Trouble calculating time to drain a capacitor
Integrating the changing current over the draining of the capacitor is correct and quite fun, but there is an easier way: use the energy formula which already has that integral done for you.

Calculate the energy stored in the capacitor, at \$V_a\$ and \$V_b\$.

$$ Energy = 0.5 \times C \times V^2 $$

Subtract the final from the initial to find the available energy.

Divide this by the power drawn to find the time it will take.

Last thought: the boost converter efficiency will drop as the voltage drops. Most of the energy is delivered while the capacitor is fairly full, so you can assume a constant efficiency near \$V_a\$. To get it more accurate, you might need to try that integral after all.