Section 5.6 Numerical Integration
ΒΆMotivating Questions
How do we accurately evaluate a definite integral such as when we cannot use the First Fundamental Theorem of Calculus because the integrand lacks an elementary algebraic antiderivative? Are there ways to generate accurate estimates without using extremely large values of in Riemann sums?
What is the Trapezoid Rule, and how is it related to left, right, and middle Riemann sums?
How are the errors in the Trapezoid Rule and Midpoint Rule related, and how can they be used to develop an even more accurate rule?
Preview Activity 5.6.1.
As we begin to investigate ways to approximate definite integrals, it will be insightful to compare results to integrals whose exact values we know. To that end, the following sequence of questions centers on
Use the applet at http://gvsu.edu/s/a9 with the function on the window of values from to to compute the left Riemann sum with three subintervals.
Likewise, use the applet to compute and the right and middle Riemann sums with three subintervals, respectively.
Use the Fundamental Theorem of Calculus to compute the exact value of
We define the error in an approximation of a definite integral to be the difference between the integral's exact value and the approximation's value. What is the error that results from using From From
In what follows in this section, we will learn a new approach to estimating the value of a definite integral known as the Trapezoid Rule. The basic idea is to use trapezoids, rather than rectangles, to estimate the area under a curve. What is the formula for the area of a trapezoid with bases of length and and height
Working by hand, estimate the area under on using three subintervals and three corresponding trapezoids. What is the error in this approximation? How does it compare to the errors you calculated in (d)?
Subsection 5.6.1 The Trapezoid Rule
So far, we have used the simplest possible quadrilaterals (that is, rectangles) to estimate areas. It is natural, however, to wonder if other familiar shapes might serve us even better. An alternative to and is called the Trapezoid Rule. Rather than using a rectangle to estimate the (signed) area bounded by on a small interval, we use a trapezoid. For example, in Figure 5.6.2, we estimate the area under the curve using three subintervals and the trapezoids that result from connecting the corresponding points on the curve with straight lines.The Trapezoid Rule.
The trapezoidal approximation, of the definite integral using subintervals is given by the rule
Moreover,
Activity 5.6.2.
In this activity, we explore the relationships among the errors generated by left, right, midpoint, and trapezoid approximations to the definite integral
Use the First FTC to evaluate exactly.
Use appropriate computing technology to compute the following approximations for and
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Let the error of an approximation be the difference between the exact value of the definite integral and the resulting approximation. For instance, if we let represent the error that results from using the trapezoid rule with 4 subintervals to estimate the integral, we have
Similarly, we compute the error of the midpoint rule approximation with 8 subintervals by the formula
Based on your work in (a) and (b) above, compute
Which rule consistently over-estimates the exact value of the definite integral? Which rule consistently under-estimates the definite integral?
What behavior(s) of the function lead to your observations in (d)?
\(\frac{1}{x^2} = x^{-2}\text{.}\)
Use a computational device.
Use a computational device.
Which estimate is larger than the true value of the definite integral?
Note that how the curve bends makes a big difference in whether the trapezoid rule over- or under-estimates the value of the definite integral.
\(\int_1^2 \dfrac{1}{x^2} dx = \dfrac{1}{2}\text{.}\)
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The table below gives values of the trapezoid rule and corresponding errors for different \(n\)-values.
\(n\) \(T_n\) \(E_{T,n}\) \(4\) \(0.50899\) \(-0.50899\) \(8\) \(0.50227\) \(-0.50227\) \(16\) \(0.50057\) \(-0.50057\) -
The table below gives values of the midpoint rule and corresponding errors for different \(n\)-values.
\(n\) \(M_n\) \(E_{M,n}\) \(4\) \(0.49555\) \(0.00445\) \(8\) \(0.49887\) \(0.00113\) \(16\) \(0.49972\) \(0.00028\) The trapezoid rule overestimates; the midpoint rule underestimates.
\(f(x) = \dfrac{1}{x^2}\) is concave up on \([1, 2]\text{.}\)
\(\int_1^2 \dfrac{1}{x^2} dx = \left. -x^{-1} \right|_1^2 = -\dfrac{1}{2} + 1 = \dfrac{1}{2}\text{.}\)
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The table below gives values of the trapezoid rule and corresponding errors for different \(n\)-values.
\(n\) \(T_n\) \(E_{T,n}\) \(4\) \(0.50899\) \(-0.50899\) \(8\) \(0.50227\) \(-0.50227\) \(16\) \(0.50057\) \(-0.50057\) -
The table below gives values of the midpoint rule and corresponding errors for different \(n\)-values.
\(n\) \(M_n\) \(E_{M,n}\) \(4\) \(0.49555\) \(0.00445\) \(8\) \(0.49887\) \(0.00113\) \(16\) \(0.49972\) \(0.00028\) From the errors in comparision to the known exact value, we see that the trapezoid rule overestimates this definite integral and the midpoint rule underestimates this definite integral.
The graph of the function given by \(f(x) = \dfrac{1}{x^2}\) is concave up on the interval \([1, 2]\text{.}\) Because of this fact, we can see graphically that the line forming the top of each trapezoid lies fully above the curve, and thus the trapezoid rule overestimates the true value of the definite integral. Later in this section we'll see graphically why this concavity makes the midpoint rule an underestimate.
Subsection 5.6.2 Comparing the Midpoint and Trapezoid Rules
We know from the definition of the definite integral that if we let be large enough, we can make any of the approximations and as close as we'd like (in theory) to the exact value of Thus, it may be natural to wonder why we ever use any rule other than or (with a sufficiently large value) to estimate a definite integral. One of the primary reasons is that as and thus in a Riemann sum calculation with a large value, we end up multiplying by a number that is very close to zero. Doing so often generates roundoff error, because representing numbers close to zero accurately is a persistent challenge for computers. Hence, we explore ways to estimate definite integrals to high levels of precision, but without using extremely large values of Paying close attention to patterns in errors, such as those observed in Activity 5.6.2, is one way to begin to see some alternate approaches. To begin, we compare the errors in the Midpoint and Trapezoid rules. First, consider a function that is concave up on a given interval, and picture approximating the area bounded on that interval by both the Midpoint and Trapezoid rules using a single subinterval.Rule | error | error | ||
Subsection 5.6.3 Simpson's Rule
When we first developed the Trapezoid Rule, we observed that it is an average of the Left and Right Riemann sums:Rule | error | error | ||
Activity 5.6.3.
A car traveling along a straight road is braking and its velocity is measured at several different points in time, as given in the following table. Assume that is continuous, always decreasing, and always decreasing at a decreasing rate, as is suggested by the data.
seconds, | Velocity in ft/sec, |
Plot the given data on the set of axes provided in Figure 5.6.8 with time on the horizontal axis and the velocity on the vertical axis.
What definite integral will give you the exact distance the car traveled on
Estimate the total distance traveled on by computing and Which of these under-estimates the true distance traveled?
Estimate the total distance traveled on by computing Is this an over- or under-estimate? Why?
Using your results from (c) and (d), improve your estimate further by using Simpson's Rule.
What is your best estimate of the average velocity of the car on Why? What are the units on this quantity?
Plot the data.
What are the units on \(v(t) \cdot \Delta t\text{?}\)
Recall the standard rules for sums that produce \(L_3\text{,}\) \(R_3\text{,}\) \(T_3\text{.}\)
Think about concavity to decide if \(M_3\) is an over- or under-estimate.
Recall how \(S_3\) is a weighted average of \(T_3\) and \(M_3\text{.}\)
Simpson's Rule gives the best estimate for a function of consistent concavity.
Plot the data.
\(\int_0^{1.8} v(t) dt\text{.}\)
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\begin{align*} L_3 \amp = 165.6 \text{ ft } \amp R_3 \amp = 105.6 \text{ ft } \amp T_3 \amp = 135.6 \text{ ft }\text{.} \end{align*}
\(R_3\) and \(T_3\) are underestimates.
\(M_3 = 143.4 \text{ ft } \) ; overestimate.
\(S_6 = 140.8 \text{ ft } \text{.}\)
Simpson's rule gives the best approximation of the distance traveled, \(\int_0^{1.8} v(t) dt \approx 140.8 \text{ ft }\text{.}\)
Plot the data.
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Since the velocity is always positive, the definite integral that will give the exact distance traveled by the car on the interval \([0, 1.8]\) is
\begin{equation*} \int_0^{1.8} v(t) dt\text{.} \end{equation*} -
The estimates of \(\int_0^{1.8} v(t) dt\) are
\begin{align*} L_3 \amp = 165.6 \text{ ft } \amp R_3 \amp = 105.6 \text{ ft } \amp T_3 \amp = 135.6 \text{ ft }\text{.} \end{align*}\(R_3\) is an underestimate of the distance traveled since \(v(t)\) is decreasing. \(T_3\) is an underestimate of the distance traveled since \(v(t)\) is concave down.
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Another estimate of the distance traveled is
\begin{equation*} M_3 = 143.4 \text{ ft }\text{.} \end{equation*}This is an overestimate since \(v(t)\) is concave down.
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For Simpson's Rule, we see that
\begin{equation*} S_6 = \frac{2}{3}M_3 + \frac{1}{3}T_3 = 140.8 \text{ ft }\text{.} \end{equation*} Simpson's rule gives the best approximation of the distance traveled since it is a weighted average of the midpoint and trapezoid rules and uses more information about the velocity than the other methods. The units on each of the estimates, including Simpson's Rule, are "feet", since ft/sec \(\cdot\) sec = ft. Thus, the best approximation we have generated is that \(\int_0^{1.8} v(t) dt \approx 140.8 \text{ ft }\text{.}\)
Subsection 5.6.4 Overall observations regarding and
As we conclude our discussion of numerical approximation of definite integrals, it is important to summarize general trends in how the various rules over- or under-estimate the true value of a definite integral, and by how much. To revisit some past observations and see some new ones, we consider the following activity.Activity 5.6.4.
Consider the functions and all on the interval For each of the questions that require a numerical answer in what follows, write your answer exactly in fraction form.
On the three sets of axes provided in Figure 5.6.9, sketch a graph of each function on the interval and compute and for each. What do you observe?
Compute for each function to approximate and respectively.
Compute for each of the three functions, and hence compute for each of the three functions.
Evaluate each of the integrals and exactly using the First FTC.
For each of the three functions and compare the results of and to the true value of the corresponding definite integral. What patterns do you observe?
For each estimate, just one function evaluation is needed.
Use the midpoint rule with \(n=1\text{.}\)
Remember that both the trapezoid and Simpson's rule can be executed using (weighted) averages of known values.
Find antiderivatives to evaluate the integrals exactly.
Think about trends in over- and under-estimates.
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For \(L_1\) and \(T_1\text{:}\)
\(f\) \(g\) \(h\) \(L_1=2\) \(L_1=2\) \(L_1=2\) \(R_1=1\) \(R_1=1\) \(R_1=1\) Table 5.6.10. Left and Trapezoid rules. The values of \(L_1\) and \(R_1\) are the same for all three.
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For the \(M_1\text{,}\)
\(f\) \(g\) \(h\) \(M_1=\frac{7}{4}\) \(M_1=\frac{15}{8}\) \(M_1=\frac{31}{16}\) Table 5.6.11. Midpoint Rule. -
For \(T_1\) and \(S_2\text{,}\)
\(f\) \(g\) \(h\) \(T_1=\frac{3}{2}\) \(T_1=\frac{3}{2}\) \(T_1=\frac{3}{2}\) \(S_2=\frac{5}{3} \approx 1.6667\) \(S_2=\frac{7}{4}\) \(S_2=\frac{43}{24} \approx 1.79167\) Table 5.6.12. Trapezoid and Simpson's Rule. - \begin{align*} \int_0^1 f(x) dx \amp = \frac{5}{3} \amp \int_0^1 g(x) dx \amp = \frac{7}{4} \amp \int_0^1 h(x) dx \amp = \frac{9}{5} \end{align*}
Left endpoint rule results are overestimates; right endpoint rules are underestimates; midpoint rules are overestimates; trapezoid rules are underestimates. Simpson's rule is exact for both \(f\) and \(g\text{,}\) while a slight overestimate of \(\int_0^1 h(x) dx\text{.}\)
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For the left and right endpoint rules, we see that
\(f\) \(g\) \(h\) \(L_1=2\) \(L_1=2\) \(L_1=2\) \(R_1=1\) \(R_1=1\) \(R_1=1\) Table 5.6.13. Left and Trapezoid rules. Thus, we observe that despite the fact the functions are all different, the values of \(L_1\) and \(R_1\) are the same for all three.
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For the midpoint rule, we find that
\(f\) \(g\) \(h\) \(M_1=\frac{7}{4}\) \(M_1=\frac{15}{8}\) \(M_1=\frac{31}{16}\) Table 5.6.14. Midpoint Rule. -
For the trapezoid rule and Simpson's rule,
\(f\) \(g\) \(h\) \(T_1=\frac{3}{2}\) \(T_1=\frac{3}{2}\) \(T_1=\frac{3}{2}\) \(S_2=\frac{5}{3} \approx 1.6667\) \(S_2=\frac{7}{4}\) \(S_2=\frac{43}{24} \approx 1.79167\) Table 5.6.15. Trapezoid and Simpson's Rule. -
The exact values of the three definite integrals are
\begin{align*} \int_0^1 f(x) dx \amp = \frac{5}{3} \amp \int_0^1 g(x) dx \amp = \frac{7}{4} \amp \int_0^1 h(x) dx \amp = \frac{9}{5}\\ \amp \approx 1.6667 \amp \amp = 1.75 \amp \amp = 1.8 \end{align*} We observe that each of the left endpoint rule results are overestimates, each of the right endpoint rules are underestimates, each of the midpoint rules are overestimates, and each of the trapezoid rules are underestimates. These results hold because each of the three functions are both decreasing and concave down. For Simpson's rule, we see that the result is exact for both \(f\) and \(g\text{,}\) while Simpson's rule is a slight overestimate of \(\int_0^1 h(x) dx\text{.}\)
Subsection 5.6.5 Summary
For a definite integral such as when we cannot use the First Fundamental Theorem of Calculus because the integrand lacks an elementary algebraic antiderivative, we can estimate the integral's value by using a sequence of Riemann sum approximations. Typically, we start by computing and for one or more chosen values of
The Trapezoid Rule, which estimates by using trapezoids, rather than rectangles, can also be viewed as the average of Left and Right Riemann sums. That is,
The Midpoint Rule is typically twice as accurate as the Trapezoid Rule, and the signs of the respective errors of these rules are opposites. Hence, by taking the weighted average we can build a much more accurate approximation to by using approximations we have already computed. The rule for is known as Simpson's Rule, which can also be developed by approximating a given continuous function with pieces of quadratic polynomials.
Exercises 5.6.6 Exercises
ΒΆ1. Various methods for numerically.
2. Comparison of methods for increasing concave down function.
3. Comparing accuracy for two similar functions.
4. Identifying and comparing methods.
5.
Consider the definite integral
Explain why this integral cannot be evaluated exactly by using either -substitution or by integrating by parts.
Using appropriate subintervals, compute and
Which of the approximations in (b) is an over-estimate to the true value of Which is an under-estimate? How do you know?
\(u\)-substitution fails since there's not a composite function present; try showing that each of the choices of \(u = x\) and \(dv = \tan(x) \, dx\text{,}\) or \(u = \tan(x)\) and \(dv = x \, dx\text{,}\) fail to produce an integral that can be evaluated by parts.
\(L_4 = 0.25892\)
\(R_4 = 0.64827\)
\(M_4 = 0.41550\)
\(T_4 = \frac{L_4 + R_4}{2} = 0.45360\)
\(S_8 = \frac{2M_4 + T_4}{3} = 0.42820\)
\(L_4\) and \(M_4\) are underestimates; \(R_4\) and \(M_4\) are overestimates.
We can't use \(u\)-substitution since there's not a function-derivative pair present (nor even a composite function). To show that integration by parts doesn't work is more complicated, but comes down to the fact that if we use the natural choice of \(u = x\) and \(dv = \tan(x) \, dx\text{,}\) to do so requires us to evaluate the integral \(\int \ln(\sec(x)) \, dx\text{,}\) which appears not to have an elementary antiderivative. If we try instead \(u = \tan(x)\) and \(dv = x \, dx\text{,}\) then we have to evaluate the integral \(\int x^2 \cdot \sec^2(x) \, dx\) which is more complicated than the integral we started with and would seem to send us back where we started if we attempt integration by parts.
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We use computational technology to generate the following estimates, which are reported to \(5\) decimal places of accuracy:
\(L_4 = 0.25892\)
\(R_4 = 0.64827\)
\(M_4 = 0.41550\)
\(T_4 = \frac{L_4 + R_4}{2} = 0.45360\)
\(S_8 = \frac{2M_4 + T_4}{3} = 0.42820\)
From the values of these estimates or from the graph of \(f(x) = x\tan(x)\text{,}\) we can observe that \(f\) is always increasing and concave up on the interval. Thus, \(L_4\) and \(M_4\) are underestimates while \(R_4\) and \(M_4\) are overestimates.
6.
For an unknown function the following information is known.
is continuous on
is either always increasing or always decreasing on
has the same concavity throughout the interval
As approximations to and
Is increasing or decreasing on What data tells you?
Is concave up or concave down on Why?
Determine the best possible estimate you can for based on the given information.
Decreasing.
Concave down.
\(\int_3^6 f(x) \approx 7.03\text{.}\)
Because \(L_4 = 7.23 \gt 6.75 = R_4\) and we know the function is always increasing or always decreasing, for the left endpoint rule to give a value greater than the right endpoint rule, \(f\) must be always decreasing.
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From the given data, we know that
\begin{equation*} T_4 = \frac{L_4 + R_4}{2} = \frac{7.23 + 6.75}{2} = 6.99\text{.} \end{equation*}Since \(M_4 = 7.05\text{,}\) it follows that
\begin{equation*} R_4 \lt T_4 \lt M_4 \lt L_4\text{.} \end{equation*}In particular, we know that the function is not only always decreasing on the interval, but it also concave down. Because the concavity is constant, it follows that if the trapezoid rule under-estimates the true value, then the midpoint rule must over-estimate the true value (or vice-versa). Since \(T_4 \lt M_4\text{,}\) this means that
\begin{equation*} T_4 \lt \int_3^6 f(x) \, dx \lt M_4 \end{equation*}and that \(f\) must be concave down on the interval.
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We use Simpson's Rule to determine the best possible estimate for the definite integral. That is,
\begin{align*} \int_3^6 f(x) \, dx &\approx S_8\\ &= \frac{2M_4 + T_4}{3}\\ &= \frac{2 \cdot 7.05 + 6.99}{3}\\ &=7.03\text{.} \end{align*}
7.
The rate at which water flows through Table Rock Dam on the White River in Branson, MO, is measured in thousands of cubic feet per second (TCFS). As engineers open the floodgates, flow rates are recorded according to the following chart.
seconds, | |||||||
flow in TCFS, |
What definite integral measures the total volume of water to flow through the dam in the 60 second time period provided by the table above?
Use the given data to calculate for the largest possible value of to approximate the integral you stated in (a). Do you think over- or under-estimates the exact value of the integral? Why?
Approximate the integral stated in (a) by calculating for the largest possible value of based on the given data.
Compute and What quantity do both of these values estimate? Which is a more accurate approximation?
\(\int_0^{60} r(t) \, dt \text{.}\)
\(\int_0^{60} r(t) \, dt \gt M_3 = 204000\text{.}\)
\(\int_0^{60} r(t) \, dt \approx S_6 = \frac{619000}{3} \approx 206333.33\text{.}\)
\(\frac{1}{60} S_6 \approx 3438.89 \text{;}\) \(\frac{2000+2100+2400+3000+3900+5100+6500}{7} = \frac{25000}{7} \approx 3571.43 \text{.}\) each estimates the average rate at which water flows through the dam on \([0,60]\text{,}\) and the first is more accurate.
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Because \(r(t)\) is measured in thousands of cubic feet per second and \(\triangle t\) is measured in seconds, the definite integral
\begin{equation*} \int_0^{60} r(t) \, dt \end{equation*}measures the total amount of water (in thousands of cubic feet) that flow through the dam in 60 seconds.
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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{align*} \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\text{.} \end{align*}Thus, approximately \(204 000 000\) cubic feet of water pass through the dam on the interval \(0 \le t \le 60\text{.}\)
We observe that \(r\) appears to be increasing at an increasing rate on \(0 \le t \le 60\text{.}\) 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,
\begin{equation*} \int_0^{60} r(t) \, dt \gt M_3 = 204000\text{.} \end{equation*} -
In order to estimate the integral using \(S_6\text{,}\) we need to know the value of \(T_3\) along with \(M_3\text{.}\) To find \(T_3\text{,}\) we compute \(L_3\) and \(R_3\text{.}\) First
\begin{equation*} L_3 = r(0) \cdot 20 + r(20) \cdot 20 + r(40) \cdot 20 = (2000 + 2400 + 3900) \cdot 20 = 166000\text{.} \end{equation*}Similarly,
\begin{equation*} R_3 = r(20) \cdot 20 + r(40) \cdot 20 + r(60) \cdot 20 = (2400 + 3900 + 6500) \cdot 20 = 256000\text{.} \end{equation*}Thus,
\begin{equation*} T_3 = \frac{L_3 + R_3}{2} = \frac{166000 + 256000}{2} = 211000\text{.} \end{equation*}Finally, we now have the information needed to compute \(S_6\text{.}\) We thus find that
\begin{equation*} \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\text{.} \end{equation*} -
First,
\begin{equation*} \frac{1}{60} S_6 = \frac{206333.33}{60} = \frac{30950}{9} \approx 3438.89\text{.} \end{equation*}This is one estimate for the average value of \(r\) on \([0,60]\) since
\begin{equation*} r_{\operatorname{AVG} [0,60]} = \frac{1}{60-0} \int_0^{60} r(t) \, dt\text{,} \end{equation*}and \(\int_0^{60} r(t) \, dt \approx S_6\text{.}\) Thus, on average, water is flowing through the dam at a rate of \(3438.89\) thousand cubic feet per second.
Next,
\begin{equation*} \frac{2000+2100+2400+3000+3900+5100+6500}{7} = \frac{25000}{7} \approx 3571.43 \end{equation*}is the average of the \(7\) known values of \(r(t)\) on the interval \([0,60]\text{,}\) and thus is one estiamte of the average value of \(r\text{.}\) 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 \(R_6\text{.}\) 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.