The power absorbed by the BOX is 2e^{-2t} W. Calculate the amount of charge that enters the BOX between 0.1 and 0.4 seconds.

Image from: J. D. Irwin and R. M. Nelms, Basic engineering circuit analysis, 10th ed. Hoboken, NJ: John Wiley, 2011.
Solution:
Remember that:
p=vi
(Where, p is power, v is voltage, and i is current)
i=\dfrac{p}{v}
Substitute our equations:
i=\dfrac{2e^{-2t}}{4e^{-t}}
i=0.5e^{-t}
i=0.5e^{-t}
To find the amount of charge that entered during the time interval, we must remember the following:
i=\dfrac{d(q(t))}{dt}
d(q(t))=i\,dt
d(q(t))=i\,dt
(Take the integral of both sides)
\,\displaystyle q(t)=\int^{t_2}_{t_1}i\,dt(Substitute our current equation)
\,\displaystyle q(t)=\int^{0.4}_{0.1}(0.5e^{-t})\,dt
q(t)=-0.5e^{-t}\Big|^{0.4}_{0.1}
q=0.117 C