AC Numerical Integration

Section 5.6 Numerical Integration

Motivating Questions

How do we accurately evaluate a definite integral such as $\int_0^1 e^{-x^2} \, dx$ 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 $n$ 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?

When we first explored finding the net signed area bounded by a curve, we developed the concept of a Riemann sum as a helpful estimation tool and a key step in the definition of the definite integral. Recall that the left, right, and middle Riemann sums of a function

f

$f$ on an interval

[a, b]

$[a,b]$ are given by

\begin{aligned} (5.6.1) & L_{n} = f (x_{0}) Δ x + f (x_{1}) Δ x + \dots + f (x_{n - 1}) Δ x & = \sum_{i = 0}^{n - 1} f (x_{i}) Δ x, \\ (5.6.2) & R_{n} = f (x_{1}) Δ x + f (x_{2}) Δ x + \dots + f (x_{n}) Δ x & = \sum_{i = 1}^{n} f (x_{i}) Δ x, \\ (5.6.3) & M_{n} = f ({\bar{x}}_{1}) Δ x + f ({\bar{x}}_{2}) Δ x + \dots + f ({\bar{x}}_{n}) Δ x & = \sum_{i = 1}^{n} f ({\bar{x}}_{i}) Δ x, \end{aligned}

$\begin{align} L_n = f(x_0) \Delta x + f(x_1) \Delta x + \cdots + f(x_{n-1}) \Delta x \amp= \sum_{i = 0}^{n-1} f(x_i) \Delta x,\label{E-Left}\tag{5.6.1}\\ R_n = f(x_1) \Delta x + f(x_2) \Delta x + \cdots + f(x_{n}) \Delta x \amp= \sum_{i = 1}^{n} f(x_i) \Delta x,\label{E-Right}\tag{5.6.2}\\ M_n = f(\overline{x}_1) \Delta x + f(\overline{x}_2) \Delta x + \cdots + f(\overline{x}_{n}) \Delta x \amp= \sum_{i = 1}^{n} f(\overline{x}_i) \Delta x\text{,}\label{E-Mid}\tag{5.6.3} \end{align}$

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where

x_{0} = a,

$x_0 = a\text{,}$

x_{i} = a + i Δ x,

$x_i = a + i\Delta x\text{,}$

x_{n} = b,

$x_n = b\text{,}$ and

Δ x = \frac{b - a}{n} .

$\Delta x = \frac{b-a}{n}\text{.}$ For the middle sum, we defined

{\bar{x}}_{i} = (x_{i - 1} + x_{i}) / 2 .

$\overline{x}_{i} = (x_{i-1} + x_i)/2\text{.}$

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A Riemann sum is a sum of (possibly signed) areas of rectangles. The value of

n

$n$ determines the number of rectangles, and our choice of left endpoints, right endpoints, or midpoints determines the heights of the rectangles. We can see the similarities and differences among these three options in Figure 5.6.1, where we consider the function

f (x) = \frac{1}{20} (x - 4)^{3} + 7

$f(x) = \frac{1}{20}(x-4)^3 + 7$ on the interval

[1, 8],

$[1,8]\text{,}$ and use 5 rectangles for each of the Riemann sums.

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Figure 5.6.1. Left, right, and middle Riemann sums for

y = f (x)

$y = f(x)$ on

[1, 8]

$[1,8]$ with 5 subintervals.

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While it is a good exercise to compute a few Riemann sums by hand, just to ensure that we understand how they work and how varying the function, the number of subintervals, and the choice of endpoints or midpoints affects the result, using computing technology is the best way to determine

L_{n},

$L_n\text{,}$

R_{n},

$R_n\text{,}$ and

M_{n} .

$M_n\text{.}$ Any computer algebra system will offer this capability; as we saw in Preview Activity 4.3.1, a straightforward option that is freely available online is the applet¹ at http://gvsu.edu/s/a9. Note that we can adjust the formula for

f (x),

$f(x)\text{,}$ the window of

x

$x$ - and

y

$y$ -values of interest, the number of subintervals, and the method. (See Preview Activity 4.3.1 for any needed reminders on how the applet works.)

Marc Renault, Shippensburg University

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In this section we explore several different alternatives for estimating definite integrals. Our main goal is to develop formulas to estimate definite integrals accurately without using a large numbers of rectangles.

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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 $\int_0^3 x^2 \, dx\text{.}$

Use the applet at http://gvsu.edu/s/a9 with the function $f(x) = x^2$ on the window of $x$ values from $0$ to $3$ to compute $L_3\text{,}$ the left Riemann sum with three subintervals.
Likewise, use the applet to compute $R_3$ and $M_3\text{,}$ the right and middle Riemann sums with three subintervals, respectively.
Use the Fundamental Theorem of Calculus to compute the exact value of $I = \int_0^3 x^2 \, dx\text{.}$
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 $L_3\text{?}$ From $R_3\text{?}$ From $M_3\text{?}$
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 $b_1$ and $b_2$ and height $h\text{?}$
Working by hand, estimate the area under $f(x) = x^2$ on $[0,3]$ 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)?

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Subsection 5.6.1 The Trapezoid Rule

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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.

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An alternative to

L_{n},

$L_n\text{,}$

R_{n},

$R_n\text{,}$ and

M_{n}

$M_n$ is called the Trapezoid Rule. Rather than using a rectangle to estimate the (signed) area bounded by

y = f (x)

$y = f(x)$ 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.

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Figure 5.6.2. Estimating

\int_{a}^{b} f (x) d x

$\int_a^b f(x) \ dx$ using three subintervals and trapezoids, rather than rectangles, where

a = x_{0}

$a = x_0$ and

b = x_{3} .

$b = x_3\text{.}$

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The biggest difference between the Trapezoid Rule and a Riemann sum is that on each subinterval, the Trapezoid Rule uses two function values, rather than one, to estimate the (signed) area bounded by the curve. For instance, to compute

D_{1},

$D_1\text{,}$ the area of the trapezoid on

[x_{0}, x_{1}],

$[x_0, x_1]\text{,}$ we observe that the left base has length

f (x_{0}),

$f(x_0)\text{,}$ while the right base has length

f (x_{1}) .

$f(x_1)\text{.}$ The height of the trapezoid is

x_{1} - x_{0} = Δ x = \frac{b - a}{3} .

$x_1 - x_0 = \Delta x = \frac{b-a}{3}\text{.}$ The area of a trapezoid is the average of the bases times the height, so we have

D_{1} = \frac{1}{2} (f (x_{0}) + f (x_{1})) \cdot Δ x .

$\begin{equation*} D_1 = \frac{1}{2}(f(x_0) + f(x_1)) \cdot \Delta x\text{.} \end{equation*}$

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Using similar computations for

D_{2}

$D_2$ and

D_{3},

$D_3\text{,}$ we find that

T_{3},

$T_3\text{,}$ the trapezoidal approximation to

\int_{a}^{b} f (x) d x

$\int_a^b f(x) \, dx$ is given by

\begin{aligned} T_{3} & = D_{1} + D_{2} + D_{3} \\ = \frac{1}{2} (f (x_{0}) + f (x_{1})) \cdot Δ x + \frac{1}{2} (f (x_{1}) + f (x_{2})) \cdot Δ x + \frac{1}{2} (f (x_{2}) + f (x_{3})) \cdot Δ x . \end{aligned}

$\begin{align*} T_3 &= D_1 + D_2 + D_3\\ &= \frac{1}{2}(f(x_0) + f(x_1)) \cdot \Delta x + \frac{1}{2}(f(x_1) + f(x_2)) \cdot \Delta x + \frac{1}{2}(f(x_2) + f(x_3)) \cdot \Delta x\text{.} \end{align*}$

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Because both left and right endpoints are being used, we recognize within the trapezoidal approximation the use of both left and right Riemann sums. Rearranging the expression for

T_{3}

$T_3$ by removing factors of

\frac{1}{2}

$\frac{1}{2}$ and

Δ x,

$\Delta x \text{,}$ grouping the left endpoint and right endpoint evaluations of

f,

$f\text{,}$ we see that

\begin{matrix} (5.6.4) & T_{3} = \frac{1}{2} [f (x_{0}) + f (x_{1}) + f (x_{2})] Δ x + \frac{1}{2} [f (x_{1}) + f (x_{2}) + f (x_{3})] Δ x . \end{matrix}

$\begin{equation} T_3 = \frac{1}{2} \left[ f(x_0) + f(x_1) + f(x_2) \right] \Delta x + \frac{1}{2} \left[ f(x_1) + f(x_2) + f(x_3) \right] \Delta x\text{.}\label{xGO}\tag{5.6.4} \end{equation}$

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We now observe that two familiar sums have arisen. The left Riemann sum

L_{3}

$L_3$ is

L_{3} = f (x_{0}) Δ x + f (x_{1}) Δ x + f (x_{2}) Δ x,

$L_3 = f(x_0) \Delta x + f(x_1) \Delta x + f(x_2) \Delta x\text{,}$ and the right Riemann sum is

R_{3} = f (x_{1}) Δ x + f (x_{2}) Δ x + f (x_{3}) Δ x .

$R_3 = f(x_1) \Delta x + f(x_2) \Delta x + f(x_3) \Delta x\text{.}$ Substituting

L_{3}

$L_3$ and

R_{3}

$R_3$ for the corresponding expressions in Equation (5.6.4), it follows that

T_{3} = \frac{1}{2} [L_{3} + R_{3}] .

$T_3 = \frac{1}{2} \left[ L_3 + R_3 \right]\text{.}$ We have thus seen a very important result: using trapezoids to estimate the (signed) area bounded by a curve is the same as averaging the estimates generated by using left and right endpoints.

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The Trapezoid Rule.

The trapezoidal approximation, $T_n\text{,}$ of the definite integral $\int_a^b f(x) \, dx$ using $n$ subintervals is given by the rule

\begin{aligned} T_{n} = & [\frac{1}{2} (f (x_{0}) + f (x_{1})) + \frac{1}{2} (f (x_{1}) + f (x_{2})) + \dots + \frac{1}{2} (f (x_{n - 1}) + f (x_{n}))] Δ x . \\ = & \sum_{i = 0}^{n - 1} \frac{1}{2} (f (x_{i}) + f (x_{i + 1})) Δ x . \end{aligned}

$\begin{align*} T_n =\mathstrut \amp \left[\frac{1}{2}(f(x_0) + f(x_1)) + \frac{1}{2}(f(x_1) + f(x_2)) + \cdots + \frac{1}{2}(f(x_{n-1}) + f(x_n)) \right] \Delta x.\\ =\mathstrut \amp \sum_{i=0}^{n-1} \frac{1}{2}(f(x_i) + f(x_{i+1})) \Delta x\text{.} \end{align*}$

Moreover, $T_n = \frac{1}{2} \left[ L_n + R_n \right]\text{.}$

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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 $\int_1^2 \frac{1}{x^2} \, dx$

Use the First FTC to evaluate $\int_1^2 \frac{1}{x^2} \, dx$ exactly.
Use appropriate computing technology to compute the following approximations for $\int_1^2 \frac{1}{x^2} \, dx\text{:}$ $T_4\text{,}$ $M_4\text{,}$ $T_8\text{,}$ and $M_8\text{.}$
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 $E_{T,4}$ represent the error that results from using the trapezoid rule with 4 subintervals to estimate the integral, we have

$\begin{equation*} E_{T,4} = \int_1^2 \frac{1}{x^2} \, dx - T_4\text{.} \end{equation*}$

Similarly, we compute the error of the midpoint rule approximation with 8 subintervals by the formula

$\begin{equation*} E_{M,8} = \int_1^2 \frac{1}{x^2} \, dx - M_8\text{.} \end{equation*}$

Based on your work in (a) and (b) above, compute $E_{T,4}\text{,}$ $E_{T,8}\text{,}$ $E_{M,4}\text{,}$ $E_{M,8}\text{.}$
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 $f(x) = \frac{1}{x^2}$ lead to your observations in (d)?

Hint

$\frac{1}{x^2} = x^{-2}\text{.}$
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.

Answer

$\int_1^2 \dfrac{1}{x^2} dx = \dfrac{1}{2}\text{.}$
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 $[1, 2]\text{.}$

Solution

$\int_1^2 \dfrac{1}{x^2} dx = \left. -x^{-1} \right|_1^2 = -\dfrac{1}{2} + 1 = \dfrac{1}{2}\text{.}$
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.

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Subsection 5.6.2 Comparing the Midpoint and Trapezoid Rules

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We know from the definition of the definite integral that if we let

n

$n$ be large enough, we can make any of the approximations

L_{n},

$L_n\text{,}$

R_{n},

$R_n\text{,}$ and

M_{n}

$M_n$ as close as we'd like (in theory) to the exact value of

\int_{a}^{b} f (x) d x .

$\int_a^b f(x) \, dx\text{.}$ Thus, it may be natural to wonder why we ever use any rule other than

L_{n}

$L_n$ or

R_{n}

$R_n$ (with a sufficiently large

n

$n$ value) to estimate a definite integral. One of the primary reasons is that as

n \to \infty,

$n \to \infty\text{,}$

Δ x = \frac{b - a}{n} \to 0,

$\Delta x = \frac{b-a}{n} \to 0\text{,}$ and thus in a Riemann sum calculation with a large

n

$n$ 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.

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Hence, we explore ways to estimate definite integrals to high levels of precision, but without using extremely large values of

n .

$n\text{.}$ 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.

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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.

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Figure 5.6.3. Estimating

\int_{a}^{b} f (x) d x

$\int_a^b f(x) \ dx$ using a single subinterval: at left, the trapezoid rule; in the middle, the midpoint rule; at right, a modified way to think about the midpoint rule.

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As seen in Figure 5.6.3, it is evident that whenever the function is concave up on an interval, the Trapezoid Rule with one subinterval,

T_{1},

$T_1\text{,}$ will overestimate the exact value of the definite integral on that interval. From a careful analysis of the line that bounds the top of the rectangle for the Midpoint Rule (shown in magenta), we see that if we rotate this line segment until it is tangent to the curve at the midpoint of the interval (as shown at right in Figure 5.6.3), the resulting trapezoid has the same area as

M_{1},

$M_1\text{,}$ and this value is less than the exact value of the definite integral. Thus, when the function is concave up on the interval,

M_{1}

$M_1$ underestimates the integral's true value.

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Figure 5.6.4. Comparing the error in estimating

\int_{a}^{b} f (x) d x

$\int_a^b f(x) \ dx$ using a single subinterval: in red, the error from the Trapezoid rule; in light red, the error from the Midpoint rule.

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These observations extend easily to the situation where the function's concavity remains consistent but we use larger values of

n

$n$ in the Midpoint and Trapezoid Rules. Hence, whenever

f

$f$ is concave up on

[a, b],

$[a,b]\text{,}$

T_{n}

$T_n$ will overestimate the value of

\int_{a}^{b} f (x) d x,

$\int_a^b f(x) \, dx\text{,}$ while

M_{n}

$M_n$ will underestimate

\int_{a}^{b} f (x) d x .

$\int_a^b f(x) \, dx\text{.}$ The reverse observations are true in the situation where

f

$f$ is concave down.

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Next, we compare the size of the errors between

M_{n}

$M_n$ and

T_{n} .

$T_n\text{.}$ Again, we focus on

M_{1}

$M_1$ and

T_{1}

$T_1$ on an interval where the concavity of

f

$f$ is consistent. In Figure 5.6.4, where the error of the Trapezoid Rule is shaded in red, while the error of the Midpoint Rule is shaded lighter red, it is visually apparent that the error in the Trapezoid Rule is more significant. To see how much more significant, let's consider two examples and some particular computations.

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If we let

f (x) = 1 - x^{2}

$f(x) = 1-x^2$ and consider

\int_{0}^{1} f (x) d x,

$\int_0^1 f(x) \,dx\text{,}$ we know by the First FTC that the exact value of the integral is

\int_{0}^{1} (1 - x^{2}) d x = x - \frac{x^{3}}{3} |_{0}^{1} = \frac{2}{3} .

$\begin{equation*} \int_0^1 (1-x^2) \, dx = x - \frac{x^3}{3} \bigg\vert_0^1 = \frac{2}{3}\text{.} \end{equation*}$

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Using appropriate technology to compute

M_{4},

$M_4\text{,}$

M_{8},

$M_8\text{,}$

T_{4},

$T_4\text{,}$ and

T_{8},

$T_8\text{,}$ as well as the corresponding errors

E_{M, 4},

$E_{M,4}\text{,}$

E_{M, 8},

$E_{M,8}\text{,}$

E_{T, 4},

$E_{T,4}\text{,}$ and

E_{T, 8},

$E_{T,8}\text{,}$ as we did in Activity 5.6.2, we find the results summarized in Table 5.6.5. We also include the approximations and their errors for the example

\int_{1}^{2} \frac{1}{x^{2}} d x

$\int_1^2 \frac{1}{x^2} \, dx$ from Activity 5.6.2.

Rule	$\int_0^1 (1-x^2) \,dx = 0.\overline{6}$	error	$\int_1^2 \frac{1}{x^2} \, dx = 0.5$	error
$T_4$	$0.65625$	$-0.0104166667$	$0.5089937642$	$0.0089937642$
$M_4$	$0.671875$	$0.0052083333$	$0.4955479365$	$-0.0044520635$
$T_8$	$0.6640625$	$-0.0026041667$	$0.5022708502$	$0.0022708502$
$M_8$	$0.66796875$	$0.0013020833$	$0.4988674899$	$-0.0011325101$

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Table 5.6.5. Calculations of

T_{4},

$T_4\text{,}$

M_{4},

$M_4\text{,}$

T_{8},

$T_8\text{,}$ and

M_{8},

$M_8\text{,}$ along with corresponding errors, for the definite integrals

\int_{0}^{1} (1 - x^{2}) d x

$\int_0^1 (1-x^2) \ dx$ and

\int_{1}^{2} \frac{1}{x^{2}} d x .

$\int_1^2 \frac{1}{x^2} \ dx\text{.}$

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For a given function

f

$f$ and interval

[a, b],

$[a,b]\text{,}$

E_{T, 4} = \int_{a}^{b} f (x) d x - T_{4}

$E_{T,4} = \int_a^b f(x) \,dx - T_4$ calculates the difference between the exact value of the definite integral and the approximation generated by the Trapezoid Rule with

n = 4 .

$n = 4\text{.}$ If we look at not only

E_{T, 4},

$E_{T,4}\text{,}$ but also the other errors generated by using

T_{n}

$T_n$ and

M_{n}

$M_n$ with

n = 4

$n = 4$ and

n = 8

$n = 8$ in the two examples noted in Table 5.6.5, we see an evident pattern. Not only is the sign of the error (which measures whether the rule generates an over- or under-estimate) tied to the rule used and the function's concavity, but the magnitude of the errors generated by

T_{n}

$T_n$ and

M_{n}

$M_n$ seems closely connected. In particular, the errors generated by the Midpoint Rule seem to be about half the size of those generated by the Trapezoid Rule.

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That is, we can observe in both examples that

E_{M, 4} \approx - \frac{1}{2} E_{T, 4}

$E_{M,4} \approx -\frac{1}{2} E_{T,4}$ and

E_{M, 8} \approx - \frac{1}{2} E_{T, 8} .

$E_{M,8} \approx -\frac{1}{2}E_{T,8}\text{.}$ This property of the Midpoint and Trapezoid Rules turns out to hold in general: for a function of consistent concavity, the error in the Midpoint Rule has the opposite sign and approximately half the magnitude of the error of the Trapezoid Rule. Written symbolically,

E_{M, n} \approx - \frac{1}{2} E_{T, n} .

$\begin{equation*} E_{M,n} \approx -\frac{1}{2} E_{T,n}\text{.} \end{equation*}$

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This important relationship suggests a way to combine the Midpoint and Trapezoid Rules to create an even more accurate approximation to a definite integral.

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Subsection 5.6.3 Simpson's Rule

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When we first developed the Trapezoid Rule, we observed that it is an average of the Left and Right Riemann sums:

T_{n} = \frac{1}{2} (L_{n} + R_{n}) .

$\begin{equation*} T_n = \frac{1}{2}(L_n + R_n)\text{.} \end{equation*}$

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If a function is always increasing or always decreasing on the interval

[a, b],

$[a,b]\text{,}$ one of

L_{n}

$L_n$ and

R_{n}

$R_n$ will over-estimate the true value of

\int_{a}^{b} f (x) d x,

$\int_a^b f(x) \, dx\text{,}$ while the other will under-estimate the integral. Thus, the errors found in

L_{n}

$L_n$ and

R_{n}

$R_n$ will have opposite signs; so averaging

L_{n}

$L_n$ and

R_{n}

$R_n$ eliminates a considerable amount of the error present in the respective approximations. In a similar way, it makes sense to think about averaging

M_{n}

$M_n$ and

T_{n}

$T_n$ in order to generate a still more accurate approximation.

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We've just observed that

M_{n}

$M_n$ is typically about twice as accurate as

T_{n} .

$T_n\text{.}$ So we use the weighted average

\begin{matrix} (5.6.5) & S_{2 n} = \frac{2 M_{n} + T_{n}}{3} . \end{matrix}

$\begin{equation} S_{2n} = \frac{2M_n + T_n}{3}\text{.}\label{CqH}\tag{5.6.5} \end{equation}$

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The rule for

S_{2 n}

$S_{2n}$ giving by Equation (5.6.5) is usually known as Simpson's Rule.² Note that we use “

S_{2 n}

$S_{2n}$ ” rather that “

S_{n}

$S_n$ ” since the

n

$n$ points the Midpoint Rule uses are different from the

n

$n$ points the Trapezoid Rule uses, and thus Simpson's Rule is using

2 n

$2n$ points at which to evaluate the function. We build upon the results in Table 5.6.5 to see the approximations generated by Simpson's Rule. In particular, in Table 5.6.6, we include all of the results in Table 5.6.5, but include additional results for

S_{8} = \frac{2 M_{4} + T_{4}}{3}

$S_8 = \frac{2M_4 + T_4}{3}$ and

S_{16} = \frac{2 M_{8} + T_{8}}{3} .

$S_{16} = \frac{2M_8 + T_8}{3}\text{.}$

Thomas Simpson was an 18th century mathematician; his idea was to extend the Trapezoid rule, but rather than using straight lines to build trapezoids, to use quadratic functions to build regions whose area was bounded by parabolas (whose areas he could find exactly). Simpson's Rule is often developed from the more sophisticated perspective of using interpolation by quadratic functions.

Rule	$\int_0^1 (1-x^2) \,dx = 0.\overline{6}$	error	$\int_1^2 \frac{1}{x^2} \, dx = 0.5$	error
$T_4$	$0.65625$	$-0.0104166667$	$0.5089937642$	$0.0089937642$
$M_4$	$0.671875$	$0.0052083333$	$0.4955479365$	$-0.0044520635$
$S_8$	$0.6666666667$	$0$	$0.5000298792$	$0.0000298792$
$T_8$	$0.6640625$	$-0.0026041667$	$0.5022708502$	$0.0022708502$
$M_8$	$0.66796875$	$0.0013020833$	$0.4988674899$	$-0.0011325101$
$S_{16}$	$0.6666666667$	$0$	$0.5000019434$	$0.0000019434$

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Table 5.6.6. Table 5.6.5 updated to include

S_{8},

$S_8\text{,}$

S_{16},

$S_{16}\text{,}$ and the corresponding errors.

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The results seen in Table 5.6.6 are striking. If we consider the

S_{16}

$S_{16}$ approximation of

\int_{1}^{2} \frac{1}{x^{2}} d x,

$\int_1^2 \frac{1}{x^2} \, dx\text{,}$ the error is only

E_{S, 16} = 0.0000019434 .

$E_{S,16} = 0.0000019434\text{.}$ By contrast,

L_{8} = 0.5491458502,

$L_8 = 0.5491458502\text{,}$ so the error of that estimate is

E_{L, 8} = - 0.0491458502 .

$E_{L,8} = -0.0491458502\text{.}$ Moreover, we observe that generating the approximations for Simpson's Rule is almost no additional work: once we have

L_{n},

$L_n\text{,}$

R_{n},

$R_n\text{,}$ and

M_{n}

$M_n$ for a given value of

n,

$n\text{,}$ it is a simple exercise to generate

T_{n},

$T_n\text{,}$ and from there to calculate

S_{2 n} .

$S_{2n}\text{.}$ Finally, note that the error in the Simpson's Rule approximations of

\int_{0}^{1} (1 - x^{2}) d x

$\int_0^1 (1-x^2) \, dx$ is zero!³

Similar to how the Midpoint and Trapezoid approximations are exact for linear functions, Simpson's Rule approximations are exact for quadratic and cubic functions. See additional discussion on this issue later in the section and in the exercises.

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These rules are not only useful for approximating definite integrals such as

\int_{0}^{1} e^{- x^{2}} d x,

$\int_0^1 e^{-x^2} \, dx\text{,}$ for which we cannot find an elementary antiderivative of

e^{- x^{2}},

$e^{-x^2}\text{,}$ but also for approximating definite integrals when we are given a function through a table of data.

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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 $v$ is continuous, always decreasing, and always decreasing at a decreasing rate, as is suggested by the data.

seconds, $t$	Velocity in ft/sec, $v(t)$
$0$	$100$
$0.3$	$99$
$0.6$	$96$
$0.9$	$90$
$1.2$	$80$
$1.5$	$50$
$1.8$	$0$

Table 5.6.7. Data for the braking car.

Figure 5.6.8. Axes for plotting the data in Activity 5.6.3.

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 $[0,1.8]\text{?}$
Estimate the total distance traveled on $[0,1.8]$ by computing $L_3\text{,}$ $R_3\text{,}$ and $T_3\text{.}$ Which of these under-estimates the true distance traveled?
Estimate the total distance traveled on $[0,1.8]$ by computing $M_3\text{.}$ 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 $[0,1.8]\text{?}$ Why? What are the units on this quantity?

Hint

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.

Answer

Plot the data.
$\int_0^{1.8} v(t) dt\text{.}$
\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, $\int_0^{1.8} v(t) dt \approx 140.8 \text{ ft }\text{.}$

Solution

Plot the data.
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.
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.
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 $\int_0^{1.8} v(t) dt \approx 140.8 \text{ ft }\text{.}$

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Subsection 5.6.4 Overall observations regarding $L_n\text{,}$ $R_n\text{,}$ $T_n\text{,}$ $M_n\text{,}$ and $S_{2n}\text{.}$

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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.

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Activity 5.6.4.

Consider the functions $f(x) = 2-x^2\text{,}$ $g(x) = 2-x^3\text{,}$ and $h(x) = 2-x^4\text{,}$ all on the interval $[0,1]\text{.}$ 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 $[0,1]\text{,}$ and compute $L_1$ and $R_1$ for each. What do you observe?
Compute $M_1$ for each function to approximate $\int_0^1 f(x) \,dx\text{,}$ $\int_0^1 g(x) \,dx\text{,}$ and $\int_0^1 h(x) \,dx\text{,}$ respectively.
Compute $T_1$ for each of the three functions, and hence compute $S_2$ for each of the three functions.
Evaluate each of the integrals $\int_0^1 f(x) \,dx\text{,}$ $\int_0^1 g(x) \,dx\text{,}$ and $\int_0^1 h(x) \,dx$ exactly using the First FTC.
For each of the three functions $f\text{,}$ $g\text{,}$ and $h\text{,}$ compare the results of $L_1\text{,}$ $R_1\text{,}$ $M_1\text{,}$ $T_1\text{,}$ and $S_2$ to the true value of the corresponding definite integral. What patterns do you observe?

Figure 5.6.9. Axes for plotting the functions in Activity 5.6.4.

Hint

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.

Answer

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.
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 $\int_0^1 h(x) dx\text{.}$

Solution

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.
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 $\int_0^1 h(x) dx\text{.}$

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The results seen in Activity 5.6.4 generalize nicely. For instance, if

f

$f$ is decreasing on

[a, b],

$[a,b]\text{,}$

L_{n}

$L_n$ will over-estimate the exact value of

\int_{a}^{b} f (x) d x,

$\int_a^b f(x) \,dx\text{,}$ and if

f

$f$ is concave down on

[a, b],

$[a,b]\text{,}$

M_{n}

$M_n$ will over-estimate the exact value of the integral. An excellent exercise is to write a collection of scenarios of possible function behavior, and then categorize whether each of

L_{n},

$L_n\text{,}$

R_{n},

$R_n\text{,}$

T_{n},

$T_n\text{,}$ and

M_{n}

$M_n$ is an over- or under-estimate.

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Finally, we make two important notes about Simpson's Rule. When T. Simpson first developed this rule, his idea was to replace the function

f

$f$ on a given interval with a quadratic function that shared three values with the function

f .

$f\text{.}$ In so doing, he guaranteed that this new approximation rule would be exact for the definite integral of any quadratic polynomial. In one of the pleasant surprises of numerical analysis, it turns out that even though it was designed to be exact for quadratic polynomials, Simpson's Rule is exact for any cubic polynomial: that is, if we are interested in an integral such as

\int_{2}^{5} (5 x^{3} - 2 x^{2} + 7 x - 4) d x,

$\int_2^5 (5x^3 - 2x^2 + 7x - 4)\, dx\text{,}$

S_{2 n}

$S_{2n}$ will always be exact, regardless of the value of

n .

$n\text{.}$ This is just one more piece of evidence that shows how effective Simpson's Rule is as an approximation tool for estimating definite integrals.⁴

One reason that Simpson's Rule is so effective is that $S_{2n}$ benefits from using $2n+1$ points of data. Because it combines $M_n\text{,}$ which uses $n$ midpoints, and $T_n\text{,}$ which uses the $n+1$ endpoints of the chosen subintervals, $S_{2n}$ takes advantage of the maximum amount of information we have when we know function values at the endpoints and midpoints of $n$ subintervals.

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Subsection 5.6.5 Summary

For a definite integral such as $\int_0^1 e^{-x^2} \, dx$ 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 $L_n\text{,}$ $R_n\text{,}$ and $M_n$ for one or more chosen values of $n\text{.}$
The Trapezoid Rule, which estimates $\int_a^b f(x) \, dx$ by using trapezoids, rather than rectangles, can also be viewed as the average of Left and Right Riemann sums. That is, $T_n = \frac{1}{2}(L_n + R_n)\text{.}$
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 $S_{2n} = \frac{2M_n + T_n}{3}\text{,}$ we can build a much more accurate approximation to $\int_a^b f(x) \, dx$ by using approximations we have already computed. The rule for $S_{2n}$ is known as Simpson's Rule, which can also be developed by approximating a given continuous function with pieces of quadratic polynomials.

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5.

Consider the definite integral $\int_0^1 x \tan(x) \, dx\text{.}$

Explain why this integral cannot be evaluated exactly by using either $u$ -substitution or by integrating by parts.
Using appropriate subintervals, compute $L_4\text{,}$ $R_4\text{,}$ $M_4\text{,}$ $T_4\text{,}$ and $S_8\text{.}$
Which of the approximations in (b) is an over-estimate to the true value of $\int_0^1 x \tan(x) \, dx\text{?}$ Which is an under-estimate? How do you know?

Answer

$u$-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.

Solution

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.
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.

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6.

For an unknown function $f(x)\text{,}$ the following information is known.

$f$ is continuous on $[3,6]\text{;}$
$f$ is either always increasing or always decreasing on $[3,6]\text{;}$
$f$ has the same concavity throughout the interval $[3,6]\text{;}$
As approximations to $\int_3^6 f(x) \, dx\text{,}$ $L_4 = 7.23\text{,}$ $R_4 = 6.75\text{,}$ and $M_4 = 7.05\text{.}$

Is $f$ increasing or decreasing on $[3,6]\text{?}$ What data tells you?
Is $f$ concave up or concave down on $[3,6]\text{?}$ Why?
Determine the best possible estimate you can for $\int_3^6 f(x) \, dx\text{,}$ based on the given information.

Answer

Decreasing.
Concave down.
$\int_3^6 f(x) \approx 7.03\text{.}$

Solution

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.
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.
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*}

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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, $t$	$0$	$10$	$20$	$30$	$40$	$50$	$60$
flow in TCFS, $r(t)$	$2000$	$2100$	$2400$	$3000$	$3900$	$5100$	$6500$

Table 5.6.16. Water flow data.

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 $M_n$ for the largest possible value of $n$ to approximate the integral you stated in (a). Do you think $M_n$ over- or under-estimates the exact value of the integral? Why?
Approximate the integral stated in (a) by calculating $S_n$ for the largest possible value of $n\text{,}$ based on the given data.
Compute $\frac{1}{60} S_n$ and $\frac{2000+2100+2400+3000+3900+5100+6500}{7}\text{.}$ What quantity do both of these values estimate? Which is a more accurate approximation?

Answer

$\int_0^{60} r(t) \, dt \text{.}$
$\int_0^{60} r(t) \, dt \gt M_3 = 204000\text{.}$
$\int_0^{60} r(t) \, dt \approx S_6 = \frac{619000}{3} \approx 206333.33\text{.}$
$\frac{1}{60} S_6 \approx 3438.89 \text{;}$ $\frac{2000+2100+2400+3000+3900+5100+6500}{7} = \frac{25000}{7} \approx 3571.43 \text{.}$ each estimates the average rate at which water flows through the dam on $[0,60]\text{,}$ and the first is more accurate.

Solution

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.
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.

\(n\)	\(T_n\)	\(E_{T,n}\)
\(4\)	\(0.50899\)	\(-0.50899\)
\(8\)	\(0.50227\)	\(-0.50227\)
\(16\)	\(0.50057\)	\(-0.50057\)

\(n\)	\(M_n\)	\(E_{M,n}\)
\(4\)	\(0.49555\)	\(0.00445\)
\(8\)	\(0.49887\)	\(0.00113\)
\(16\)	\(0.49972\)	\(0.00028\)

\(n\)	\(T_n\)	\(E_{T,n}\)
\(4\)	\(0.50899\)	\(-0.50899\)
\(8\)	\(0.50227\)	\(-0.50227\)
\(16\)	\(0.50057\)	\(-0.50057\)

\(n\)	\(M_n\)	\(E_{M,n}\)
\(4\)	\(0.49555\)	\(0.00445\)
\(8\)	\(0.49887\)	\(0.00113\)
\(16\)	\(0.49972\)	\(0.00028\)

\(f\)	\(g\)	\(h\)
\(L_1=2\)	\(L_1=2\)	\(L_1=2\)
\(R_1=1\)	\(R_1=1\)	\(R_1=1\)

\(f\)	\(g\)	\(h\)
\(M_1=\frac{7}{4}\)	\(M_1=\frac{15}{8}\)	\(M_1=\frac{31}{16}\)

\(f\)	\(g\)	\(h\)
\(T_1=\frac{3}{2}\)	\(T_1=\frac{3}{2}\)	\(T_1=\frac{3}{2}\)
\(S_2=\frac{5}{3} \approx 1.6667\)	\(S_2=\frac{7}{4}\)	\(S_2=\frac{43}{24} \approx 1.79167\)

\(f\)	\(g\)	\(h\)
\(M_1=\frac{7}{4}\)	\(M_1=\frac{15}{8}\)	\(M_1=\frac{31}{16}\)

\(f\)	\(g\)	\(h\)
\(T_1=\frac{3}{2}\)	\(T_1=\frac{3}{2}\)	\(T_1=\frac{3}{2}\)
\(S_2=\frac{5}{3} \approx 1.6667\)	\(S_2=\frac{7}{4}\)	\(S_2=\frac{43}{24} \approx 1.79167\)

Section 5.6 Numerical Integration

Motivating Questions

Preview Activity 5.6.1.

Subsection 5.6.1 The Trapezoid Rule

The Trapezoid Rule.

Activity 5.6.2.

Subsection 5.6.2 Comparing the Midpoint and Trapezoid Rules

Subsection 5.6.3 Simpson's Rule

Activity 5.6.3.

Subsection 5.6.4 Overall observations regarding Ln,Ln,L_n\text{,} Rn,Rn,R_n\text{,} Tn,Tn,T_n\text{,} Mn,Mn,M_n\text{,} and S2n.S2n.S_{2n}\text{.}

Activity 5.6.4.

Subsection 5.6.5 Summary

Exercises 5.6.6 Exercises

1. Various methods for exexe^x numerically.

2. Comparison of methods for increasing concave down function.

3. Comparing accuracy for two similar functions.

4. Identifying and comparing methods.

5.

6.

7.

Subsection 5.6.4 Overall observations regarding $L_n\text{,}$ $R_n\text{,}$ $T_n\text{,}$ $M_n\text{,}$ and $S_{2n}\text{.}$

1. Various methods for $e^x$ numerically.