Average Value of Functions

[paulxzt]

If a cup of coffee has temperature 95*C in a room where the temperature is 20*C, then, according to Newton's Law of Cooling, the temperature of the coffee after t minutes is T(t) = 20 + 75e^(-t/50). What is the avg. temperature of the coffee during the first half hour.

(1/30) * Integral of T(t) from 0 to 30.
Is this right? Do the first two temperatures matter in this problem?
Also, i forgot but how do u integrate 75+e^(-t/50) by hand?

thanks.

 

[skeeter]

yes ... the average value of a function is \(\displaystyle \L f_{avg} = \frac{1}{b-a} \int_a^b f(x) dx\)

\(\displaystyle \L \int 75 + e^{\frac{-t}{50}} dt = 75t - 50e^{\frac{-t}{50}} + C\)

 

[paulxzt]

yes ... the average value of a function is \(\displaystyle \L f_{avg} = \frac{1}{b-a} \int_a^b f(x) dx\)

\(\displaystyle \L \int 75 + e^{\frac{-t}{50}} dt = 75t - 50e^{\frac{-t}{50}} + C\)

So the 95*C and 20*C do not matter? the integral is from 0 to 30?
Also, I see how the integral of e^-t/50 is -50e^(-t/50) but could you show me how you do it by hand? Would I use integration by parts?

 

[galactus]

No, you don't need parts. A simple u sub will do.

Let \(\displaystyle \L\\u=\frac{-t}{50}, \;\ du=\frac{-1}{50}dt, \;\ -50du=dt\)

 

[skeeter]

If a cup of coffee has temperature 95*C in a room where the temperature is 20*C, then, according to Newton's Law of Cooling, the temperature of the coffee after t minutes is T(t) = 20 + 75e^(-t/50). What is the avg. temperature of the coffee during the first half hour.

(1/30) * Integral of T(t) from 0 to 30.
Is this right? Do the first two temperatures matter in this problem?
Also, i forgot but how do u integrate 75+e^(-t/50) by hand?

thanks.

the two temperatures are T(0) = 95 and limit{t->infinity} T(t) = 20.

why do you want to integrate 75+e^(-t/50) when the function is T(t) = 20 + 75e^(-t/50)?

the average temperature for the first 30 minutes is ...

\(\displaystyle \L T_{avg} = \frac{1}{30} \int_0^{30} 20 + e^{-\frac{t}{50}} dt\)