certainly the more time passes, the more the microcontroller consumes, but it seems to me that even making the integral of many signal periods, the maximum peak value of the integral is always the same, each ramp will have approximately the same consumption as the other ramps; but I don't concentrate on how much the summative value of several periods could be, but it was enough for me to understand what the max value of the integral represents, for example relative to a single half-period; in fact the peak of the integral of your half period represents the value of that area (636).
For the unit of measure, Dave seems to be telling us that the integral value represents that value in mV or uV I believe in the time of 1 second .. he explains it about 11 to 13 minutes of the video, but with subtitles maybe yes understands wrong.
It would be good to know, in your example, if 636 is uV per second, or is it us in 1 volt ....
Maybe I better go to bed, the neurons are dancing in my head
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The unit is <whatever> seconds. Microvolts, millivolts, megavolts, <whatever> seconds. Always seconds. Yes, we can play with the math to get everything aligned but AFAIK, we always want to wind up with seconds.
Maybe other units of time are appropriate. Small batteries are rated in mA hours and this is exactly where Dave wants to wind up. He may need to convert the integral from mA seconds to mA hours...
There is is no condition under which the integral of battery current in Dave's example will diminish. The CPU may go to sleep and the integral not change very much for some time (a nearly straight line on the integral graph) but it never reverses. The only way the integral of current can diminish is if there is a battery charger recharging the battery. In Dave's case, pumping current back into the battery which the circuit clearly doesn't do.
When you think of the sine wave being integrated over 0 to 2pi, the last half of the waveform is at a negative voltage and subtracts from the previously accumulated value. At 2pi, the result is 0 uV s. We integrated over pi radians to keep the integral increasing until it became asymptotic at pi radians. In the very next delta t, the value would start to decline because sin(pi+delta t) is negative. Try integrating from 0 to 3*pi/2 -- 270 degrees. It won't be back down to 0 but the value is clearly decreasing.
In my example, the AREA is in terms of uV seconds. Microvolts on the Y axis and time on the X axis. It is NOT the same as uV/second. Y is uV, X is time (some variation of seconds) and the result is an AREA of Y * X or uV * time or uV second (implicit multiply symbol omitted by convention).
We expect to get an AREA when we integrate. Converting that to the real world is a math exercise.
In Dave's example, he is continually draining water out of the previously full bucket. Sooner or later the bucket is empty. Unless somebody shows up with a hose...