Coordinated Calculus

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 $\text{LEFT}(n)\text{,}$ $\text{RIGHT}(n)\text{,}$ and $\text{MID}(n)$ is called the Trapezoid Rule. Rather than using a rectangle to estimate the (signed) area bounded by $y=f(x)$ on a small interval, we use a trapezoid. For example, in Figure 5.77, 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.

We now observe that two familiar sums have arisen. The left Riemann sum $\text{LEFT}(3)$ is $\text{LEFT}(3)=f(x_0)\mathrm{\Delta }x+f(x_1)\mathrm{\Delta }x+f(x_2)\mathrm{\Delta }x\text{,}$ and the right Riemann sum is $\text{RIGHT}(3)=f(x_1)\mathrm{\Delta }x+f(x_2)\mathrm{\Delta }x+f(x_3)\mathrm{\Delta }x\text{.}$ Substituting $\text{LEFT}(3)$ and $\text{RIGHT}(3)$ for the corresponding expressions in Equation (5.18), it follows that $\text{TRAP}(3)=\frac{\text{LEFT}(3)+\text{RIGHT}(3)}{2}\text{.}$ 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.

We know from the definition of the definite integral that if we let $n$ be large enough, we can make any of the approximations $\text{LEFT}(n)\text{,}$ $\text{RIGHT}(n)\text{,}$ and $\text{MID}(n)$ as close as we’d like (in theory) to the exact value of ${\int }_a^{b}f(x)dx\text{.}$ Thus, it may be natural to wonder why we ever use any rule other than $\text{LEFT}(n)$ or $\text{RIGHT}(n)$ (with a sufficiently large $n$ value) to estimate a definite integral. One of the primary reasons is that as $n\to \mathrm{\infty }\text{,}$ $\mathrm{\Delta }x=\frac{b-a}{n}\to 0\text{,}$ and thus in a Riemann sum calculation with a large $n$ 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.

Hence, we explore ways to estimate definite integrals to high levels of precision, but without using extremely large values of $n\text{.}$ Paying close attention to patterns in errors, such as those observed in Example 5.78, 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.

As seen in Figure 5.79, it is evident that whenever the function is concave up on an interval, the Trapezoid Rule with one subinterval, $\text{TRAP}(1)\text{,}$ 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 5.79), the resulting trapezoid has the same area as $\text{MID}(1)\text{,}$ and this value is less than the exact value of the definite integral. Thus, when the function is concave up on the interval, $\text{MID}(1)$ underestimates the integral’s true value.

The preceding discussion explores how to determine if $\text{MID}(n)$ and $\text{TRAP}(n)$ are underestimates or overestimates of ${\int }_a^{b}f(x)dx$ if $f(x)$ is concave up. Similar statements can be made when $f(x)$ is concave down. These are summarized next.

Next, we compare the size of the errors between $\text{MID}(n)$ and $\text{TRAP}(n)\text{.}$ Again, we focus on $\text{MID}(1)$ and $\text{TRAP}(1)$ on an interval where the concavity of $f$ is consistent. In Figure 5.80, 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.

Using appropriate technology to compute $\text{MID}(4)\text{,}$ $\text{MID}(8)\text{,}$ $\text{TRAP}(4)\text{,}$ and $\text{TRAP}(8)\text{,}$ as well as the corresponding errors $\text{ERROR}(M,4)\text{,}$ $\text{ERROR}(M,8)\text{,}$ $\text{ERROR}(T,4)\text{,}$ and $\text{ERROR}(T,8)\text{,}$ as we did in Example 5.78, we find the results summarized in Table 5.81. We also include the approximations and their errors for the example ${\int }_1^2\frac{1}{x^2}dx$ from Example 5.78.

For a given function $f$ and interval $[a,b]\text{,}$ $\text{ERROR}(T,4)={\int }_a^{b}f(x)dx-\text{TRAP}(4)$ calculates the difference between the exact value of the definite integral and the approximation generated by the Trapezoid Rule with $n=4\text{.}$ If we look at not only $\text{ERROR}(T,4)\text{,}$ but also the other errors generated by using $\text{TRAP}(n)$ and $\text{MID}(n)$ with $n=4$ and $n=8$ in the two examples noted in Table 5.81, 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 $\text{TRAP}(n)$ and $\text{MID}(n)$ 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.

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

If a function is always increasing or always decreasing on the interval $[a,b]\text{,}$ one of $\text{LEFT}(n)$ and $\text{RIGHT}(n)$ will over-estimate the true value of ${\int }_a^{b}f(x)dx\text{,}$ while the other will under-estimate the integral. Thus, the errors found in $\text{LEFT}(n)$ and $\text{RIGHT}(n)$ will have opposite signs; so averaging $\text{LEFT}(n)$ and $\text{RIGHT}(n)$ eliminates a considerable amount of the error present in the respective approximations. In a similar way, it makes sense to think about averaging $\text{MID}(n)$ and $\text{TRAP}(n)$ in order to generate a still more accurate approximation.

We’ve just observed that $\text{MID}(n)$ is typically about twice as accurate as $\text{TRAP}(n)\text{.}$ This leads to an approximation method known as Simpson’s Rule

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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.

which is a weighted average of the Midpoint and Trapezoid approximations.

Note that we use “$\text{SIMP}(2n)$” rather that “$\text{SIMP}(n)$” since the $n$ points the Midpoint Rule uses are different from the $n$ points the Trapezoid Rule uses, and thus Simpson’s Rule is using $2n$ points at which to evaluate the function. We build upon the results in Table 5.81 to see the approximations generated by Simpson’s Rule. In particular, in Table 5.82, we include all of the results in Table 5.81, but include additional results for $\text{SIMP}(8)=\frac{2\text{MID}(4)+\text{TRAP}(4)}{3}$ and $\text{SIMP}(16)=\frac{2\text{MID}(8)+\text{TRAP}(8)}{3}\text{.}$

The results seen in Table 5.82 are striking. If we consider the $\text{SIMP}(16)$ approximation of ${\int }_1^2\frac{1}{x^2}dx\text{,}$ the error is only $\text{ERROR}(S,16)=0.0000019434\text{.}$ By contrast, $\text{LEFT}(8)=0.5491458502\text{,}$ so the error of that estimate is $\text{ERROR}(L,8)=-0.0491458502\text{.}$ Moreover, we observe that generating the approximations for Simpson’s Rule is almost no additional work: once we have $\text{LEFT}(n)\text{,}$ $\text{RIGHT}(n)\text{,}$ and $\text{MID}(n)$ for a given value of $n\text{,}$ it is a simple exercise to generate $\text{TRAP}(n)\text{,}$ and from there to calculate $\text{SIMP}(2n)\text{.}$ Finally, note that the error in the Simpson’s Rule approximations of ${\int }_0^1(1-x^2)dx$ is zero!

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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.

These rules are not only useful for approximating definite integrals such as ${\int }_0^1{e}^{-x^2}dx\text{,}$ for which we cannot find an elementary antiderivative of $e^{-x^2}\text{,}$ but also for approximating definite integrals when we are given a function through a table of data.

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 example.

The results seen in Example 5.86 generalize nicely. For instance, if $f$ is decreasing on $[a,b]\text{,}$ $\text{LEFT}(n)$ will overestimate the exact value of ${\int }_a^{b}f(x)dx\text{,}$ and if $f$ is concave down on $[a,b]\text{,}$ $\text{MID}(n)$ will overestimate 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 $\text{LEFT}(n)\text{,}$ $\text{RIGHT}(n)\text{,}$ $\text{TRAP}(n)\text{,}$ and $\text{MID}(n)$ is an over- or under-estimate.

Finally, we make two important notes about Simpson’s Rule. When T. Simpson first developed this rule, his idea was to replace the function $f$ on a given interval with a quadratic function that shared three values with the function $f\text{.}$ 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 ${\int }_2^5(5x^3-2x^2+7x-4)dx\text{,}$ the approximation $\text{SIMP}(2n)$ will always be exact, regardless of the value of $n\text{.}$ This is just one more piece of evidence that shows how effective Simpson’s Rule is as an approximation tool for estimating definite integrals.

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