Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 5 - The Integral - 5.6 Net Change as the Integral of a Rate of Change - Exercises - Page 268: 2


$682.25 \ insects$

Work Step by Step

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.
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