a、前25分钟水杯的平均温度？
平均温度*1500=
从0到1500得定积分T(t) = 18t - 80*50e^(-t/50)
=18*1500-[400*exp(-1500/50)-(0-80*50*exp^0]=31000
平均温度=31000/1500=20.66666667
t1=0
t2=2500
b、什么时候水杯降低到平均温度？
20.66666667=18 + 80e^(-t/50)
t=-50*ln[(20.667-18)/80]=170s

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