the rate (in liters per minute) at which water drains from a tank is recorded at half - minute intervals…

Explanation:

Step1: Calculate left - endpoint approximation

The time intervals are of length $\Delta t = 0.5$ minutes. The left - endpoint values for the rate $r(t)$ are $r(0)=56$, $r(0.5) = 54$, $r(1)=52$, $r(1.5)=50$, $r(2)=48$, $r(2.5)=46$. The left - endpoint approximation $L$ is given by $L=\sum_{i = 0}^{5}r(t_i)\Delta t$, where $\Delta t=0.5$. So $L = 0.5\times(56 + 54+52+50+48+46)=0.5\times296 = 148$.

Step2: Calculate right - endpoint approximation

The right - endpoint values for the rate $r(t)$ are $r(0.5)=54$, $r(1)=52$, $r(1.5)=50$, $r(2)=48$, $r(2.5)=46$, $r(3)=44$. The right - endpoint approximation $R$ is given by $R=\sum_{i = 1}^{6}r(t_i)\Delta t$, where $\Delta t = 0.5$. So $R=0.5\times(54 + 52+50+48+46+44)=0.5\times294=147$.

Step3: Calculate the average of left - and right - endpoint approximations

The average $A=\frac{L + R}{2}=\frac{148+147}{2}=\frac{295}{2}=147.5$.

Answer:

$147.5$