The temperature of a chemical reaction,

t
seconds after the reaction starts, is given
by:
T(t)= 20+5.25t^.75 degrees F
,
What is the average temperature during the whole process, if the reaction lasts for 4
minutes and 16 seconds ?

integral of T dt from t = 0 to t = 256

divided by 256

[ 20 t + (5.25/1.75) t^1.75 ] at 256 - at 0 divided by 256

at 0 it is 0
[ 20(256) + 3 (256^1.75) ] / 256

20 + 3 (256)^.75

= 20 + 3(64)

= 212 interesting, boiling point of water

Thanks! I get why you did every thing in the numerator but why did you have to divide by 256?

oh is it because it is the average?

To find the average temperature during the entire process, we need to find the average of the temperature function over the given time interval.

To do this, we need to calculate the definite integral of the temperature function over the given time interval and then divide it by the length of the interval.

The given temperature function is:
T(t) = 20 + 5.25t^0.75

We need to integrate this function over the time interval from 0 to 4 minutes and 16 seconds, which can be written as 4.27 minutes (or 4.27/60 hours).

The integral of the temperature function T(t) over the given time interval is denoted as follows:

∫(0 to 4.27/60) (20 + 5.25t^0.75) dt

Integrating this expression, we get:
[20t + 7t^(1.75)/1.75] evaluated from 0 to 4.27/60

Now, substitute the upper limit (4.27/60) and lower limit (0) into the expression and evaluate:

[20∙(4.27/60) + 7(4.27/60)^(1.75)/1.75] - [20∙0 + 7(0)^(1.75)/1.75]

Simplifying further:
[85.4/60 + 7(4.27/60)^(1.75)/1.75]

Calculate the numerical value of this expression to find the average temperature during the whole process.