3.6 Numerical Integration

1. Numerical Integration

Title

Calculus II · Section 3.6

When there is no antiderivative, add up heights, and know how far off you are

2. What this lesson gives you

Objectives

Every technique in this chapter so far has ended with an antiderivative. Many integrals have none you can write down. This lesson approximates them, and, just as important, says how good the approximation is.

• Approximate an integral with the midpoint rule and the trapezoidal rule
• Use concavity to say which rule overestimates and which underestimates, and trap the answer between them
• Compute the absolute and relative error of an estimate
• Use the error-bound formulas to choose the number of subintervals before computing
• Derive and apply Simpson's rule, and explain why it needs an even number of subintervals

OpenStax Calculus Volume 2, §3.6 Numerical Integration §3.6, p. 316 — learning objectives 3.6.1 to 3.6.5

Everything in Chapter 3 so far has been about finding antiderivatives: by parts, trigonometric integrals, substitutions, partial fractions. This section admits the uncomfortable truth that for many integrands there is no antiderivative you can write down, and for data from the real world there is not even a formula to integrate.

So the question changes from what is the exact value to how close can you get, and how do you know. The first half of that question has three answers, the midpoint, trapezoidal and Simpson rules, which are all ways of adding up a handful of function values with sensible weights.

The second half is where the real skill lies. You will learn to say which direction an estimate errs in, to measure its error when you know the answer, and, most usefully, to guarantee its accuracy in advance when you do not.

3. Before anything new: what is a Riemann sum?

Warm-up

OpenStax Calculus Volume 2, §3.6 Numerical Integration §3.6, p. 316 — Riemann sums recalled

Write your answer before revealing it. This is the definition from Volume 1 that the whole lesson rests on, and it is worth being able to state precisely.

A Riemann sum needs two choices: a partition of the interval into pieces, and one sample point in each piece. Multiply the function's value at each sample point by the width of its piece and add. As the pieces shrink, every such sum approaches the integral.

Notice what that means: any single Riemann sum is already an approximation to the integral. The rules in this lesson do not replace the definition; they are particular, cleverly chosen Riemann sums, or averages of them, picked so that the approximation is good even when the pieces are not very small.

4. Why approximate?

Section

Part 1

5. Some integrals have no formula

Concept

\[ \int_0^1 e^{-x^2}\,dx \]

The integrand is continuous, so this integral exists. But no function built from powers, exponentials, logarithms and trigonometric functions has derivative e to the minus x squared, so the Fundamental Theorem gives you nothing to evaluate.

The same happens with data: a surveyor or a sensor gives you heights at a few points and no formula at all.

OpenStax Calculus Volume 2, §3.6 Numerical Integration §3.6, p. 316 — introduction

Look at the shaded region. It is a perfectly ordinary area under a smooth, positive curve, and it certainly has a value. Yet no function built from the ones you know has e to the minus x squared as its derivative. That is a theorem, not a failure of cleverness, so no technique from this chapter will ever crack it.

This integral is not an oddity. It is the heart of the normal distribution in statistics, and tables of its values have been computed for centuries by exactly the methods of this lesson.

The second situation is even more common in practice. An engineer measuring a flow rate every minute, or a surveyor measuring a width every hundred metres, has numbers but no formula. The Fundamental Theorem is useless there too. You need methods that use only values of the function.

6. Replace the curve by something you can integrate

Intuition

Cut the interval into pieces and, on each piece, swap the curve for a shape whose area you know: a flat top, a sloped top, or a parabola. Each rule in this lesson is one of those choices.

\[ \int_a^b f(x)\,dx \approx \sum_{i} w_i\, f(x_i) \]

All three end up as a weighted sum of function values. Only the sample points and the weights differ, and that difference decides the accuracy.

The three panels show the single idea behind every rule in this lesson. On a stretch of the interval, swap the curve for a simpler shape whose area you can compute exactly, then add those areas.

On the left the shape is flat: a rectangle as tall as the curve at the middle. In the middle it is a trapezoid, whose slanted top joins the curve's values at the two ends. On the right it is a parabola through three points of the curve. With one wide piece the flat and sloped shapes miss badly, while the parabola is already close to the true area of 3.65.

Look at the formula underneath. Each rule ends up as a weighted sum of heights. Once you see them that way, comparing the rules is just comparing where they sample and how they weight what they find.

7. The midpoint rule

Section

Part 2

8. The midpoint rule

Concept

Use equal widths and sample each subinterval at its midpoint.

\[ \Delta x = \frac{b-a}{n}, \qquad m_i = \frac{x_{i-1} + x_i}{2} \]

\[ M_n = \sum_{i=1}^{n} f(m_i)\,\Delta x \]

Midpoint rule (Theorem 3.3) — If f is continuous on the interval from a to b, the midpoint sums tend to the integral as n grows without bound.

\[ \lim_{n\to\infty} M_n = \int_a^b f(x)\,dx \]

OpenStax Calculus Volume 2, §3.6 Numerical Integration §3.6, p. 316 — Theorem 3.3 and Figure 3.13

The midpoint rule is the Riemann sum with the most natural choices: equal widths, and the sample point in the middle of each piece. You already met it in Volume 1; here it becomes a tool with a name and a theorem.

Look at the picture. Each rectangle's top crosses the curve at its midpoint, so on one side of the midpoint the rectangle rises above the curve and on the other side it falls below. On a smooth curve those two slivers are nearly the same size, which is why the midpoint rule is surprisingly accurate.

Theorem 3.3 says that for a continuous function the midpoint sums approach the integral as the number of pieces grows. That is reassuring, but it does not tell you how many pieces are enough. The error bounds in Part 6 will.

9. Example 3.39: four midpoint rectangles under a parabola

Worked example

Estimate the integral with four subintervals, then compare with the exact value.

\[ \int_0^1 x^2\,dx, \qquad n = 4 \]

\[ \Delta x = \frac{1 - 0}{4} = \frac14 \]

\[ m_i = \frac18,\; \frac38,\; \frac58,\; \frac78 \]

\[ f(m_i) = \frac{1}{64},\; \frac{9}{64},\; \frac{25}{64},\; \frac{49}{64} \]

\[ \frac{1 + 9 + 25 + 49}{64} = \frac{84}{64} \]

\[ M_4 = \frac14 \cdot \frac{84}{64} = \frac{21}{64} \approx 0.3281 \]

\[ \int_0^1 x^2\,dx = \frac13, \qquad \left|\frac13 - \frac{21}{64}\right| = \frac{1}{192} \approx 0.0052 \]

OpenStax Calculus Volume 2, §3.6 Numerical Integration §3.6, p. 317 — Example 3.39

This example deliberately uses an integral you can do exactly, so that you can see how good the estimate is. Work through it one line at a time: width, midpoints, heights, sum, and multiply by the width.

Keeping fractions makes the arithmetic clean. The heights are the squares of one eighth, three eighths, five eighths and seven eighths, all over 64, and their numerators add to 84. One quarter of 84 over 64 is 21 over 64.

In the picture you can see why the answer is slightly low: the parabola bends upward, so on each rectangle the part poking above the curve is a little smaller than the part falling short. The check shows the shortfall is exactly one part in 192, about half a hundredth. Keep that number; the next part of the lesson explains exactly where it comes from.

10. Example 3.40: the length of a parabola by M₆

Worked example

Estimate the length of the curve below with six midpoint rectangles.

\[ y = \tfrac12 x^2, \qquad 1 \le x \le 4 \]

\[ L = \int_1^4 \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx \]

\[ \frac{dy}{dx} = x \;\Longrightarrow\; L = \int_1^4 \sqrt{1 + x^2}\,dx \]

\[ \Delta x = \tfrac12, \quad m_i = \tfrac54,\, \tfrac74,\, \tfrac94,\, \tfrac{11}{4},\, \tfrac{13}{4},\, \tfrac{15}{4} \]

\[ 1.6008,\; 2.0156,\; 2.4622,\; 2.9262,\; 3.4004,\; 3.8810 \]

\[ M_6 \approx \tfrac12(16.2862) = 8.1431 \]

\[ L = \left[\tfrac{x}{2}\sqrt{1+x^2} + \tfrac12\ln\left(x + \sqrt{1+x^2}\right)\right]_1^4 \approx 8.1458 \]

OpenStax Calculus Volume 2, §3.6 Numerical Integration §3.6, pp. 317-318 — Example 3.40

The arc-length formula from Chapter 2 often produces integrals that are hard or impossible to evaluate, which makes it a natural customer for numerical methods. Here the integrand is the square root of one plus x squared.

The work is organised the same way as before. Three units cut into six gives width one half, and the midpoints run from five quarters to fifteen quarters in steps of one half. Evaluate the square root at each, add the six values, and halve the total.

This integral can in fact be done exactly with a trigonometric substitution, which is how the check works: the exact length is about 8.1458, so six rectangles came within three thousandths. The picture of the arc gives a quick sanity test too. An arc must be longer than the straight chord between its ends, and the chord is about 8.078, so an answer near 8.15 is believable.

11. Six rectangles on an almost straight curve

Picture it

The integrand of Example 3.40 is nearly a straight line over this interval. On a straight piece, the part of a midpoint rectangle above the line exactly matches the part below it, so the midpoint rule is exact for every linear function. Here the curve bends only slightly, and the estimate is correspondingly close.

This is the integrand from Example 3.40 with its six midpoint rectangles. Look at how nearly straight the curve is over this stretch. That is the reason the estimate is so good.

Think about a truly straight line first. On each rectangle, the line rises above the rectangle's top on one side of the midpoint and dips below it on the other, by exactly the same triangle. The two triangles cancel, so the midpoint rule gives the exact area under any straight line, with any number of rectangles.

Errors therefore come only from bending. A curve that barely bends, like this one, leaves only small errors. That observation will turn into the error bounds of Part 6, where the second derivative, which measures bending, controls everything.

12. Checkpoint 3.22: M₂ for one over x

Worked example

\[ \int_1^2 \frac{1}{x}\,dx, \qquad n = 2 \]

\[ \Delta x = \frac12, \qquad m_1 = \frac54, \quad m_2 = \frac74 \]

\[ f\left(\tfrac54\right) = \frac45, \qquad f\left(\tfrac74\right) = \frac47 \]

\[ \frac45 + \frac47 = \frac{28 + 20}{35} = \frac{48}{35} \]

\[ M_2 = \frac12 \cdot \frac{48}{35} = \frac{24}{35} \approx 0.6857 \]

\[ \int_1^2 \frac{dx}{x} = \ln 2 \approx 0.6931, \qquad 0.6931 - 0.6857 = 0.0074 \]

OpenStax Calculus Volume 2, §3.6 Numerical Integration §3.6, p. 318 — Checkpoint 3.22

With only two rectangles this is a short computation, and the exact answer, the natural log of 2, is known, so you can judge the result.

The midpoints of the two halves are five quarters and seven quarters, and their reciprocals are four fifths and four sevenths. Adding those over the common denominator 35 gives 48 over 35, and half of that is 24 over 35, about 0.6857.

The check shows the estimate is low by about seven thousandths. That is not an accident. The curve one over x bends upward, just like the parabola in Example 3.39, and in both cases the midpoint rule came out low. Hold on to that pattern; Part 4 turns it into a rule.

13. The trapezoidal rule

Section

Part 3

14. One trapezoid per subinterval

Concept

Join the curve's heights at the two ends of a subinterval by a straight chord. The region under the chord is a trapezoid: its parallel sides are the two heights and its width is the subinterval's width.

\[ \text{area} = \tfrac12 h(b_1 + b_2) \]

\[ \text{first trapezoid} = \tfrac12\,\Delta x\,\bigl(f(x_0) + f(x_1)\bigr) \]

OpenStax Calculus Volume 2, §3.6 Numerical Integration §3.6, pp. 318-319 — Figure 3.14

The midpoint rule replaces the curve by a flat top on each piece. The trapezoidal rule replaces it by a slanted top: the straight chord joining the curve's values at the two ends of the piece.

The region under a chord is a trapezoid lying on its side. Its two parallel sides are the vertical heights at the ends, and its width is the width of the piece. The area of a trapezoid is its width times the average of its parallel sides, which gives one half times the width times the sum of the two end values.

Look at the picture. Where the curve bends upward the chord rides above it, and where it bends downward the chord cuts beneath. Unlike the midpoint rectangle, the chord's error has one sign on each piece. That will matter shortly.

15. Adding the trapezoids: why the inside points count twice

Concept

Write out all four trapezoids for n equal to 4 and add.

\[ \tfrac12\Delta x\bigl(f_0 + f_1\bigr) + \tfrac12\Delta x\bigl(f_1 + f_2\bigr) + \tfrac12\Delta x\bigl(f_2 + f_3\bigr) + \tfrac12\Delta x\bigl(f_3 + f_4\bigr) \]

Take out the common factor one half of the width. Each interior value is the right side of one trapezoid and the left side of the next, so it appears twice; the two end values appear once.

\[ = \tfrac12\Delta x\bigl(f_0 + 2f_1 + 2f_2 + 2f_3 + f_4\bigr) \]

Trapezoidal rule (Theorem 3.4) — With equal widths and endpoints from a to b, the trapezoidal sums tend to the integral as n grows without bound.

OpenStax Calculus Volume 2, §3.6 Numerical Integration §3.6, p. 319 — derivation and Theorem 3.4

Write out the four trapezoid areas in full, as on the slide, before simplifying. The pattern is much easier to understand than to memorise.

Every term carries the same factor, one half of the width, so take it out. Then look at which function values remain. The first value appears only in the first trapezoid and the last only in the last. But each interior value is the right side of one trapezoid and the left side of the next, so it appears twice.

That is the whole reason for the weights one, two, two, two, one. If you ever forget the formula, draw three trapezoids side by side and count how many of them touch each point. This is Theorem 3.4, and like the midpoint rule it converges to the integral as the number of pieces grows.

16. Reading the trapezoidal rule piece by piece

Notation

OpenStax Calculus Volume 2, §3.6 Numerical Integration §3.6, p. 319 — equation 3.11

Step through the annotations and connect each part of the formula to the picture of trapezoids.

The factor in front is the width divided by two. The two comes from the average in the trapezoid area formula, and it is the part most often lost. The end values carry weight one because each belongs to a single trapezoid; every interior value carries weight two because two trapezoids share it.

The last annotation compares the rules' costs. The trapezoidal rule samples at the endpoints, one more point than the number of pieces; the midpoint rule samples in between, exactly one point per piece. Both are cheap, and later you will see that Simpson's rule uses exactly the trapezoidal rule's points, just with different weights.

17. The trapezoidal rule is the average of two Riemann sums

Concept

\[ L_n = \sum_{i=1}^{n} f(x_{i-1})\,\Delta x, \qquad R_n = \sum_{i=1}^{n} f(x_i)\,\Delta x \]

The left sum uses every point except the last; the right sum every point except the first. Add them and every interior point appears twice, the ends once: exactly the trapezoidal weights.

\[ T_n = \tfrac12\bigl(L_n + R_n\bigr) \]

OpenStax Calculus Volume 2, §3.6 Numerical Integration §3.6, p. 319 — the observation after Theorem 3.4

The two panels show the left and right Riemann sums from Volume 1 for x squared with four pieces. On this rising curve the left rectangles all sit under the curve and the right rectangles all poke above it.

Now add the two sums. The left sum uses every point except the last, and the right sum every point except the first. So in the total, each interior point appears twice and each end point once. Halve the total and you have exactly the trapezoidal rule.

This is more than a curiosity. It shows why the trapezoidal rule is better than either one-sided sum: when one is too low and the other too high, their average lands between them. For this example the left sum is 0.21875 and the right sum 0.46875, and their average is 0.34375, which is the next example's answer.

18. Example 3.41: four trapezoids under a parabola

Worked example

Estimate the same integral as Example 3.39, now with trapezoids.

\[ \int_0^1 x^2\,dx, \qquad n = 4 \]

\[ P = \left\{0,\, \tfrac14,\, \tfrac12,\, \tfrac34,\, 1\right\}, \qquad \Delta x = \tfrac14 \]

\[ T_4 = \tfrac18\Bigl(f(0) + 2f\bigl(\tfrac14\bigr) + 2f\bigl(\tfrac12\bigr) + 2f\bigl(\tfrac34\bigr) + f(1)\Bigr) \]

\[ T_4 = \tfrac18\Bigl(0 + 2\cdot\tfrac1{16} + 2\cdot\tfrac14 + 2\cdot\tfrac9{16} + 1\Bigr) \]

\[ 0 + \tfrac18 + \tfrac48 + \tfrac98 + \tfrac88 = \tfrac{22}{8} = \tfrac{11}{4} \]

\[ T_4 = \tfrac18 \cdot \tfrac{11}{4} = \tfrac{11}{32} = 0.34375 \]

\[ \tfrac12\left(\tfrac{7}{32} + \tfrac{15}{32}\right) = \tfrac{11}{32} \]

OpenStax Calculus Volume 2, §3.6 Numerical Integration §3.6, p. 320 — Example 3.41

This is the same integral as Example 3.39, so you can compare the two rules directly. The five endpoints are zero, one quarter, one half, three quarters and one, and the factor in front is one half of one quarter, which is one eighth.

Substituting the squares and doubling the three interior ones, the bracket becomes a sum of eighths: zero, one, four, nine and eight eighths, which is 22 eighths or eleven quarters. One eighth of that is 11 over 32.

The check uses the previous slide. The left sum is 7 over 32 and the right sum 15 over 32, and their average is exactly 11 over 32. In the picture, notice the thin red slivers between each chord and the curve: the chords all lie above this upward-bending curve, so the trapezoidal estimate comes out too high, while the midpoint estimate came out too low.

19. Checkpoint 3.23: T₂ for one over x

Worked example

\[ \int_1^2 \frac{1}{x}\,dx, \qquad n = 2 \]

\[ x_0 = 1,\; x_1 = \tfrac32,\; x_2 = 2, \qquad \Delta x = \tfrac12 \]

\[ T_2 = \tfrac14\Bigl(f(1) + 2f\bigl(\tfrac32\bigr) + f(2)\Bigr) \]

\[ T_2 = \tfrac14\Bigl(1 + \tfrac43 + \tfrac12\Bigr) \]

\[ 1 + \tfrac43 + \tfrac12 = \tfrac{17}{6} \]

\[ T_2 = \tfrac{17}{24} \approx 0.7083 \]

\[ M_2 = 0.6857 < \ln 2 = 0.6931 < 0.7083 = T_2 \]

OpenStax Calculus Volume 2, §3.6 Numerical Integration §3.6, p. 320 — Checkpoint 3.23

Two trapezoids on the interval from one to two. The points are one, three halves and two, and the factor in front is one half of one half, one quarter.

The middle value is two thirds, doubled to four thirds. Adding one, four thirds and one half gives seventeen sixths, and a quarter of that is 17 over 24, about 0.7083.

The check places this beside the midpoint estimate from Checkpoint 3.22. The midpoint estimate was low, the trapezoidal estimate is high, and the exact value, the natural log of 2, sits between them. You have now seen this happen twice, on two different upward-bending curves. The next part of the lesson explains why it must happen, and turns it into a way of trapping an unknown integral.

20. Trap: the missing one half

Trap

21. Which over, which under

Section

Part 4

22. Concavity decides the direction

Concept

On a subinterval where the curve bends upward, every chord lies above the curve, so the trapezoid holds too much area.

The midpoint rectangle has the same area as the trapezoid under the tangent line at the midpoint, because the tangent tilts area from one half to the other without changing the total. An upward-bending curve lies above its tangents, so the rectangle holds too little.

\[ f'' > 0 \;\Longrightarrow\; M_n < \int_a^b f(x)\,dx < T_n \]

\[ f'' < 0 \;\Longrightarrow\; T_n < \int_a^b f(x)\,dx < M_n \]

OpenStax Calculus Volume 2, §3.6 Numerical Integration §3.6, p. 319 — Figure 3.15 and the discussion before it

This magnified picture explains everything you have noticed so far. Start with the chord, the top of the trapezoid. On a curve that bends upward every chord lies above the curve, so each trapezoid includes an extra lens of area, shaded red. The trapezoidal rule overestimates.

The midpoint rectangle needs one more idea. Tilt the flat top of the rectangle about its midpoint until it becomes the tangent line there. The triangle you add on one side equals the triangle you remove on the other, so the area does not change. The rectangle has the same area as the trapezoid under the tangent. And an upward-bending curve lies above its tangent lines, so the midpoint rule underestimates.

Flip the curve over and both directions flip. That gives the two lines at the bottom of the slide: with a positive second derivative the integral is trapped between the midpoint estimate below and the trapezoidal estimate above, and with a negative second derivative the order reverses.

23. Where each rule's error lives

Picture it

Every trapezoid on an upward-bending curve overshoots, so its errors pile up. Each midpoint rectangle overshoots on one half of its subinterval and falls short on the other, so its errors largely cancel. That is why the book expects the midpoint rule to be the more accurate of the two.

Both panels show the same upward-bending curve with two pieces, and the red regions are the errors.

On the left, every red lens lies between a chord and the curve, and every one of them is extra area. The trapezoid's errors all have the same sign, so they simply add up.

On the right, each midpoint rectangle has a red sliver above the curve on one half and a red sliver below the curve on the other half. One counts as too much and the other as too little, so they largely cancel, leaving only a small net error. That is the picture behind the book's remark that the midpoint rule tends to be more accurate than the trapezoidal rule, and the next slide makes the comparison exact.

24. For a parabola, the midpoint error is exactly half

Concept

Take one subinterval of width h, starting at a, under y equal to x squared, and compute all three areas.

\[ \int_a^{a+h} x^2\,dx = a^2h + ah^2 + \tfrac13 h^3 \]

\[ h\left(a + \tfrac h2\right)^2 = a^2h + ah^2 + \tfrac14 h^3 \]

\[ \tfrac h2\bigl(a^2 + (a+h)^2\bigr) = a^2h + ah^2 + \tfrac12 h^3 \]

Subtract: the midpoint rectangle is short by one twelfth of h cubed, the trapezoid long by one sixth. Opposite signs, and the trapezoid's error is twice the midpoint's.

\[ \int - M = \tfrac{1}{12}h^3, \qquad T - \int = \tfrac16 h^3 \]

With four pieces of width one quarter, the midpoint error is four times one twelfth of one sixty-fourth: one over 192, exactly what Example 3.39 found.

This short calculation turns the picture into numbers. Take a single piece of width h under the parabola, starting at some point a, and compute the exact area, the midpoint rectangle and the trapezoid.

Expanding each one, the terms with a squared and with a match in all three. They differ only in the last term: one third of h cubed for the exact area, one quarter for the rectangle, one half for the trapezoid. So the rectangle falls short by one twelfth of h cubed and the trapezoid overshoots by one sixth. Opposite signs, and the trapezoid's error is exactly twice the rectangle's.

The final line checks this against Example 3.39. Four pieces, each short by one twelfth of one sixty-fourth, total one over 192, exactly what you found there. For general curves the factor of two is only approximate, but it explains the 24 and the 12 you will see in the error bounds.

25. A curve that bends downward

Prediction

\[ \int_0^{\pi} \sin x\,dx = 2, \qquad n = 4 \]

\[ T_4 \approx 1.8961 < 2 < 2.0523 \approx M_4 \]

Commit to one option before revealing. You have the tools to answer without any arithmetic.

The sine curve over the interval from zero to pi is an arch, bending downward everywhere. Every chord cuts beneath an arch, so the trapezoids leave out area and the trapezoidal rule is too low. Every tangent line lies above an arch, and the midpoint rectangle has the same area as the region under its tangent, so the midpoint rule is too high.

The numbers confirm it: about 1.896 for the trapezoids and about 2.052 for the midpoint rule, on either side of the exact 2. If you chose the other order, you probably remembered the concave-up rule without the reason. Rebuild it from the chord and the tangent each time and it cannot come out backwards.

26. Does the trapezoidal rule overshoot?

Sorting

For each item, decide the sign of the second derivative on the given interval, then use it. You are sorting by concavity, and the concavity decides the trapezoidal rule's direction.

The exponential, one over x on positive values and x cubed on positive values all bend upward, so their chords lie above them and the trapezoidal rule overestimates. The natural log, the square root and cosine on the first quarter-turn all bend downward, so the trapezoidal rule underestimates.

Notice that the interval matters. x cubed bends upward for positive x but downward for negative x, and cosine bends downward only where cosine is positive. The sign of the second derivative must hold on the WHOLE interval before the direction is guaranteed, which is exactly what the error-analysis slide in a moment will test.

27. Exercises 317 and 318: trapping the logarithmic integral

Worked example

No antiderivative exists for one over ln x. Trap the integral between two estimates.

\[ \int_2^4 \frac{dx}{\ln x}, \qquad n = 4 \]

\[ f''(x) = \frac{\ln x + 2}{x^2(\ln x)^3} > 0 \]

\[ M_4 = \tfrac12(1.23315 + 0.98853 + 0.84842 + 0.75657) \approx 1.9133 \]

\[ T_4 = \tfrac14\bigl(1.44270 + 2(2.79984) + 0.72135\bigr) \approx 1.9409 \]

\[ 1.9133 < \int_2^4 \frac{dx}{\ln x} < 1.9409 \]

\[ T_8 \approx 1.9271, \qquad S_{1000} \approx 1.9224 \]

OpenStax Calculus Volume 2, §3.6 Numerical Integration §3.6, p. 327 — Exercises 317 and 318

One over ln x is another integrand with no elementary antiderivative, so this is a genuine use of the method: you do not know the answer, and you want two numbers that certainly surround it.

The concavity comes first, because without it the two estimates are just two estimates. The second derivative has numerator ln x plus 2 and denominator x squared times ln x cubed, and for x above one every one of those pieces is positive. So the curve bends upward on the whole interval.

Then the two rules are routine arithmetic, and the concavity tells you which is which: the midpoint estimate, about 1.9133, is too low, and the trapezoidal estimate, about 1.9409, too high. The check shows both finer estimates landing inside that interval. The width of the trap, under three hundredths, is itself an honest error estimate that needed no knowledge of the true value.

28. Find the error: concavity checked at one point

Error analysis

Read the claim and try to find the flaw before revealing the annotations. The arithmetic is not where the problem lies.

The second derivative was checked at a single point. For the bell curve it is negative near zero but positive beyond about 0.71, as the picture shows with its two shaded regions. A bracket needs one concavity on the whole interval, and this interval has both. The chords of the two trapezoids show it: the first lies under the curve, the second over it.

Here the conclusion happens to come out true, which makes the mistake dangerous. Stretch the interval to three and the same reasoning puts the numbers in the wrong order. The fix is simple: split the integral at the inflection point, where the concavity changes, and trap each piece separately.

29. Is the midpoint rule always better?

Counterexample

OpenStax Calculus Volume 2, §3.6 Numerical Integration §3.6, p. 319 — tends to be more accurate

Try to build the counterexample yourself before revealing it. The hint is in the pictures you have seen: the midpoint rule depends on a single value in the middle of each piece.

A wiggly curve with one wide piece does the job. On the interval from zero to pi, sine of 3x has a trough exactly at the midpoint, so the single midpoint rectangle has height minus one and area minus pi. The trapezoid uses the two end values, both zero, and gets zero. The true value is two thirds, so the trapezoid is much closer.

The lesson is about the word tends. The comparison between the rules comes from the behaviour of a smooth curve on a small piece where the concavity is fixed. With pieces too coarse to follow the curve, either rule can be terrible and either can win. Enough pieces restores the usual order.

30. Absolute and relative error

Section

Part 5

31. Measuring how far off an estimate is

Concept

Suppose the true value is A and your estimate is B.

Absolute error — The distance between the estimate and the true value, in the same units as the quantity.

\[ \text{absolute error} = |A - B| \]

Relative error — The absolute error as a fraction, or percentage, of the size of the true value.

\[ \text{relative error} = \left|\frac{A - B}{A}\right| \cdot 100\% \]

An error of 0.01 is excellent on an answer near 100 and useless on an answer near 0.02. The relative error is the figure that tells you which.

OpenStax Calculus Volume 2, §3.6 Numerical Integration §3.6, p. 320 — definition of absolute and relative error

Two different questions hide behind how far off. The absolute error answers how many units away the estimate is. The relative error answers how large that is compared with the thing being measured.

The relative error divides the absolute error by the size of the true value, and is usually quoted as a percentage. That makes it independent of units, and it lets you compare the quality of estimates of very different quantities.

The example in the last line is the reason both exist. An error of one hundredth on an answer near one hundred is a hundredth of a percent, excellent. The same error on an answer near two hundredths is fifty percent, useless. When someone quotes an absolute error without the size of the answer, you cannot tell which case you are in.

32. Example 3.42: the error in M₄

Worked example

\[ A = \int_0^1 x^2\,dx = \frac13, \qquad B = M_4 = \frac{21}{64} \]

\[ \frac13 - \frac{21}{64} = \frac{64 - 63}{192} = \frac{1}{192} \]

\[ |A - B| = \frac{1}{192} \approx 0.0052 \]

\[ \frac{1/192}{1/3} = \frac{3}{192} = \frac{1}{64} \]

\[ \frac{1}{64} = 0.015625 \approx 1.6\% \]

\[ 4 \cdot \frac{1}{12}\left(\frac14\right)^3 = \frac{4}{768} = \frac{1}{192} \]

OpenStax Calculus Volume 2, §3.6 Numerical Integration §3.6, pp. 320-321 — Example 3.42

This returns to the midpoint estimate of Example 3.39 and measures its error both ways. The true value is one third and the estimate 21 over 64.

Over the common denominator 192 the difference is one over 192, about 0.0052. For the relative error divide by one third, which is the same as multiplying by three, giving one over 64, about 1.6 percent.

The check reconnects this to the parabola calculation. There, each piece of width h was short by one twelfth of h cubed. With four pieces of width one quarter, that is four times one twelfth of one sixty-fourth, which is one over 192. The error is not just measured; it is explained.

33. Example 3.43: the error in T₄

Worked example

\[ A = \frac13, \qquad B = T_4 = \frac{11}{32} \]

\[ \frac13 - \frac{11}{32} = \frac{32 - 33}{96} = -\frac{1}{96} \]

\[ |A - B| = \frac{1}{96} \approx 0.0104 \]

\[ \frac{1/96}{1/3} = \frac{3}{96} = \frac{1}{32} \]

\[ \frac{1}{32} = 0.03125 \approx 3.1\% \]

\[ \frac{1}{96} = 2 \cdot \frac{1}{192} \]

OpenStax Calculus Volume 2, §3.6 Numerical Integration §3.6, p. 321 — Example 3.43

The same measurement for the trapezoidal estimate, 11 over 32. The subtraction comes out negative, and that sign carries information: the true value minus the estimate is negative, so the estimate is too high, exactly as the concavity predicted.

The absolute error is one over 96, about 0.0104, and dividing by one third gives one over 32, a relative error of about 3.1 percent.

The check compares the two examples. One over 96 is exactly twice one over 192, and the two errors have opposite signs. That is the factor of two you derived for parabolas, now seen in the book's own numbers, and it points straight at Simpson's rule: if the two errors are in the ratio two to one with opposite signs, a suitable weighted average of the two estimates should cancel them.

34. Checkpoint 3.24: the error in estimating ln 2

Worked example

\[ A = \ln 2 \approx 0.6931, \qquad B = \frac{24}{35} \approx 0.6857 \]

\[ |A - B| \approx |0.6931 - 0.6857| = 0.0074 \]

\[ \frac{0.0074}{0.6931} \approx 0.0107 \]

\[ \text{relative error} \approx 1.1\% \]

\[ \left|\ln 2 - \tfrac{17}{24}\right| \approx 0.0152 \approx 2.2\% \]

OpenStax Calculus Volume 2, §3.6 Numerical Integration §3.6, p. 321 — Checkpoint 3.24

Here the true value is ln 2 and the estimate is 24 over 35, both rounded to four places as the book gives them. The absolute error is about 0.0074, and dividing by 0.6931 gives about 0.0107, a relative error of a little over one percent.

The last step deserves attention. The book says this estimate came from the trapezoidal rule with two pieces, but look back at the two checkpoints: 24 over 35 was the midpoint estimate, and the trapezoidal estimate was 17 over 24. It is a small slip in the book, and noticing it is good practice.

The trapezoidal estimate's own error is about 0.0152, roughly twice the midpoint error again, and on the high side. Once more the two rules err in opposite directions, with the trapezoid about twice as far off.

35. Three rules on one over x

Comparison

\[ \int_1^2 \frac{dx}{x} = \ln 2 \approx 0.693147 \]

The third row is Simpson's rule with two subintervals, coming in Part 7. It uses the same three points as the trapezoidal estimate.

Fill in each blank before checking. You have the true value, the natural log of 2 to six places, and each estimate; the absolute error is the distance between them and the relative error that distance divided by 0.693.

The midpoint and trapezoidal rows repeat the last three slides. The third row previews Simpson's rule, which you are about to meet. It uses exactly the same three points as the two-piece trapezoidal rule, yet its error is about 0.0013, around a tenth of the trapezoid's, a relative error under two tenths of a percent.

Nothing about Simpson's rule costs more: same points, same number of function evaluations. The improvement comes entirely from the weights, which is the strongest argument in this lesson for understanding where weights come from.

36. Error bounds

Section

Part 6

37. You rarely know the true value

Concept

Examples 3.42 and 3.43 cheated: they knew the exact answer. If you knew it, you would not be approximating. What you need is a guarantee computed from the function alone.

Error bounds (Theorem 3.5) — If f has a second derivative on the interval, and M is the largest value of its absolute value there, then the errors of the midpoint and trapezoidal rules are at most the amounts below.

\[ \text{Error in } M_n \le \frac{M(b-a)^3}{24n^2} \]

\[ \text{Error in } T_n \le \frac{M(b-a)^3}{12n^2} \]

OpenStax Calculus Volume 2, §3.6 Numerical Integration §3.6, p. 321 — Theorem 3.5, equations 3.12 and 3.13

The examples so far measured errors against known answers, which is a little like checking a map by visiting the place. In real use you are approximating precisely because you cannot find the exact value, so the error cannot be computed. What you can do is bound it.

Theorem 3.5 gives the bounds. They need one number from the function: the largest value of the absolute value of the second derivative on the interval, called M. The second derivative measures how sharply the curve bends, and bending is the only thing that makes these rules wrong, as the straight-line slide showed.

Notice the two bounds differ only by the 24 and the 12. The midpoint bound is half the trapezoidal bound, matching the factor of two you have now seen several times. The book states the theorem without proof, but the parabola calculation shows where both constants come from.

38. Reading the error bound piece by piece

Notation

Take the pieces one at a time; each one tells you something practical.

M is the worst-case bending on the interval. A curve that bends sharply somewhere makes the bound large, even if it is gentle elsewhere. The cube of the interval's length means long intervals are expensive: doubling the length at a fixed number of pieces multiplies the bound by eight.

The n squared in the denominator is the payoff for work. Double the number of pieces and the bound drops to a quarter; multiply by ten and it drops to a hundredth. That is why a modest n often suffices. And because the bound is an inequality, the actual error is usually smaller still; the bound is a guarantee, not a prediction.

39. Checkpoint 3.25: bounding the error in M₄

Worked example

\[ \int_0^1 x^2\,dx, \qquad n = 4 \]

\[ f''(x) = 2 \;\Longrightarrow\; M = 2 \]

\[ \text{Error in } M_4 \le \frac{2(1)^3}{24(4)^2} = \frac{2}{384} = \frac{1}{192} \]

\[ \text{Error in } T_4 \le \frac{2(1)^3}{12(4)^2} = \frac{1}{96} \]

\[ \left|\tfrac13 - M_4\right| = \tfrac{1}{192}, \qquad \left|\tfrac13 - T_4\right| = \tfrac{1}{96} \]

OpenStax Calculus Volume 2, §3.6 Numerical Integration §3.6, p. 322 — Checkpoint 3.25

The book asks for Equation 3.13, the trapezoidal bound, but applies it to the midpoint estimate. The slide does both, because comparing them is instructive.

For x squared the second derivative is 2 everywhere, so M is 2. The midpoint bound, with 24 in the denominator, gives one over 192, and the trapezoidal bound, with 12, gives one over 96.

Now the check. These are exactly the actual errors from Examples 3.42 and 3.43. The bounds are usually generous, but here they are exact, because the second derivative of a parabola is constant: the worst-case bending is the bending everywhere. That also confirms the constants 24 and 12 are the right ones; no smaller numbers would work.

40. Example 3.44: how many subintervals for 0.01?

Worked example

Choose n so the midpoint rule is guaranteed to be within 0.01.

\[ \int_0^1 e^{x^2}\,dx \]

\[ f'(x) = 2x\,e^{x^2} \]

\[ f''(x) = 2e^{x^2} + 4x^2e^{x^2} = 2e^{x^2}(1 + 2x^2) \]

\[ |f''(x)| \le 2 \cdot e \cdot 3 = 6e \]

\[ \frac{6e(1)^3}{24n^2} \le 0.01 \]

\[ n^2 \ge \frac{600e}{24} = 25e \;\Longrightarrow\; n \ge 5\sqrt e \approx 8.24 \]

\[ n = 9 \]

\[ |1.46265 - 1.45987| \approx 0.0028 < 0.01 \]

OpenStax Calculus Volume 2, §3.6 Numerical Integration §3.6, p. 322 — Example 3.44

This is the most practical use of the error bound: deciding how much work to do before doing it. The integrand e to the x squared has no elementary antiderivative, so there is nothing to compare with; the bound is your only guarantee.

Finding M takes two derivatives. The chain rule gives the first, and the product rule the second, which factors as two e to the x squared times one plus two x squared. Both factors grow with x, so on the interval the largest value is at x equal to one, as the graph shows: two times e times three, or six e.

Put M into the midpoint bound, require it to be at most one hundredth, and solve: n squared must be at least 25e, so n is at least about 8.24. Round up to 9. The check computes the actual estimate with nine rectangles and compares it with an accurate value: the error is under three thousandths, comfortably inside the guarantee.

41. Trap: rounding n down

Trap

42. Before solving for n

Step zero

\[ \int_0^1 \sqrt{1+x^2}\,dx, \quad \text{trapezoidal rule, error} \le 0.01 \]

Answer before revealing. This slide exists because the previous example trains an expectation that the maximum is at the right-hand end, and here it is not.

Differentiating the square root of one plus x squared once gives x over the square root; differentiating again, and simplifying, gives one over one plus x squared to the three halves. That is positive, and it decreases as x grows, because its denominator grows.

So on the interval from zero to one, the largest value is at the left end, x equal to zero, where it equals one. Finding M always means finding the maximum of the absolute value of the derivative on the whole interval, by whatever method that takes: checking where it increases or decreases, or checking critical points and endpoints.

43. Exercise 330: n for the trapezoidal rule

Worked example

\[ \int_0^1 \sqrt{1+x^2}\,dx, \qquad \text{error} \le 0.01 \]

\[ M = f''(0) = 1 \]

\[ \frac{1 \cdot 1^3}{12n^2} \le 0.01 \]

\[ n^2 \ge \frac{1}{0.12} \approx 8.33 \]

\[ n \ge 2.89 \;\Longrightarrow\; n = 3 \]

\[ T_3 \approx 1.15435, \quad \int_0^1\sqrt{1+x^2}\,dx \approx 1.14779, \quad \text{error} \approx 0.0066 \]

OpenStax Calculus Volume 2, §3.6 Numerical Integration §3.6, p. 328 — Exercise 330

With M equal to one from the previous slide, the rest is the same routine as Example 3.44, now with the trapezoidal bound and its 12.

Setting one over twelve n squared at most one hundredth gives n squared at least about 8.33, so n is at least about 2.89. Round up: three trapezoids are enough.

The check is possible here because this integral can be done exactly by a trigonometric substitution: it is one half of root 2 plus the log of one plus root 2, about 1.14779. Three trapezoids give about 1.15435, off by about 0.0066, inside the promised hundredth. Notice that the estimate is too high, which you could have predicted: the second derivative is positive, so the curve bends upward and the chords sit above it.

44. Slide n and watch the chords close in

Tweak it

The curve on this slide is not the function itself but the gap between the sine arch and the chords of the trapezoidal rule, piece by piece. Each bump is one piece's error, seen from the side.

Start at n equal to 2 and move the slider up slowly. The bumps multiply, but each gets much lower. At 2 the tallest bump is about 0.21, at 4 about 0.07, at 8 about 0.019. Doubling n cuts the height to roughly a quarter.

That quartering is the n squared in the error bound, made visible. The formula in the readout is the height bound behind it: the largest gap is at most the maximum of the second derivative, pi squared, times the square of the width over eight. Because both the height and the width of each bump shrink, the total error shrinks like one over n squared.

45. Simpson's rule

Section

Part 7

46. Flat, then sloped, then curved

Concept

The midpoint rule replaces the curve by constants on each piece; the trapezoidal rule by straight lines. Simpson's rule goes one step further: through each pair of neighbouring subintervals it fits the parabola through three points of the curve, and integrates that.

\[ \int_{x_0}^{x_2} f(x)\,dx \approx \int_{x_0}^{x_2} (Ax^2 + Bx + C)\,dx \]

OpenStax Calculus Volume 2, §3.6 Numerical Integration §3.6, pp. 322-323 — Figure 3.16

The midpoint rule fits constants and the trapezoidal rule fits straight lines. The natural next step is to fit parabolas, and that is Simpson's rule.

A straight line is fixed by two points, which is why each trapezoid uses one piece. A parabola needs three points, so each parabola spans two neighbouring pieces: the two ends and the point where they meet. Look at the picture. The first parabola covers the first pair of pieces, the second covers the next pair, and each hugs the curve far more closely than a chord could.

Two consequences follow at once. You integrate each parabola instead of the curve, and parabolas are easy to integrate. And because the pieces are used in pairs, the number of pieces must be even.

47. The area under one parabola

Concept

Slide the pair so its middle point is at 0; sliding does not change an area. The three points are at minus h, 0 and h, where h is the width.

\[ \int_{-h}^{h}(Ax^2 + Bx + C)\,dx = \left[\tfrac{A}{3}x^3 + \tfrac{B}{2}x^2 + Cx\right]_{-h}^{h} \]

\[ = \tfrac{2A}{3}h^3 + 2Ch = \tfrac{h}{3}\bigl(2Ah^2 + 6C\bigr) \]

Now add the three heights, counting the middle one four times:

\[ f(-h) + 4f(0) + f(h) = (Ah^2 - Bh + C) + 4C + (Ah^2 + Bh + C) \]

\[ = 2Ah^2 + 6C, \quad \text{the same bracket as the area} \]

\[ \int_{x_0}^{x_2} p(x)\,dx = \tfrac{\Delta x}{3}\bigl(f(x_0) + 4f(x_1) + f(x_2)\bigr) \]

OpenStax Calculus Volume 2, §3.6 Numerical Integration §3.6, p. 324 — the derivation of Simpson's rule

This derivation looks like it will be messy, and the book's version with general points is. Sliding the pair sideways so that its middle point sits at zero makes it short, and sliding does not change any area.

With the points at minus h, zero and h, integrate the parabola. The odd power, B times x, integrates to zero over a symmetric interval, leaving two A over three times h cubed plus two C h. Factor out h over three to get h over three times two A h squared plus six C.

Now evaluate the parabola at the three points and weight the middle one by four. The B terms cancel between the two ends, and what remains is exactly the same bracket, two A h squared plus six C. So the area is h over three times the first value plus four times the middle value plus the last. The parabola's coefficients have disappeared entirely; only the three heights remain.

48. Stacking the parabolas: Simpson's rule

Concept

Add the formula over each pair. Where two pairs meet, the shared endpoint collects a 1 from each side.

\[ \tfrac{\Delta x}{3}(f_0 + 4f_1 + f_2) + \tfrac{\Delta x}{3}(f_2 + 4f_3 + f_4) = \tfrac{\Delta x}{3}(f_0 + 4f_1 + 2f_2 + 4f_3 + f_4) \]

Simpson's rule (Theorem 3.6) — For a positive EVEN number of subintervals of equal width, the Simpson sums tend to the integral as n grows.

OpenStax Calculus Volume 2, §3.6 Numerical Integration §3.6, p. 324 — Theorem 3.6

One parabola covers two pieces. For more pieces, add the one-parabola formula over each pair, as in the first line of the slide.

Every point in the middle of a pair collects weight four. The points where two pairs meet belong to two parabolas, the end of one and the start of the next, so they collect a one from each, weight two. The very first and last points belong to one parabola each, weight one.

The bar charts compare the result with the trapezoidal weights on the same nine points. The trapezoidal rule weights every interior point equally; Simpson's rule alternates four and two. Same points, same work, different weights, and as you will see, a dramatically smaller error. That is Theorem 3.6, and its hypothesis includes the word even.

49. Reading Simpson's rule piece by piece

Notation

OpenStax Calculus Volume 2, §3.6 Numerical Integration §3.6, p. 324 — equation 3.14

Connect each annotation to the parabola derivation you just saw.

The factor in front is a third of the width, not a half, because it came from the area under one parabola. The fours sit at the odd-numbered points, which are the middles of the parabolas, and the twos at the even-numbered interior points, where two parabolas meet.

The last annotation is the one to check every time. The pattern must begin with one, four and end with four, one. That only happens when the number of pieces is even. If you count your weights and find the pattern ending two, one or four, four, the number of pieces is odd and the formula you have written is not Simpson's rule.

50. Simpson's rule is a weighted average of M and T

Concept

On one pair of width 2h, the midpoint rule with one rectangle and the trapezoidal rule with one trapezoid give:

\[ M_1 = 2h\,f_1, \qquad T_1 = h\,(f_0 + f_2) \]

Weight the midpoint twice as heavily, as its smaller error deserves:

\[ \tfrac23 M_1 + \tfrac13 T_1 = \tfrac{4h}{3}f_1 + \tfrac{h}{3}(f_0 + f_2) = \tfrac{h}{3}(f_0 + 4f_1 + f_2) \]

\[ S_{2n} = \tfrac23 M_n + \tfrac13 T_n \]

Check with Examples 3.39 and 3.41: two thirds of 21 over 64 plus one third of 11 over 32 is 64 over 192, exactly one third. The two errors, in the ratio two to one with opposite signs, cancel.

OpenStax Calculus Volume 2, §3.6 Numerical Integration §3.6, p. 325 — the weighted-average relation

Here is a second route to Simpson's rule, and it connects everything in the lesson. Take one pair of pieces, total width two h. As a single piece, its midpoint rectangle has area two h times the middle value, and its trapezoid has area h times the sum of the end values.

You know the trapezoid's error tends to be about twice the midpoint's, with the opposite sign. So weight the midpoint estimate twice as heavily: two thirds of one plus one third of the other. The arithmetic on the slide shows that this weighted average is exactly the Simpson formula.

The check uses the book's numbers from Examples 3.39 and 3.41. Two thirds of 21 over 64 plus one third of 11 over 32 comes to exactly one third, the true value. For a parabola the ratio of errors is exactly two to one, so the cancellation is perfect.

51. Simpson's error bound and the fourth derivative

Concept

Error bound for Simpson's rule — If M is the largest absolute value of the fourth derivative of f on the interval, the error in the Simpson estimate is at most the amount below.

\[ \text{Error in } S_n \le \frac{M(b-a)^5}{180n^4} \]

Two consequences. Doubling n divides the bound by sixteen, not four. And every cubic has fourth derivative zero, so Simpson's rule is EXACT for cubics, even though it only fits parabolas.

\[ f(x) = Ax^3 + Bx^2 + Cx + D \;\Longrightarrow\; f^{(4)} = 0 \;\Longrightarrow\; S_n = \int_a^b f(x)\,dx \]

OpenStax Calculus Volume 2, §3.6 Numerical Integration §3.6, p. 325 — equation 3.15

Simpson's bound looks like the others but with three important changes: the fourth derivative instead of the second, the fifth power of the interval's length, and the fourth power of n.

The fourth power of n is the practical headline. Doubling the number of pieces divides the bound by sixteen, where the midpoint and trapezoidal bounds only fell by four. For demanding accuracy that difference is enormous.

The fourth derivative carries a surprise. Every cubic has fourth derivative zero, so the bound is zero and Simpson's rule gives the exact integral of every cubic, not just every parabola. That is one degree better than the rule was designed for, and the next example shows why.

52. Example 3.45: Simpson's rule on a cubic

Worked example

Use Simpson's rule with two subintervals, and bound its error.

\[ \int_0^1 x^3\,dx \]

\[ \Delta x = \tfrac12, \qquad x_0 = 0,\; x_1 = \tfrac12,\; x_2 = 1 \]

\[ S_2 = \tfrac13 \cdot \tfrac12\Bigl(f(0) + 4f\bigl(\tfrac12\bigr) + f(1)\Bigr) \]

\[ S_2 = \tfrac16\left(0 + 4\cdot\tfrac18 + 1\right) = \tfrac16 \cdot \tfrac32 = \tfrac14 \]

\[ \text{Error in } S_2 \le \frac{0 \cdot 1^5}{180 \cdot 2^4} = 0 \]

\[ \int_0^1 x^3\,dx = \left[\tfrac{x^4}{4}\right]_0^1 = \tfrac14 \]

OpenStax Calculus Volume 2, §3.6 Numerical Integration §3.6, p. 325 — Example 3.45

This example tests the surprise from the previous slide. With two pieces the points are zero, one half and one, the factor in front is one sixth, and the weights are one, four, one. The value at one half is one eighth, four times that is one half, and the bracket totals three halves. One sixth of three halves is one quarter.

The error bound needs the fourth derivative of x cubed, which is zero, so the bound is zero: Simpson's rule claims to be exact. The Fundamental Theorem confirms it.

The picture shows how this is possible. The parabola through the three points is a different curve from the cubic: it dips below the cubic on the left half and rises above it on the right. But the two gaps have exactly the same area, one sixty-fourth each, and they cancel. That happens for every cubic, because the difference between a cubic and its parabola is symmetric about the middle point in just this way.

53. Example 3.46: the parabola's length again, by S₆

Worked example

\[ L = \int_1^4 \sqrt{1+x^2}\,dx, \qquad n = 6 \]

\[ \Delta x = \tfrac12, \qquad x_i = 1,\; \tfrac32,\; 2,\; \tfrac52,\; 3,\; \tfrac72,\; 4 \]

\[ S_6 = \tfrac16\bigl(f_0 + 4f_1 + 2f_2 + 4f_3 + 2f_4 + 4f_5 + f_6\bigr) \]

\[ \tfrac16\bigl(1.41421 + 4(1.80278) + 2(2.23607) + 4(2.69258) \]

\[ \qquad + 2(3.16228) + 4(3.64005) + 4.12311\bigr) \]

\[ S_6 \approx \tfrac{48.87566}{6} \approx 8.14594 \]

\[ |8.14577 - 8.14594| \approx 0.00017, \qquad |8.14577 - 8.14307| \approx 0.0027 \]

OpenStax Calculus Volume 2, §3.6 Numerical Integration §3.6, pp. 325-326 — Example 3.46

This is the arc-length integral from Example 3.40, now with Simpson's rule, so the two rules can be compared on the same problem with the same six pieces.

The points run from one to four in steps of one half. Write the weights one, four, two, four, two, four, one first, then substitute the values, then add. The long line is split over two lines on the slide; the weighted total is 48.87566, and dividing by six gives about 8.14594.

The check compares both rules with the exact length, 8.14577. Simpson's rule misses by about two ten-thousandths; the midpoint rule with six pieces missed by about 27 ten-thousandths, more than ten times as much. Simpson used seven function values and the midpoint rule six, so the extra accuracy cost almost nothing.

54. Checkpoint 3.26: S₂ for one over x

Worked example

\[ \int_1^2 \frac{dx}{x}, \qquad n = 2 \]

\[ S_2 = \tfrac16\Bigl(f(1) + 4f\bigl(\tfrac32\bigr) + f(2)\Bigr) \]

\[ S_2 = \tfrac16\left(1 + \tfrac83 + \tfrac12\right) \]

\[ 1 + \tfrac83 + \tfrac12 = \tfrac{25}{6} \]

\[ S_2 = \tfrac{25}{36} \approx 0.69444 \]

\[ \tfrac23\cdot\tfrac23 + \tfrac13\cdot\tfrac34 = \tfrac{25}{36}, \qquad |0.69444 - 0.69315| \approx 0.0013 \]

OpenStax Calculus Volume 2, §3.6 Numerical Integration §3.6, p. 326 — Checkpoint 3.26

Two pieces again on the interval from one to two, so the factor is one sixth and the weights one, four, one. The middle value is two thirds, and four times it is eight thirds.

Adding one, eight thirds and one half gives twenty-five sixths, and one sixth of that is 25 over 36, about 0.69444.

The check does two things. It confirms the weighted-average relation: the one-rectangle midpoint estimate is two thirds, the one-trapezoid estimate three quarters, and two thirds of the first plus one third of the second is 25 over 36 again. And it compares with the natural log of 2: the error is about 0.0013, against about 0.0074 for the midpoint rule and 0.0152 for the trapezoidal rule with the same number of pieces.

55. Simpson's rule on the bell curve, with a guarantee

Worked example

Return to the integral with no antiderivative. Estimate it with Simpson's rule on four subintervals and bound the error.

\[ \int_0^1 e^{-x^2}\,dx, \qquad n = 4 \]

\[ 1,\; 0.939413,\; 0.778801,\; 0.569783,\; 0.367879 \]

\[ S_4 = \tfrac{1}{12}(8.962265) \approx 0.746855 \]

\[ f^{(4)}(x) = (16x^4 - 48x^2 + 12)\,e^{-x^2} \]

\[ M = 12 \;\Longrightarrow\; \text{Error} \le \frac{12}{180 \cdot 4^4} = \frac{1}{3840} \approx 0.00026 \]

\[ |0.746855 - 0.746824| \approx 0.000031 < 0.00026 \]

OpenStax Calculus Volume 2, §3.6 Numerical Integration §3.6, p. 327 — Exercise 311's integrand on a new interval

This brings the lesson back to where it started, the integral with no antiderivative, and does the complete job: an estimate and a guarantee.

The estimate is routine: five heights, weights one, four, two, four, one, and a factor of one twelfth. The guarantee is where the work is. Differentiating e to the minus x squared four times, the chain rule produces a polynomial factor, 16 x to the fourth minus 48 x squared plus 12. On the interval from zero to one its absolute value is largest at zero, where it equals 12. That value goes into Simpson's bound.

The bound is about 0.00026, so the estimate 0.746855 is certainly correct to three decimal places. The check uses an accurate value from tables: the actual error is about 0.00003, nearly ten times smaller than guaranteed. Bounds are pessimistic by design.

56. How fast each rule improves

Picture it

The errors for the integral of one over x from 1 to 2, with n doubling from 2 to 32. The midpoint and trapezoidal errors fall by a factor of four per doubling, the midpoint always about half the trapezoid. Simpson's errors fall by sixteen: at n equal to 32 it is correct to seven decimal places.

Both axes of this plot are logarithmic, so equal steps across mean doubling n, and equal steps down mean dividing the error by the same factor. On such axes an error that behaves like a power of n becomes a straight line.

The trapezoidal and midpoint lines are parallel, one step down per doubling equal to a factor of four, which is the n squared in their bounds. The midpoint line sits just below the trapezoidal one, a factor of two, all the way along.

The Simpson line is twice as steep, a factor of sixteen per doubling, the n to the fourth. At 32 pieces Simpson's rule is correct to about seven decimal places, while the other two are still wrong in the fifth. If you need high accuracy, the steepness of the line matters far more than where it starts.

57. Trap: Simpson's rule with an odd n

Trap

58. How many subintervals for six decimal places?

Estimation

\[ \int_0^1 e^{-x^2}\,dx, \quad \text{error} \le 10^{-6} \]

\[ \frac{2}{12n^2} \le 10^{-6} \iff n \ge 408.2, \qquad \frac{12}{180n^4} \le 10^{-6} \iff n \ge 16.07 \]

Commit to a guess first. The trapezoidal count, 409, sets your expectation, and most people guess something in the hundreds.

Work Simpson's bound through. The fourth derivative's largest value is 12, the interval has length one, so twelve over 180 times n to the fourth must be at most one millionth. That gives n to the fourth at least about 66,667, and n at least about 16.07.

Now two roundings. Up, because n must be at least 16.07, which gives 17. And then up again to an even number, because Simpson's rule needs one, so 18. At 16 the bound is just above one millionth, so rounding really matters. Eighteen function evaluations against four hundred and ten for the same guarantee: that is what the fourth power buys.

59. Surveying a lot with no formula

Real world

OpenStax Calculus Volume 2, §3.6 Numerical Integration §3.6, p. 329 — Exercise 343

Here the function is a set of measurements: the width of a lot every hundred metres. There is no formula, no antiderivative, and no derivative to bound, only numbers.

The trapezoidal rule needs only those numbers: the ends weighted once, the nine interior widths twice, and the whole multiplied by half the spacing. With eleven points there are ten pieces, an even number, so Simpson's rule also applies: weight four on the odd-numbered widths and two on the even-numbered interior ones, times a third of the spacing.

The two estimates, 89,250 and 89,500 square metres, differ by about a quarter of a percent. With no formula there is no error bound to compute, so agreement between two different rules is the best evidence of accuracy available. It is also a reminder that the surveyor's measurements themselves are probably not accurate to better than that.

60. Putting it together

Section

Part 8

61. Pattern: approximating an integral you can trust

Pattern

1. Try for an antiderivative first. If the Fundamental Theorem works, no approximation is needed.
2. Pick the rule. Simpson's rule is the default; it costs the same function values as the trapezoidal rule.
3. Bound the right derivative on the whole interval: the second for M and T, the fourth for Simpson. Find where its absolute value is largest.
4. Solve the bound for n, and round up (to an even number for Simpson). Then compute.
5. Use concavity when the second derivative keeps one sign: M and T then trap the answer between them.

This is the order to work in. The flow diagram shows the decisions and the list gives the details.

First, try for an antiderivative; an exact answer beats any approximation. If there is none, Simpson's rule is almost always the right choice, because it uses the same function values as the trapezoidal rule and is far more accurate.

To guarantee accuracy, bound the right derivative, the second for the midpoint and trapezoidal rules and the fourth for Simpson, over the whole interval, then solve the bound for n and round up, to an even number for Simpson. Finally, whenever the second derivative keeps one sign, use the midpoint and trapezoidal estimates together to trap the answer. With data only, there is no derivative to bound, and comparing two rules is the practical check.

62. Match each rule to its weights

Matching

Match each rule before checking. The aim is to see all four as members of one family: weighted sums of function values.

The midpoint rule is the only one that samples between the endpoints, and it weights each sample equally. The left Riemann sum samples at the endpoints but ignores the last one entirely. The trapezoidal rule uses every endpoint, halving the ends. Simpson's rule uses every endpoint too, with the alternating four and two, and a factor of a third of the width instead of a half.

If you can reconstruct these weights from pictures, rectangles, trapezoids and parabolas, you never need to memorise a formula from this section.

63. Rank the rules on the bell curve

Ranking

\[ \int_0^1 e^{-x^2}\,dx \approx 0.746824, \qquad n = 4 \]

All four estimates use four pieces on the same integral, so the ranking isolates the effect of the rule. Order them before checking, using what you know rather than subtracting.

Simpson's rule should be best by far, and it is: its error is about three hundred-thousandths. The midpoint rule should beat the trapezoidal rule by about a factor of two, and it does, about 0.0019 against 0.0038. The left sum, which has no cancellation of any kind, is worst, off by 0.075.

Notice that the midpoint rule overestimates and the trapezoidal rule underestimates here, even though the bell curve changes concavity at about 0.71. On this interval the downward-bending part dominates. The ranking is reliable; the direction, as the error-analysis slide warned, is not guaranteed.

64. Check: which way the rules err

Check

Answer from the pictures rather than from memory: chord above an upward-bending curve, tangent below it.

The trapezoid's top is a chord, which lies above the curve, so the trapezoidal rule is too big. The midpoint rectangle has the same area as the region under the tangent at the midpoint, which lies below the curve, so the midpoint rule is too small. The integral is trapped between them.

The last option is worth a thought. The whole value of this fact is that it needs no knowledge of the integral, only the sign of the second derivative, which is usually easy to find.

65. Check: a trapezoidal error bound

Check

Three ingredients: the second derivative's maximum, the interval's length, and the constant 12. The second derivative of x cubed is 6x, which on the interval from zero to two is largest at two, so M is 12.

Twelve times two cubed is 96; twelve times four squared is 192; the bound is one half. The actual error, 0.25, sits inside it, as it must.

Each distractor is a real slip. One uses the midpoint constant 24. One confuses the trapezoidal bound with Simpson's, which would be zero for a cubic. And one puts the maximum of the function in place of the maximum of its second derivative.

66. Check: doubling n in Simpson's rule

Check

Look at the power of n in the denominator of Simpson's bound: it is four. Doubling n multiplies n to the fourth by two to the fourth, which is sixteen.

Each of the other options belongs to a different rule. One over n errors, the factor two, belong to the one-sided Riemann sums; one over n squared, the factor four, to the midpoint and trapezoidal rules. The eight is a mix-up with the fifth power of the interval length.

This is why the log-scale plot showed Simpson's errors falling twice as steeply. For any serious accuracy requirement, the power of n is the most important number in an error bound.

67. Explain the exactness

Explain it to yourself

Write your explanation before revealing. Both halves rest on the same idea: an error that is positive on one half of a piece and negative on the other, by exactly the same amount.

For the midpoint rule and a straight line, the two triangles between the line and the rectangle's top are mirror images, so they cancel. For Simpson's rule and a cubic, the difference between the cubic and the parabola through three points is a cubic that vanishes at all three points, and it is odd about the middle point: positive on one side, negative on the other, with equal areas. You saw exactly that in the picture for Example 3.45.

The error bounds say the same thing in one line: they vanish when the relevant derivative is zero. The second derivative of a line is zero, and so is the fourth derivative of a cubic.

68. Exit ticket

Exit ticket

\[ \int_1^3 \frac{dx}{x} \]

This combines the two skills from the second half of the lesson: predicting the direction of an error without computing, and choosing n from a bound.

The second derivative of one over x is two over x cubed, positive on the whole interval, so the curve bends upward and the trapezoidal rule overestimates. Its largest value is at the left end, x equal to one, so M is two.

The interval has length two, so the bound is two times eight over twelve n squared, which simplifies to four over three n squared. Requiring that to be at most a thousandth gives n squared at least about 1333, so n at least about 36.5, and rounding up gives 37. At 36 the bound is just over a thousandth, so the rounding direction decides the answer.

69. Recap

Recap

\[ f'' > 0: \; M_n < \int_a^b f < T_n, \qquad S_{2n} = \tfrac23 M_n + \tfrac13 T_n \]

Next, Section 3.7 lets an interval or an integrand run off to infinity: the improper integral.

Stewart, Calculus: Early Transcendentals 8e, §7.7 Approximate Integration §7.7, pp. 514-526 — the same material in Stewart

Three rules, each a weighted sum of function values: rectangles at the midpoints, trapezoids on the endpoints, and parabolas over pairs of pieces. The table puts each rule's formula, its error bound and the functions it gets exactly right side by side.

Two relations tie them together. When the second derivative keeps one sign, the midpoint and trapezoidal estimates trap the integral between them. And Simpson's rule is two thirds of the midpoint estimate plus one third of the trapezoidal estimate, which is why it cancels so much error.

The practical skill is using the bounds before computing: find the largest value of the right derivative, solve for n, and round up. The next section keeps the integral but lets the interval or the integrand become infinite, and asks when the area is still finite.