Questions: 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 five-hour time intervals are shown in the table. t(hr) r(t)(L / h) 0 9.5 5 8.7 10 7.9 15 7.3 20 6.6 25 6.1 Find lower and upper estimates for the total amount of liquid that leaked out. lower estimate = liters upper estimate = liters

Solution Steps

To estimate the total amount of liquid that leaked out, we can use the trapezoidal rule and the left and right Riemann sums. The left Riemann sum will give us a lower estimate, and the right Riemann sum will provide an upper estimate. The trapezoidal rule will give us a more accurate estimate by averaging the left and right sums.

To find the lower estimate of the total amount of liquid that leaked out, we use the left Riemann sum. The formula is given by:

\[ \text{Lower Estimate} = \sum_{i=0}^{n-1} r(t_i) \Delta t \]

where \( \Delta t = 5 \) hours and \( r(t_i) \) are the rates at the left endpoints of the intervals. Thus, we have:

\[ \text{Lower Estimate} = (9.5 + 8.7 + 7.9 + 7.3 + 6.6) \times 5 = 200.0 \text{ liters} \]

To find the upper estimate, we use the right Riemann sum. The formula is:

\[ \text{Upper Estimate} = \sum_{i=1}^{n} r(t_i) \Delta t \]

Again, with \( \Delta t = 5 \) hours and \( r(t_i) \) being the rates at the right endpoints of the intervals, we have:

\[ \text{Upper Estimate} = (8.7 + 7.9 + 7.3 + 6.6 + 6.1) \times 5 = 183.0 \text{ liters} \]

Final Answer