I feel dubious about the use of “integral” for this - you have Watts, measured
once per minute. That’s either going to be an instantaneous measurement (such
as “7.8 Watts right now, no idea what it was 5 seconds ago”), or it’s going to
be an average over the previous minute (such as “468 Joules used in the past
60 seconds = an average of 7.8 Watts”) - which it is depends on the instrument
making the measurements.
If you want to convert Watts (measured once per minute) back into Joules (or
kiloWatt-hours; same thing except for a multiplication factor of 3.6 x 10^6)
then I think you need a simple “sum” operator:
Minute 1: 7.8 Watts
Minute 2: 10.2 Watts
Minute 3: 13.3 Watts
Minute 4: 6.4 Watts
Minute 5: 9.6 Watts
assume this pattern repeats (mysteriously) 12 times to create 1 hour’s data.
The sum of the above 5 values is 47.3; multiply this by 60 (seconds) to get
2838 Joules (or Watt-seconds). Since the pattern repeats 12 times in an hour,
multiply by 12 to get 34056 Joules of energy used in that hour.
34056 Jouleṡ ÷ 3.6 x 10^6 = 9.46 x 10-³ kWh
So, I think you should be using “sum” to add up the Watt power values, and
then dividing by ( 3.6 x 10^6 ) ÷ ( 60 x 12 ) = 2500 to convert into kWh.
Note that 2500 is approximately twice 1440 (which is what you’re dividing by
at present), hence explaining the values you’re currently getting as being
about twice what you expected.