Net change in population via integration of a rate

Let $u = 0.05t$, so $du = 0.05\,dt$, which means $dt = du/0.05$. Substituting gives $\int 200e^{u} \cdot \frac{du}{0.05} = 4000\int e^u\,du = 4000e^u + C$. Back-substituting $u = 0.05t$ gives the antiderivative $4000e^{0.05t} + C$, so $P(t) = 4000e^{0.05t} + C$ (an antiderivative of $P'(t)$).

By the Net Change Theorem, $\int_0^{10} 200e^{0.05t}\,dt = \left[4000e^{0.05t}\right]_0^{10} = 4000e^{0.5} - 4000e^{0}$. This simplifies to $4000(e^{0.5} - 1) \approx 4000(1.6487 - 1) \approx 2594.9$. Thus the population increases by about 2595 people over the 10-year period.