Liquid leaked from a damaged tank at a rate of $r(t)$ liters per hour. The rate decreased as time passed and values of the rate at four-hour time intervals are shown in the table. \begin{tabular}{|r|r|} \hline$t(\mathrm{hr})$ & $r(t)(\mathrm{L} / \mathrm{h})$ \\ \hline 0 & 8.1 \\ \hline 4 & 7.5 \\ \hline 8 & 6.9 \\ \hline 12 & 6.4 \\ \hline 16 & 6 \\ \hline 20 & 5.5 \\ \hline \end{tabular} Find lower and upper estimates for the total amount of liquid that leaked out. lower estimate $=$ liters upper estimate $=$ liters Question Help: Video Calculator Submit Question

Step 1 ：We are given the rate of leakage from a tank at different time intervals. The rate is decreasing over time. We are asked to find the lower and upper estimates for the total amount of liquid that leaked out.

Step 2 ：We can use numerical integration methods to estimate the total amount of liquid leaked. Specifically, we can use the trapezoidal rule and the rectangle rule.

Step 3 ：The trapezoidal rule approximates the definite integral of a function by approximating the region under the graph of the function as a trapezoid and calculating its area.

Step 4 ：The rectangle rule approximates the definite integral of a function by approximating the region under the graph of the function as a series of rectangles and calculating their area.

Step 5 ：We can find the lower estimate by using the rectangle rule with the right endpoints of each interval, and the upper estimate by using the rectangle rule with the left endpoints of each interval.

Step 6 ：Given the time intervals t = [ 0 4 8 12 16 20] and the corresponding rates r = [8.1 7.5 6.9 6.4 6. 5.5], we calculate the differences in time delta_t = [4 4 4 4 4].

Step 7 ：Using these values, we calculate the lower estimate to be 129.2 liters and the upper estimate to be 139.6 liters.

Step 8 ：Final Answer: The lower estimate for the total amount of liquid that leaked out is \(\boxed{129.2}\) liters and the upper estimate is \(\boxed{139.6}\) liters.