Chapter 5 - The Integral - 5.6 Net Change as the Integral of a Rate of Change - Exercises - Page 268: 1

$15,250 \ gallons$

Let $V$ be the volume of the reservoir. Therefore, $\dfrac{dV}{dt}=3000+20 t$ At the initial point, the reservoir is empty. So, $V(0)=0$ and $dV=(3000+20 t) \ dt$ Integrate the above equation to obtain: $\int_0^5 dV=\int_0^5 (3000+20 t) \ dt$ $\implies V(5)=[3000t+10t^2]_0^5$ $\implies V(5)=[3000(5)+10(5^2)]-[3000(0)+10(0)]$ $\implies V(5)=15, 250 \ gallons$ Therefore, the quantity of water in the reservoir after five hours is $15, 250 \ gallons$