Average Value of a Function - temperature

[Slimsta]

Homework Statement

If a cup of tea has temperature 98C in a room where the temperature is 20C, then according to Newton's Law of Cooling the temperature of the tea after t minutes is
$\displaystyle \Large T(t)=20 + 78 e^{-t/50}.$
What is the average temperature of the tea during the first 38 minutes?

Homework Equations

$ fave ={1}/{b-a} \int _a^bf(x)dx$

The Attempt at a Solution

$ fave ={1}/{98-20} \int _{20}^{98} T(t)dt$ ${fave} =\frac{1}{78}\int _{20}^{98} 20 + 78 e^{-t/50}dt?$

 

[Dick]

Well, no. It's integral from t=0 to t=38. Why are you putting the bounds for T in instead of t? It's an average temperature over a time interval.

 

[Slimsta]

Well, no. It's integral from t=0 to t=38. Why are you putting the bounds for T in instead of t? It's an average temperature over a time interval.

so <br /> ${fave} =\frac{1}{38}\int _{0}^{38} 20 + 78 e^{-t/50}dt?$<br />

i don't understand your question.. how should i construct the question?
in my textbook they wrote like 2 sentences about temperature and i don't get it.. they don't even have an example for this type of questions

 

[Dick]

That's fine. What's your question? You want to average T(t) from t=0 to t=38. That's correct. You just put the T limits in instead of t. That's wrong.

 

[Slimsta]

That's fine. What's your question? You want to average T(t) from t=0 to t=38. That's correct. You just put the T limits in instead of t. That's wrong.

${fave} =\frac{1}{38}\int _{0}^{38} 20 + 78 e^{-t/50}dt$
my problem is to find the antiderivative of this i think.
would it be:
${fave} =\frac{1}{38} (\int _{0}^{38} 20dt + \int _{0}^{38}78 e^{-t/50}dt )$
${fave} =\frac{1}{38} (20t+C|^3^8 + \int _{0}^{38}78 e^{-t/50}dt )$
${fave} =\frac{1}{38} (760 + 78 \int _{0}^{38}e^{-t/50}dt )$

let u = -t/50
dt = -50du
${fave} =\int _{0}^{38}e^{u}dt $
${fave} =\int _{0}^{38}e^{u}(-50)du $
${fave} =(-50)* \int _{0}^{38}e^{u}du $
${fave} =(-50)* e^{-t/50} + C |^3^8 $

-50 * e^(-38/50) = -23.38

${fave} \frac{1}{38} (760 + 78 * (-23.38) )$ = -27.997

im sure its wrong.. but where did i f.. up?

 

[Dick]

Yeah, it's definitely wrong. The mean should be somewhere between 98C and 20C. You are just being sloppy. As to where it happened, I would look at where you have a du integral and you still have t limits. If you change the variable from t to u aren't you supposed to change the limits as well?

 

[CalculusSandwich]

not quite what you did was forget to subtract e^ (-t/50) from e^0.
<br /> ${fave} =\int _{0}^{38}e^{u}(-50)du $ <br />

is

evaluates to

e^ (-38/50) - e ^0
which is something like -.54
times your constant which was -78*50/38
gives you a positive number, somewhere around 55.
add that to

<br /> ${fave} =\int _{0}^{38}du $ <br />

The correct exact answer is 74.63, which makes perfect sense.

 

[Dick]

You are forgetting to change the t limits to u limits as well.

 

[Slimsta]

cant believe how stupid my mistake was this small e^0 changed everything...
well.. at least i get how to do this kind of question now! its actually not that hard when i think about it now

thanks once more guys