수치적분

(a) Because $r(t)$ is measured in cubic feet per second and $\triangle t$ is measured in seconds, the definite integral $$\int_0^{60} r(t) \, dt$$ measures the total amount of water (in cubic feet) that flow through the dam in 60 seconds. (b) With $7$ function values and thus $6$ subintervals as the maximum number of intervals we can use, we can compute $M_3$ since we need to use every other data point to evaluate the function. Doing so, $$\begin{aligned}\int_0^{60} r(t) \, dt &\approx M_3 \\ &= r(10) \cdot 20 + r(30) \cdot 20 + r(50) \cdot 20 \\ &= (2100 + 3000 + 5100) \cdot 20 \\ &= 204000\end{aligned}\,.$$ Thus, approximately $204 000$ cubic feet of water pass through the dam on the interval $0 \le t \le 60$. We observe that $r$ appears to be increasing at an increasing rate on $0 \le t \le 60$. In particular, the data suggests that $r$ is concave up on the interval. As such, the trapezoid rule will over-estimate the true value of the definite integral and the midpoint rule will under-estimate the value. Thus, $$\int_0^{60} r(t) \, dt > M_3 = 204000\,.$$ (c) In order to estimate the integral using $S_6$, we need to know the value of $T_3$ along with $M_3$. To find $T_3$, we compute $L_3$ and $R_3$. First $$L_3 = r(0) \cdot 20 + r(20) \cdot 20 + r(40) \cdot 20 = (2000 + 2400 + 3900) \cdot 20 = 166000\,.$$ Similarly, $$R_3 = r(20) \cdot 20 + r(40) \cdot 20 + r(60) \cdot 20 = (2400 + 3900 + 6500) \cdot 20 = 256000\,.$$ Thus, $$T_3 = \frac{L_3 + R_3}{2} = \frac{166000 + 256000}{2} = 211000\,.$$ Finally, we now have the information needed to compute $S_6$. We thus find that $$\int_0^{60} r(t) \, dt \approx S_6 = \frac{2M_3 + T_3}{3} = \frac{2 \cdot 204000 + 211000}{3} = \frac{619000}{3} \approx 206333.33\,.$$ (d) First, $$\frac{1}{60} S_6 = \frac{206333.33}{60} = \frac{30950}{9} \approx 3438.89\,.$$ This is one estimate for the average value of $r$ on $[0,60]$ since $$r_{\operatorname{AVG} [0,60]} = \frac{1}{60-0} \int_0^{60} r(t) \, dt\,,$$ and $\int_0^{60} r(t) \, dt \approx S_6$. Thus, on average, water is flowing through the dam at a rate of $3438.89$ cubic feet per second. Next, $$\frac{2000+2100+2400+3000+3900+5100+6500}{7} = \frac{25000}{7} \approx 3571.43$$ is the average of the $7$ known values of $r(t)$ on the interval $[0,60]$, and thus is one estimate of the average value of $r$. This average treats all of them equally, as opposed to $S_6$ which counts midpoints more than endpoints and is developed in such a way to remove some of the error inherent in using simple estimates like $L_6$ and $R_6$. This simple average of $7$ rate values is analogous to using a left or right Riemann sum, and thus we expect that the estimate of the average value of $r$ that comes from $S_6$ is more accurate for computing the average rate at which water flows through the dam.