The function `f` is the derivative of the function `F`, so `f(t) = 0.032 e^(-0.032t)`, and the integral is
`0.032 int_0^200 t e^(-0.032t) dt`.
We may evaluate this by integration by parts. Let `u=t` and `dv = e^(-0.032t)dt`. Then
| `0.032 int_0^200 te^(-0.032t) dt` | `=0.032 [t((-1)/0.032 e^(-0.032t))|_0^200+ int_0^200 1/0.032 e^(-0.032t) dt]` |
| `= int_0^200 e^(-0.032t) dt - 200 e^(-6.4)` | |
| `= (-1)/0.032 e^(-0.032t) |_0^200 - 200 e^(-6.4) ~~ 30.87` days. |