How do plants absorb different "rates" of light? Context
I'm trying to understand the broader question of why plants are green despite our Sun's "green" star status. This Wired article has a nice explanation, but I don't understand the key argument:


the pigments of the photosystem had to be very finely tuned in a
certain way. The pigments needed to absorb light at similar
wavelengths to reduce the internal noise. But they also needed to
absorb light at different rates to buffer against the external noise
caused by swings in light intensity. The best light for the pigments
to absorb, then, was in the steepest parts of the intensity curve
for the solar spectrum—the red and blue parts of the spectrum.

Question(s)
Doesn't a steep gradient denote drastically changing (and potentially noise) signal? And isn't that bad?
The only way I can rationalize this is if energy production is the sum of an expression where the rate-based term is larger than the "standard" term:
energy ~ A * d/dt(light) + B * light + ... where A() > B()
 A: The details are explained in the paper, especially in the supplementary material. Since I'm not a biologist, let's just accept the assumptions about plants made in the paper. These are:

*

* Plants function best if the input power is as close as possible to a constant used power $\Omega$. 

* Input power is fed in from two nodes $A$ and $B$, which deliver powers $\mathcal{P}_A$ and $\mathcal{P}_B$ respectively, with $\mathcal{P}_A > \Omega > \mathcal{P}_B$. 

* To achieve an average power of $\Omega$ the plant switches between nodes $A$, $B$ and $\text{none}$, such that $A$ is on for a fraction $p_A$ and $B$ for a fraction $p_B$ of the time. 

* To be able to stabilize external power fluctuations it is desirable to have a large spread of powers $\Delta = \mathcal{P}_A - \mathcal{P}_B$. 

* The powers $\mathcal{P}_A$ and $\mathcal{P}_B$ are given by the intensity in the spectrum within narrow absorption peaks. These peaks must be spectrally close to each other. 

To maximize $\Delta$ one could choose the absorption peaks at the maximum ($A$) and far from the maximum ($B$) of the spectrum, if there wasn't condition 5. Therefore, the best one can do is to place the two peaks at the steepest slope of the spectrum. In the figure you posted the two blueish peaks and the two reddish peaks are such a pair each.
