# Thread: average value of a function

1. ## average value of a function

the temperature (in Celsius) $t$ hours after 9am was modeled by the function $T(t) = 20 + 6\sin \frac{\pi t}{12}$ find the average temperature from 9am to 9pm.

what I did: $\frac{1}{21-9} \int_9^{21} + 6\sin \frac{\pi t}{12}dt$ which gives me about 17.299

2. Hello superdude
Originally Posted by superdude
the temperature (in Celsius) $t$ hours after 9am was modeled by the function $T(t) = 20 + 6\sin \frac{\pi t}{12}$ find the average temperature from 9am to 9pm.

what I did: $\frac{1}{21-9} \int_9^21 + 6\sin \frac{\pi t}{12}dt$ which gives me about 17.299

the upper limit of integration is suppose to be 21 but (I believe) Latex is limited to 1 character
Read the question carefully! The values of $t$ are measured after 9 am, so the limits of the integral are 0 to 12, not 9 to 21.

Incidentally you can get LaTeX to accept more that one character for the limit of an integral by enclosing the character string in {...} as you have done with \frac{...}{...}. Like this:

$\int_0^{12}(20 +6\sin \frac{\pi t}{12})dt$

Hover the mouse over the LaTeX image, and the code will appear; or click on it and it'll pop up in a separate window. Try it!

3. got it, thanks.
Is the reason I integrate from 0 to 12 because the function/data is undefined every where else? I understand why it should be this, I am having trouble coming up with a rule so I don't make the same mistake in another question.

4. Hello superdude
Originally Posted by superdude
got it, thanks.
Is the reason I integrate from 0 to 12 because the function/data is undefined every where else? ...
No, it's because the value of t is defined as follows:
Originally Posted by superdude
the temperature (in Celsius) $t$ hours after 9am was modeled by the function $T(t) = 20 + 6\sin \frac{\pi t}{12}$ ...
and the question asks for
Originally Posted by superdude
... the average temperature from 9am to 9pm.
So the values of t you need are from 0 to 12 (9am + 0 hours to 9am + 12 hours).