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[Omnibus]

@broli,

I think Mathematica can help in doing this.

[blueplanet]

This condition is always satisfied because:

d/dt( I*2V) is always greater than d/dt(I*V)
                     
The term I*2V has nothing to do with the power dissipated in the load resistor R.

                     

slope(int_0^T IVdt vs. t) < slope(int_0^T I*2Vdt vs. t).

[Omnibus]

This condition is always satisfied because:

d/dt( I*2V) is always greater than d/dt(I*V)
                     
The term I*2V has nothing to do with the power dissipated in the load resistor R.

                   

That's not the condition. Read above what the condition is.

[blueplanet]

I am afraid you are wasting other people's time here.
Your criterion does not need any numerical computation!

What we need to do now is find the expressions for the integrals int_0^T IVdt and int_0^T I*2Vdt and then find the first derivatives over time of these expressions. Many of us, myself included, are so spoiled by the numerical methods that the real good old math is put slightly on the side (it's true that numerical methods are very powerful and can dwell in territories inaccessible by the 'good old math' as far as engineering is concerned).

[Omnibus]

I am afraid you are wasting other people's time here.
Your criterion does not need any numerical computation!

You're the one wasting the time and the bandwidth of this forum with your confusion. Don't clog the discussion with you gibberish.