FFT and Power Spectrum Normalization
I would say most of your confusion arises from conflating the continuous case with the discrete case of the FT.  The are similar, and related, but not equivalent.  "Area" is a continuous concept.  "Sum" is discrete.  

I normalize my DFTs by 1/N principally because it makes the bin values essentially sample count independent.  Other reasons too.  For the sum of squares to hold, you should normalize with $1/\sqrt{N}$.

It would be helpful to know if you have a Linear Algebra background in order to better explain some of these things.  Also if you are talking about the continuous case or the discrete one.



How much energy is there in the fluctuation of a color pattern in a fabric?  Meaningless question, right?  The concept of energy is an application issue and is not inherent in the DFT.  What is inherent is the ability to sum squares of values.  With a $1/\sqrt{N}$ normalization, the discrete fourier transform can be represented as the multiplication of a unitary matrix, thus the sum of squares is preserved.

$$ Z = F S $$

Suppose Z is a complex vector of the DFT bins, F is the tranformation matrix, and S is a complex vector with your signal.  Of course, a real valued signal will fit too.

Let ($^*$) stand for Hermitian, aka, the conjugate transpose.

$$ Z^* = S^* F^* $$

$$ Z^* \cdot Z = ( S^* F^*  ) \cdot F S $$

$F^*$ also happens to be the inverse DFT so $ F^* \cdot F = I $, the identity matrix.

$$ Z^* \cdot Z = S^* \cdot I \cdot S  = S^* \cdot S $$

The left and right parts of these equations are simply the sum of the squares of the magnitudes.

Between the forward and inverse DFTs, a factor of $1/N$ needs to be included.  To split it evenly, it becomes $1/\sqrt{N}$ in each direction.

The "energy" term is associated with the sum of squares.  The units are dependent on the application.  The DFT doesn't care.