Since \(r(t) \ge 0\text{,}\) the value of \(\displaystyle \int\limits_4^{10} r(t) \, dt\) is the area under the curve on the interval \([4,10]\text{.}\) A Riemann sum for this area will have rectangles with heights measured in gallons per day and widths measured in days, so the area of each rectangle will have units of
\begin{equation*}
\frac{\text{gallons} }{\text{day} } \cdot \text{days} = \text{gallons}\text{.}
\end{equation*}
Thus, the definite integral tells us the total number of gallons of pollutant that leak from the tank from day 4 to day 10. The Total Change Theorem tells us the same thing: if we let \(R(t)\) denote the total number of gallons of pollutant that have leaked from the tank up to day \(t\text{,}\) then \(R'(t) = r(t)\text{,}\) and
\begin{equation*}
\int\limits_4^{10} r(t) \, dt = R(10) - R(4)\text{,}
\end{equation*}
the number of gallons that have leaked from day 4 to day 10.
To compute the exact value of the integral, we use the Fundamental Theorem of Calculus. Antidifferentiating \(r(t) = 0.0069t^3 -0.125t^2+11.079\text{,}\) we find that
\begin{align*}
\int\limits_4^{10} 0.0069t^3 -0.125t^2+11.079 \, dt =\mathstrut \amp \left. 0.0069 \cdot \frac{1}{4} \, t^4 - 0.125 \cdot \frac{1}{3} t^3 + 11.079t \right|_4^{10}\\
\approx\mathstrut \amp 44.282\text{.}
\end{align*}
Thus, approximately 44.282 gallons of pollutant leaked over the six day time period.
To find the average rate at which pollutant leaked from the tank over \(4 \le t \le 10\text{,}\) we compute the average value of \(r\) on \([4,10]\text{.}\) Thus,
\begin{equation*}
r_{\operatorname{AVG} [4,10]} = \frac{1}{10-4} \int\limits_4^{10} r(t) \, dt \approx \frac{44.282}{6} = 7.380
\end{equation*}
gallons per day.