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: 3

Answer

$3,660,000 $

Work Step by Step

Let $N$ be the number of supporters for the candidate. Therefore, $\dfrac{dN}{dt}=2000t+1000$ At the initial point, there are no supporters. So, $N(0)=0$ and $dN=(2000t+1000)$ Integrate the above equation to obtain: $\int_0^{60} dN=\int_0^{60} (2000t+1000) \ dt$ or, $ N(60)-N(0)=(1000t^2+1000t)_0^{60} $ or, $N(60)=36,000,00+60000-0$ or, $N(60)=3660000 $ Therefore, the number of supporters the candidate has after 60 days is: $3,660,000 $
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