Integral Calculus 201-SN3-RE (201-NYB) • CEGEP, Montreal

Revision sheet: numerical integration, 201-SN3-RE (201-NYB)

Q: Why do you have to give an error bound?

Because an approximation without a bound does not say how many decimal places can be trusted. The bound turns a number into a bracket around the exact value, which is the only usable result. The marking scheme always gives it marks of its own, separate from those for computing the approximation.

Q: Why does Simpson's rule need an even number of subintervals?

Because the method replaces the curve with a parabola on each pair of consecutive subintervals. An even number of them is therefore needed for the partition to close up exactly. With an odd number, the last pair is incomplete and the coefficients do not close the pattern.

Q: How do you find the constant K in the error bound?

It is an upper bound for the absolute value of the derivative concerned over the whole interval, so its maximum. Study whether this derivative increases or decreases, which is often obvious, and take its largest value. Choosing a value at random gives a wrong bound, sometimes smaller than the actual error.

Q: Does the trapezoidal rule always overestimate the integral?

Only when the function is convex, that is when its second derivative is positive. The chords then lie above the curve. On a concave function it is the other way round, and the midpoint rule always errs in the opposite direction, which gives a free bracket.

Q: How many subintervals do you need for a given accuracy?

Write the inequality stating that the error bound is less than the accuracy required, then isolate the number of subintervals. Always round up, and for Simpson move up again to the next even integer, since the method requires one.

This sheet does not copy out the three formulas of the chapter: they are in your notes. It deals with what the marking scheme penalizes in 201-SN3-RE (formerly 201-NYB), namely the error bound, the coefficients that get swapped and the number of subintervals, three places where an otherwise correct approximation loses half its marks.

It is written for CEGEP students in the Science program (Sciences de la nature). Once the sheet is read, the corrected exercise set on the same chapter puts every reflex to the test.

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The thread of the chapter

A numerical method does not give a number, it gives a number AND an ERROR BOUND. Without the bound, nothing tells you how many decimal places can be trusted, and the answer is then worth no more than the first number that comes to mind.

This chapter is part of 201-SN3-RE (201-NYB), Integral Calculus

The essentials

The three formulas, with the same step

• Step: $h=\frac{b-a}{n}$ , where $n$ is the number of SUBINTERVALS, not the number of points.
• MIDPOINT: $M_{n}=h\left[f(\bar{x}_{1})+\cdots+f(\bar{x}_{n})\right]$ , where $\bar{x}_{i}$ is the midpoint of the $i$ -th subinterval.
• TRAPEZOIDS: $T_{n}=\frac{h}{2}\left[f(x_{0})+2f(x_{1})+\cdots+2f(x_{n-1})+f(x_{n})\right]$ . Coefficients $1, 2, 2,\ldots,2, 1$ .
• SIMPSON: $S_{n}=\frac{h}{3}\left[f(x_{0})+4f(x_{1})+2f(x_{2})+4f(x_{3})+\cdots+4f(x_{n-1})+f(x_{n})\right]$ , with $n$ EVEN.
• There are always $n+1$ points $x_{0},\ldots,x_{n}$ for $n$ subintervals.

Four trapezoids on

\left[1, 3\right]

: each roof is a CHORD of the curve, and since

\frac{1}{x}

is convex, every roof lies above it, so

T_{4}

overestimates the integral.

Simpson's coefficients start and end with $1$ , and alternate $4, 2, 4, 2,\ldots,4$ in between: the $4$ always sits on the ODD indices.

The error bounds, and where to look for the maximum

• Trapezoids: $\left|E_{T}\right|\le\frac{K_{2}\left(b-a\right)^{3}}{12n^{2}}$ , where $K_{2}$ bounds $\left|f''\right|$ on $\left[a,b\right]$ .
• Midpoint: $\left|E_{M}\right|\le\frac{K_{2}\left(b-a\right)^{3}}{24n^{2}}$ , twice as good as the trapezoids.
• Simpson: $\left|E_{S}\right|\le\frac{K_{4}\left(b-a\right)^{5}}{180n^{4}}$ , where $K_{4}$ bounds $\left|f^{(4)}\right|$ .
• $K_{2}$ and $K_{4}$ are looked for over the WHOLE interval, by studying whether the derivative concerned is increasing or decreasing.
• Doubling $n$ divides the trapezoidal error by $4$ and Simpson's error by $16$ .

The bound is an UPPER BOUND: the actual error is almost always much smaller, and that does not prove the bound is wrong.

The direction of the error, given by concavity

• $f$ CONVEX, that is $f''>0$ : the trapezoids OVERESTIMATE, the midpoint rule UNDERESTIMATES.
• $f$ CONCAVE, $f''<0$ : the trapezoids underestimate, the midpoint rule overestimates.
• The two methods therefore BRACKET the integral, which gives a free check.
• Simpson is exact on every polynomial of degree at most THREE, which explains its high order.

On the slice

\left[1, 3\right]

, the chord lies ABOVE the convex curve; the tangent line at the midpoint, which has the area of the midpoint rectangle, lies BELOW it: the two methods bracket the integral.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Handing in an approximation without an error bound

half the question

What not to write

“ $T_{4}=1.1167$ .”

What to write

“ $T_{4}=1.1167$ , with $\left|E_{T}\right|\le\frac{2(2)^{3}}{12(16)}=0.0833$ : the exact value is therefore between $1.033$ and $1.200$ .”

Why: An approximation without a bracket says nothing: it could be wrong from the first decimal place. The bound is what turns a number into a result, and the marking scheme always gives it marks of its own, separate from the computation.

2. Confusing the number of subintervals with the number of points

the whole question, since every evaluation point is shifted

What not to write

“I have five points, so $n=5$ and $h=\frac{3-1}{5}=0.4$ .”

What to write

“Five points make FOUR subintervals: $n=4$ and $h=\frac{3-1}{4}=0.5$ .”

Five points

x_{0},\ldots,x_{4}

mark off FOUR subintervals:

n=4

and

h=\frac{3-1}{4}=0.5

Why: The points run from $x_{0}$ to $x_{n}$ , so there are $n+1$ of them. The immediate check: the last point computed must land exactly on $b$ . If it lands anywhere else, the step is wrong.

3. Using Simpson with an odd number of subintervals

the whole question

What not to write

“ $n=5$ , I apply Simpson.”

What to write

“Simpson requires an EVEN $n$ : I take $n=6$ .”

Each parabola covers TWO subintervals. With

n=5

, the fifth one is left alone (in red): Simpson does not apply.

Why: The method replaces the curve with a parabola on EACH PAIR of subintervals: an even number of them is needed for the partition to close up. With an odd $n$ , the last pair is incomplete and the coefficients do not end on a $1$ .

4. Swapping Simpson's coefficients

the whole question

What not to write

$\frac{h}{3}\left[f(x_{0})+2f(x_{1})+4f(x_{2})+2f(x_{3})+f(x_{4})\right]$

What to write

$\frac{h}{3}\left[f(x_{0})+4f(x_{1})+2f(x_{2})+4f(x_{3})+f(x_{4})\right]$

Why: The $4$ goes on the ODD indices, the middle point of each parabola. The foolproof check: the sum of the coefficients must equal $3n$ , here $12$ ; with $1+4+2+4+1=12$ ✓ and $1+2+4+2+1=10$ ✗.

5. Looking for the maximum of the second derivative in the wrong place

the whole bound, which becomes nearly thirty times too small

What not to write

“ $f''(x)=\frac{2}{x^{3}}$ , I take $K_{2}=f''(3)=\frac{2}{27}$ .”

What to write

“ $\left|f''\right|$ is DECREASING on $\left[1, 3\right]$ , so its maximum is at $x=1$ : $K_{2}=2$ .”

\left|f''(x)\right|=\frac{2}{x^{3}}

decreases on

\left[1, 3\right]

: its maximum is at

x=1

K_{2}=2

. The value at

3

is the MINIMUM.

Why: $K_{2}$ bounds $\left|f''\right|$ over the whole interval, so it is a MAXIMUM, not a value picked at random. Study whether $f''$ is increasing or decreasing in one line, or compare its values at the two endpoints when it is monotonic. An underestimated bound is worse than no bound at all: it claims a false accuracy.

6. Forgetting the $\frac{h}{2}$ or the $\frac{h}{3}$

the whole question, since the answer is four times too large

What not to write

$T_{4}=1+2(0.667)+2(0.5)+2(0.4)+0.333=4.467$

What to write

$T_{4}=\frac{0.5}{2}\times 4.467=1.1167$

Why: The factor in front of the brackets carries the whole dimension: without it, you are adding heights, not areas. The check is immediate: the integral of $\frac{1}{x}$ on $\left[1, 3\right]$ is clearly of the order of $1$ , since the function lies between $\frac{1}{3}$ and $1$ there, over a width of $2$ .

7. Thinking that a wide bound signals a mistake

time wasted redoing a correct computation

What not to write

“The bound is $0.083$ but the actual error is $0.018$ : my bound is wrong.”

What to write

“A bound is an UPPER BOUND on the error: $0.018\le 0.083$ is exactly what is expected.”

Why: The bound assumes the worst case, where $\left|f''\right|$ would reach its maximum everywhere. In practice it is usually three to five times too generous. What would be alarming is the reverse: an actual error LARGER than the announced bound.

8. Looking for $n$ without solving the inequality

1 to 2 marks, and an answer that cannot be checked

What not to write

“For an accuracy of $10^{-4}$ , I take $n=100$ at random.”

What to write

“I solve $\frac{2(8)}{12n^{2}}\le 10^{-4}$ , that is $n^{2}\ge 13\,333$ , so $n\ge 116$ : I take $n=116$ .”

Why: The question “how many subintervals” expects a solved inequality, not a guess. Isolate $n$ , take the square root, ROUND UP, and for Simpson move up again to the next even integer.

Which method to choose

Which method, depending on what the problem asks

The three methods serve the same purpose with different accuracies. The problem almost always imposes which one, but you need to know why.

If the problem gives a table of values at the endpoints of the subintervals → TRAPEZOIDS or SIMPSON, the only ones that use just these points

Example: flow rate readings every two hours
If the problem gives the values at the MIDPOINTS → the MIDPOINT rule, twice as accurate as the trapezoids for the same work

Example: measurements taken at the centre of each slice
If the problem requires high accuracy with few evaluations → SIMPSON, whose error decreases like $\frac{1}{n^{4}}$ instead of $\frac{1}{n^{2}}$

Example: four correct decimal places with $n=8$
If the antiderivative does not exist in elementary form → any of the three, but with its error bound

Example: $\int_{0}^{1}e^{-x^{2}}dx$
If the integrand is a polynomial of degree at most three → SIMPSON is EXACT: the error is zero, whatever $n$

Example: $\int_{0}^{2}x^{3}dx$ gives exactly $4$ from $n=2$ on

The midpoint rule and the trapezoids bracket the integral when the concavity does not change: computing both gives a bracket without any theoretical bound.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

An approximation with its bound, written up for the marker

When to use it: Every numerical integration question, even when the bound is not explicitly asked for.

1 Compute $h=\frac{b-a}{n}$ and LIST the $n+1$ points $x_{0},\ldots,x_{n}$ .
2 Draw up a table of the $f(x_{i})$ , with enough decimal places for the accuracy required.
3 Apply the formula writing out the coefficients, then multiply by the factor in front of the brackets.
4 Compute the required derivative, study whether it increases or decreases, and give $K$ as a justified MAXIMUM.
5 Apply the bound formula and conclude with a BRACKET.

Concluding sentence

$T_{4}=1.1167$ with $\left|E_{T}\right|\le 0.0833$ , so $\int_{1}^{3}\frac{dx}{x}$ lies between $1.033$ and $1.200$ .

The trap: Giving $K$ without justifying that it is a maximum. The sentence “ $\left|f''\right|=\frac{2}{x^{3}}$ is decreasing on $\left[1, 3\right]$ , so its maximum is $f''(1)=2$ ” is worth a mark on its own.

Marking: Usually 1 mark for the step and the points, 2 marks for the approximation, 1 mark for a justified $K$ , 1 mark for the bound and the bracket.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

Check that the last point lands on $b$

Compute $x_{0}+nh$ : the result must be exactly $b$ . This is the check that catches the confusion between points and subintervals.

$1+4\times 0.5=3$ ✓. With $h=0.4$ , you would get $2.6$ , which signals a wrong step.
Check the sum of the coefficients

For the trapezoids, the coefficients add up to $2n$ ; for Simpson, to $3n$ . Any other total signals a misplaced coefficient.

Simpson with $n=4$ : $1+4+2+4+1=12=3\times 4$ ✓. The sequence $1, 2, 4, 2, 1$ gives $10$ and gives the swap away.
Bracket with the midpoint rule and the trapezoids

When the concavity does not change, compute both approximations: the integral lies between them, without any error formula.

On $\frac{1}{x}$ , which is convex, $M_{n}\le\int\le T_{n}$ : a result outside this bracket signals a computational mistake.
Compare the actual error with the bound

When the exact value is known, check that the gap is SMALLER than the announced bound. A larger gap proves a mistake, not a pessimistic bound.

$\left|1.1167-1.0986\right|=0.0181\le 0.0833$ ✓.
Check the order of magnitude with an average rectangle

Multiply the width of the interval by a typical value of the function: the result must be of the same order as the approximation.

$\frac{1}{x}$ lies between $0.33$ and $1$ over a width of $2$ : the integral is between $0.67$ and $2$ , and $1.12$ falls inside ✓.

The typical problem, taken apart

Trapezoids and Simpson on the same integral, with the bounds

Approximate $\displaystyle\int_{1}^{3}\frac{dx}{x}$ with $n=4$ , by the trapezoidal rule and then by Simpson's rule, and give an error bound in each case.

The parabola through

(1, 1)

(2, 0.5)

and

(3, \frac{1}{3})

hugs the curve far better than the chord: this substitution is what gives Simpson its order four.

Step 1

$h=\frac{3-1}{4}=0.5$ , and the FIVE points are $1$ , $1.5$ , $2$ , $2.5$ , $3$ . Values: $1$ , $0.6667$ , $0.5$ , $0.4$ , $0.3333$ .

Why

Five points for four subintervals, and the last one lands exactly on $3$ : both checks of the step are done before any computation.

Step 2

Trapezoids: $T_{4}=\frac{0.5}{2}\left[1+2(0.6667)+2(0.5)+2(0.4)+0.3333\right]=0.25\times 4.4667=1.1167$ .

Why

The coefficients $1, 2, 2, 2, 1$ add up to $8=2n$ ✓, and the factor $\frac{h}{2}$ is applied at the end. Both checks take five seconds.

Step 3

Trapezoidal bound: $f''(x)=\frac{2}{x^{3}}$ , decreasing on $\left[1, 3\right]$ , so $K_{2}=f''(1)=2$ . Hence $\left|E_{T}\right|\le\frac{2(3-1)^{3}}{12(4)^{2}}=\frac{16}{192}=0.0833$ .

Why

The fact that $\left|f''\right|$ is decreasing is written out explicitly: it is what justifies that $2$ really is a MAXIMUM, and it is the point most papers lose.

Step 4

Simpson: $S_{4}=\frac{0.5}{3}\left[1+4(0.6667)+2(0.5)+4(0.4)+0.3333\right]=0.16667\times 6.6=1.1000$ .

Why

Coefficients $1, 4, 2, 4, 1$ , sum $12=3n$ ✓, with the $4$ on the odd indices. The factor is $\frac{h}{3}$ and not $\frac{h}{2}$ : the two are easily confused.

Step 5

Simpson bound: $f^{(4)}(x)=\frac{24}{x^{5}}$ , decreasing, so $K_{4}=24$ . Hence $\left|E_{S}\right|\le\frac{24(2)^{5}}{180(4)^{4}}=\frac{768}{46\,080}=0.0167$ .

Why

Same method, one derivative higher. Simpson's bound is five times tighter than the trapezoidal one for the same number of evaluations, which is the whole point of the method.

Step 6

Check: the exact value is $\ln 3=1.0986$ . Trapezoidal error $0.0181\le 0.0833$ ✓; Simpson error $0.0014\le 0.0167$ ✓. Simpson is thirteen times more accurate.

Why

Both actual errors are well below their bounds, as expected of an upper bound. The ratio between the two confirms Simpson's higher order, which is the conclusion the question is after.

The conclusion, written out

$T_{4}=1.1167$ to within $0.0833$ and $S_{4}=1.1000$ to within $0.0167$ , while $\ln 3=1.0986$ : both brackets contain the exact value, and Simpson's is much tighter.

The classic mistake on this problem: Taking $K_{2}=f''(3)=\frac{2}{27}$ because it is the last value computed. The bound would become $0.0031$ , smaller than the actual error of $0.0181$ : it would claim an accuracy the computation does not have.

Learn by heart

• $h=\frac{b-a}{n}$ , with $n$ SUBINTERVALS and $n+1$ points.
• Trapezoids: coefficients $1, 2,\ldots,2, 1$ and factor $\frac{h}{2}$ . Sum of the coefficients $2n$ .
• Simpson: coefficients $1, 4, 2,\ldots,4, 1$ and factor $\frac{h}{3}$ , with $n$ EVEN. Sum $3n$ .
• $\left|E_{T}\right|\le\frac{K_{2}(b-a)^{3}}{12n^{2}}$ , $\left|E_{M}\right|\le\frac{K_{2}(b-a)^{3}}{24n^{2}}$ , $\left|E_{S}\right|\le\frac{K_{4}(b-a)^{5}}{180n^{4}}$ .
• $K$ is a MAXIMUM over the whole interval, justified by whether the derivative increases or decreases.
• Convex: trapezoids above, midpoint below. The two bracket the integral.
• Simpson is EXACT on every polynomial of degree at most three.
• To find $n$ : solve the inequality and ROUND UP.

Frequently asked questions

Why do you have to give an error bound?

Because an approximation without a bound does not say how many decimal places can be trusted. The bound turns a number into a bracket around the exact value, which is the only usable result. The marking scheme always gives it marks of its own, separate from those for computing the approximation.

Why does Simpson's rule need an even number of subintervals?

Because the method replaces the curve with a parabola on each pair of consecutive subintervals. An even number of them is therefore needed for the partition to close up exactly. With an odd number, the last pair is incomplete and the coefficients do not close the pattern.

How do you find the constant K in the error bound?

It is an upper bound for the absolute value of the derivative concerned over the whole interval, so its maximum. Study whether this derivative increases or decreases, which is often obvious, and take its largest value. Choosing a value at random gives a wrong bound, sometimes smaller than the actual error.

Does the trapezoidal rule always overestimate the integral?

Only when the function is convex, that is when its second derivative is positive. The chords then lie above the curve. On a concave function it is the other way round, and the midpoint rule always errs in the opposite direction, which gives a free bracket.

How many subintervals do you need for a given accuracy?

Write the inequality stating that the error bound is less than the accuracy required, then isolate the number of subintervals. Always round up, and for Simpson move up again to the next even integer, since the method requires one.

Practise it

Corrected exercises: Numerical integration, 201-SN3-RE (201-NYB)

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

15 corrected exercises
150 points
225 minutes

Do the exercises

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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-calc2-numerical-integration. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: numerical integration, 201-SN3-RE (201-NYB)

The essentials

The three formulas, with the same step

The error bounds, and where to look for the maximum

The direction of the error, given by concavity

The mistakes that cost marks

1. Handing in an approximation without an error bound

2. Confusing the number of subintervals with the number of points

3. Using Simpson with an odd number of subintervals

4. Swapping Simpson's coefficients

5. Looking for the maximum of the second derivative in the wrong place

6. Forgetting the h2\frac{h}{2}2h​ or the h3\frac{h}{3}3h​

7. Thinking that a wide bound signals a mistake

8. Looking for nnn without solving the inequality

Which method to choose

Which method, depending on what the problem asks

How the answer is expected to be written

An approximation with its bound, written up for the marker

Check before you hand in

The typical problem, taken apart

Trapezoids and Simpson on the same integral, with the bounds

Learn by heart

Frequently asked questions

Why do you have to give an error bound?

Why does Simpson's rule need an even number of subintervals?

How do you find the constant K in the error bound?

Does the trapezoidal rule always overestimate the integral?

How many subintervals do you need for a given accuracy?

Corrected exercises: Numerical integration, 201-SN3-RE (201-NYB)

See also

Looking for a 201-SN3-RE (201-NYB) tutor in Montreal?

6. Forgetting the $\frac{h}{2}$ or the $\frac{h}{3}$

8. Looking for $n$ without solving the inequality