Single Variable Calculus, Chapter 5, 5.4, Section 5.4, Problem 60

Find the amount of $H_2 O$ that flows from the tank during the first 10min. Suppose that the $H_2 O$ flows from the bottom of a storage tank at a rate of $r(t) = 200 - 4t$ $\displaystyle \frac{\text{liters}}{\text{min}}$, where $0 \leq t \leq 50$.
Using Net Change Theorem


$
\begin{equation}
\begin{aligned}
\int^b_a r(t) dt &= \int^{10}_0 (200 -4t) dt\\
\\
\int^b_a r(t) dt &= \left[ 200t - 2t^2 \right]^{10}_0\\
\\
\int^b_a r(t) dt &= 200(10) - 2(10)^2 - \left[ 200(0) - 2(0)^2 \right]\\
\\
\int^b_a r(t) dt &= 2000 - 200 - 0\\
\\
\int^b_a r(t) dt &= 1800 \text{ liters}

\end{aligned}
\end{equation}
$


This means that during the first 10 mins. the amount of water that flows inside the tank is 1800 liters.