The dual of Faraday's Law is Ampere's Law but, while Faraday's Law is fundamental to the physics of an inductor, Ampere's Law is not fundamental to the physics of a capacitor.
Now, it is true that, in circuit theory, the capacitor and inductor are duals:
$$i_C = C\frac{dv_C}{dt} \leftrightarrow v_L = L \frac{di_L}{dt}$$
However, we have to be more careful outside the context of circuit theory.
In physics, the fundamental relationship
$$Q = CV$$
clearly requires the existence of electric charge and an electric scalar potential due to a conservative electric field. This equation relates electric charge and electric scalar potential.
The closest we can get to a dual of this is
$$\Phi = LI $$
which relates magnetic flux and electric current. But magnetic flux is not the dual of electric charge.
The missing ingredient here is the hypothetical magnetic charge (magnetic monopole) which is the dual of electric charge.* Were magnetic charge \$Q_m\$ (measured in webers) to exist, it would be a source or sink of a conservative magnetic field (measured in amperes per meter) and there would be an associated scalar magnetic potential (measured in amperes).
We could thus relate magnetic charge and magnetic scalar potential with a magnetic "capacitance" measured in henrys.
Further, we could relate electric flux to magnetic current (measured in volts) with an electric "inductance" measured in farads.
To summarize, while electric flux and magnetic flux are duals, and changing magnetic flux is fundamental to the physics of an inductor, changing electric flux is not fundamental to the physics of a capacitor. Indeed, it is the electric field itself, not the electric flux, that is fundamental.
*Assuming magnetic charge exists, Maxwell's equations become
$$\nabla \cdot \vec D = \rho_e$$
$$\nabla \cdot \vec B = \rho_m$$
$$\nabla \times \vec E = - (\vec J_m + \frac{\partial \vec B}{\partial t})$$
$$\nabla \times \vec H = \vec J_e + \frac{\partial \vec D}{\partial t}$$
I would use an easier approach: the capacitor is only needed to act as buffer for the high-current spikes. Just get the duration of the (biggest) spike (t_duration) and the height of the (biggest) spike (I_max) from the data you got ("For the time period of interest, I have the load's current draw over time.").
The capacitor's voltage should never drop below 90%, so dv = 10% * Vs.
Use
$$i = C*dv/dt $$
$$ C = \frac{ I_{max} * t_{duration} } { 0.1 * V_s}$$
This will yield a bigger capacitor than needed, so the voltage drop will be quite less than 10% of Vs.
Another approach is to simulate the circuit is Spice, putting the data you have ("For the time period of interest, I have the load's current draw over time.") in a Piece-wise linear current source. Pick a starting value for C, like e.g. 100uF (or use the value given in the first approach) and tweak by running a few simulations.
If you really want to solve your approach, I would suggest iteratively picking values for C and evaluate the definite integral.
Best Answer
https://www.omron.com.au/service_support/FAQ/FAQ02804/index.asp OMRON were the best suppliers of contact relays for reliability. Now they specialize in solid-state.
Heed their advice de-ratings as max and add margin.
When contacts close on a capacitor, the ESR of the Cap determines the initial surge current and if there is contact bounce, this causes a 5000'K arc in air which gradually wears out the silver content,
When contacts open with dry contacts dI/dt is extremely high, and the motor current creates a bigger arc than the small Mylar Cap. But this is a tradeoff to choose the right size. Exceed the nominal peak voltage by at least 50~100% and add a series R to limit contact closure current.
Use the following as guides for C and R values:
C: 0.5 to 1 μF per 1 A of contact current (A)
R: 0.5 to 1 Ω per 1 V of contact voltage (V)
Ref:
https://www.omron.com.au/service_support/FAQ/FAQ02804/index.asp