How to calculate power from this diagram? (Another one)
This is easiest done in two parts, for the first and second halves of the repeating waveform.  This time the average power will obviously be negative since the power magnitude in the second half clearly is larger than in the first half, and the first half is obviously positive and the second negative.  This means the load is supplying net power.

The total power magnitude is the RMS voltage times the RMS current.  The RMS voltage is obviously 1 V.  The current is a ramp from 0 to 1, or I(t) = t.  That squared is t^2, the average of that is 1/3, and the square root of that is 0.577.  The power magnitude is therefore 1 V * 577 mA = 577 mW.

The real power is the integral of the instantaneous voltage and current over one repeating cycle.  From inspection, this is 250 mW in the first half and -750 mW in the second half, for a total of -500 mW.

The power factor is -500 mW / 577 mW = -0.866.  Often the absolute value of the power factor is quoted with a statement as to whether the reactive load is inductive or capacitive.  You can take the arc cosine of the power factor to get the phase angle of -30 deg in this case, but that has little meaning when the voltage and current are so non-sinusoidal.

The total complex power is the vector sum of the real and reactive power.  Since we know the real power and the power magnitude, we can calculate the reactive power, which is -289 mW.  The sign only indicates whether it is capacitive or inductive (whether it is a +90 deg or -90 deg).