## Calculus (3rd Edition)

$682.25 \ insects$
Let $P$ be the population of insects.Therefore, $\dfrac{dP}{dt}=200+10 t+0.25t^2$ At the initial point, there are $35$ insects. So, $P(0)=35$ and $dP=200+10 t+0.25t^2$ Integrate the above equation to obtain: $\int_0^3 dP=\int_0^3 200+10 t+0.25t^2 \ dt$ or, $P(3)-P(0)=(200t+5t^2+\dfrac{t^3}{12})_0^3$ or, $P(3)-35=600+45+\dfrac{9}{4}-0$ or, $P(3)=682.25 \ insects$ Therefore, the population of insects after 3 days is: 682.25 insects.