# Average Value of a Function - temperature

1. Feb 9, 2010

### Slimsta

1. The problem statement, all variables and given/known data
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?

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

3. The attempt at a solution
$$fave ={1}/{98-20} \int _{20}^{98} T(t)dt$$ $${fave} =\frac{1}{78}\int _{20}^{98} 20 + 78 e^{-t/50}dt?$$

2. Feb 9, 2010

### 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.

3. Feb 9, 2010

### Slimsta

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

i dont understand your question.. how should i construct the question?
in my textbook they wrote like 2 sentences about temperature and i dont get it.. they dont even have an example for this type of questions

4. Feb 9, 2010

### 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.

5. Feb 9, 2010

### Slimsta

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

Last edited: Feb 9, 2010
6. Feb 9, 2010

### 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?

7. Feb 9, 2010

### CalculusSandwich

not quite what you did was forget to subtract e^ (-t/50) from e^0.

$${fave} =\int _{0}^{38}e^{u}(-50)du$$

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.

$${fave} =\int _{0}^{38}du$$

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

Last edited: Feb 9, 2010
8. Feb 9, 2010

### Dick

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

9. Feb 9, 2010

### 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