I have a sensor that draws 20 mA for 20 µs and draws 60 µA sleep current for the remaining time. The period can be 5 minutes, 10 minutes, or 15 minutes. Can someone help me calculate the average current for each case?
Back of the envelope calculations, several grains of salt recommended:
Period: 5 sec average: 60.080 uA
Period: 10 sec average: 60.040 uA
Period: 15 sec average: 60.027 uA
First, convert all the quantities into the same units. For example all the currents into uA and all the times into us.
Then multiply each current by its corresponding time. Add those results together.
Finally, divide by the total time.
If you convert 10min to microseconds, you find out that you don't have to go further since the wake up time is proportionally so low that it's becoming insignificant.
Avg = Duty Cycle * Amplitude
Duty Cycle = ON time / Period
Lets find the average for the 20mA pulse for 5 minute period
Amplitude = 20mA
ON time = 20µs
Period = 5 minutes
Convert the Period to microseconds:
Period = 5 * 60 * 1E+6 = 300000000µs
Duty Cycle = 20µs / 300000000µs = 66.66E-9
Avg = 66.66E-9 * 20mA =1.333nA
Do the same for the 60µA pulse, then add the two
check this
average value.pdf (36,5 KB)
What is that big squiggly line that looks like a big 'S'
this?
Yes
Insignificantly different from 60 uA. For 10 minutes = 600E6us:
(20E3uA*20us + 60uA*(600E6us-20us))/600E6us = 60.0007uA
See post #10
So do you konow what is is?
So what's an integral.
The Stewart
text will get you past some of the beginner concepts like that.
It's over a thousand pages long!
You can just read page 2, or for more detail, pages 360–367. The formal definition is on page 372. 
The definite integral, yes.