What are reading the ordinates?

There are two ways the ordinates y_0, , y_n reach you.

What is wrong strip width?

h = (b - a)/(n), the spacing between adjacent x-values. Reading h off a table means checking the x-values really are evenly spaced.

§-Syllabus dot point

NSWMaths AdvancedSyllabus dot point

How can we estimate the area under a curve, or a definite integral, when we only have a table of values or an integral we cannot evaluate exactly?

Use the trapezoidal rule to estimate areas and definite integrals, and determine whether the estimate is an over- or under-estimate

A focused answer to the HSC Maths Advanced dot point on the trapezoidal rule. The single-application and multi-strip formulas, reading values from a table or a function, estimating a definite integral, and using concavity to decide over- or under-estimate.

Generated by Claude Opus 4.816 min answerUpdated 2026-06-21

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

What this dot point is asking

NESA wants you to estimate the area under a curve, or equivalently the value of a definite integral, by slicing the region into vertical strips and treating the top of each strip as a straight line rather than a curve. Each strip becomes a trapezium, you add the trapezium areas, and that total approximates $\int_a^b f(x)\,dx$ . You also need to say whether the estimate is too big or too small by looking at the concavity of the curve.

This is the tool for the situation where exact integration is unavailable: either you are only given a table of measured values (a surveyor's offsets, a car's speed at fixed times) and have no formula at all, or you have a formula whose integral is beyond the course. The trapezoidal rule turns either case into a few multiplications and additions.

The answer

The idea is to replace the curved top of each strip with the straight chord joining its two end points. A strip with vertical sides of height $f(a)$ and $f(b)$ and width $b - a$ is then a trapezium, and the area of a trapezium is the average of the two parallel sides times the distance between them.

Why the interior ordinates are doubled

The multi-strip formula is not a new rule, just the single rule applied to each strip and the results added. Take three strips with ordinates $y_0, y_1, y_2, y_3$ and common width $h$ . The three trapezia have areas

\frac{h}{2}(y_0 + y_1), \qquad \frac{h}{2}(y_1 + y_2), \qquad \frac{h}{2}(y_2 + y_3).

Add them and factor out $\frac{h}{2}$ :

\frac{h}{2}\big[(y_0 + y_1) + (y_1 + y_2) + (y_2 + y_3)\big] = \frac{h}{2}\big[y_0 + 2y_1 + 2y_2 + y_3\big].

Every interior ordinate is shared by two adjacent trapezia, so it appears twice; the two outermost ordinates belong to only one trapezium each, so they appear once. That is the whole reason for the "ends once, middles twice" pattern, and seeing it this way means you never have to memorise which values get doubled. The single-application rule is just the $n = 1$ case, where there are no interior ordinates at all.

The multi-strip rule, strip by strip

The figure shows four strips ( $n = 4$ ). The five vertical lines are the ordinates $y_0$ through $y_4$ , equally spaced $h$ apart along the $x$ -axis. Each shaded trapezium has a flat top (the chord), and stacking the four areas with the "ends once, middles twice" weighting gives the estimate. The more strips you use, the more closely the chords hug the curve, so a larger $n$ gives a more accurate estimate. In the exam you use exactly the number of strips the question (or its table) dictates: "two applications of the trapezoidal rule" means $n = 2$ , and a table with five columns of values means $n = 4$ .

Reading the ordinates: from a table or from a formula

There are two ways the ordinates $y_0, \dots, y_n$ reach you.

From a table. The question hands you a table of $x$ -values and matching $f(x)$ -values. Read the heights straight off it; you do not need (and may not have) a formula. This is the common case when the data comes from measurement, and it is exactly how the 2024 exam supplied $\ln(1 + x^2)$ to 4 decimal places. Check first that the $x$ -values are equally spaced, because the rule in this form requires a constant strip width $h$ .

From a formula. The question gives $y = f(x)$ , and you choose (or are told) how many strips to use, work out $h = \frac{b - a}{n}$ , list the $x$ -values $a, a + h, a + 2h, \dots, b$ , and substitute each into $f$ to get its ordinate. Set the working out as a small table of $x$ against $y$ so you do not mismatch a height with the wrong strip.

Either way, the arithmetic of the rule is identical once you have the ordinates.

Estimating a definite integral

Because the area under $y = f(x)$ from $a$ to $b$ is the definite integral $\int_a^b f(x)\,dx$ (for $f(x) \ge 0$ ), the trapezoidal rule is also a way to estimate an integral you cannot evaluate exactly. The phrasing "use the trapezoidal rule to estimate $\int_a^b f(x)\,dx$ " is asking for precisely the same calculation as "estimate the area under the curve": list the ordinates, weight the ends once and the middles twice, multiply by $\frac{h}{2}$ .

Over-estimate or under-estimate: read the concavity

The trapezoidal rule replaces the curve with straight chords, so the only question is whether each chord sits above or below the curve it is cutting across, and that is decided by concavity.

If the curve is concave up ( $f''(x) > 0$ , the curve bends upward like a valley) on the interval, every chord lies above the curve, so each trapezium has a little extra area between the chord and the curve. The rule overestimates.

If the curve is concave down ( $f''(x) < 0$ , the curve bends like a hill) on the interval, every chord lies below the curve, so each trapezium misses a sliver of area under the curve. The rule underestimates.

A clean way to remember which is which: a chord always joins two points on the curve, so it is a straight shortcut between them. On a valley (concave up) the shortcut runs over the dip, so it is too high and the area comes out too big. On a hill (concave down) the shortcut cuts under the bulge, so it is too low and the area comes out too small.

How exam questions ask about the trapezoidal rule

The wording is fairly fixed, and it tells you exactly which form of the rule to use:

"Use the trapezoidal rule to estimate the area / the shaded region / $\int_a^b f(x)\,dx$ ." The default request. Read or compute the ordinates and apply $\frac{h}{2}[\,\text{ends} + 2(\text{middles})\,]$ .
"Using the function values provided ..." or a table is given. Read $y_0, \dots, y_n$ straight off the table; the number of columns fixes $n$ (five columns means four strips). Check the $x$ -values are evenly spaced before using $h$ .
"Use two applications of the trapezoidal rule" (or one, or $n$ applications). "Applications" counts the strips: two applications means $n = 2$ , so $h = \frac{b - a}{2}$ and three ordinates. One application is the single-trapezium rule.
"Use the trapezoidal rule with $n = 4$ strips" or "... with sub-intervals of width $0.5$ ." Either form pins down $h$ ; if width is given, $n = \frac{b - a}{h}$ .
"Is your answer an over-estimate or an under-estimate? Give a reason." Decide from concavity: concave up over, concave down under. If a graph is shown or $f''$ was found earlier in the question, quote that as your reason.
"Hence" after estimating an area exactly and approximately. The two values are sometimes combined to bound a constant (the 2025 paper used the area to deduce an inequality for $e$ ), so keep both your exact and estimated answers.

The trap to watch is the meaning of "applications" or "strips" versus "ordinates": $n$ strips always need $n + 1$ ordinates, and the strip width is $h = \frac{b - a}{n}$ , not $\frac{b - a}{n+1}$ .

Worked examples

These run through the single-application rule on a real area, then the multi-strip rule from a table and from a formula, the two examined past questions in full, and a distance-from-speed estimate.

Estimate the area of a block of land with one trapezium

A rectangular survey baseline runs $40$ metres along one straight edge of a block of land. The opposite, slightly sloping boundary is $18$ metres from the baseline at the start and $26$ metres from it at the end. Estimate the area of the block using a single application of the trapezoidal rule.

Set up the single-application rule. The two ends of the region are the offsets $f(a) = 18$ m and $f(b) = 26$ m, and the width is $b - a = 40$ m. One trapezium treats the boundary between them as straight:

A \approx \frac{b - a}{2}\big[\,f(a) + f(b)\,\big] = \frac{40}{2}\,(18 + 26).

Evaluate. This is $20 \times 44 = 880$ square metres.

State with units. The block has an area of approximately $880 \text{ m}^2$ . (The single strip assumes the boundary is a straight line; if the boundary curves, more offsets and more strips would sharpen the estimate.)

Estimate an area from a table of offsets

A surveyor measures the perpendicular distance from a straight $60$ m baseline to an irregular creek boundary every $15$ m. The offsets, in metres, are $12, 17, 21, 20, 14$ . Estimate the area between the baseline and the creek.

Identify $h$ , $n$ and the ordinates. The offsets are $15$ m apart, so $h = 15$ and there are five ordinates, hence $n = 4$ strips: $y_0 = 12$ , $y_1 = 17$ , $y_2 = 21$ , $y_3 = 20$ , $y_4 = 14$ .

Apply the multi-strip rule, ends once and middles twice.

A \approx \frac{h}{2}\big[\,y_0 + 2(y_1 + y_2 + y_3) + y_4\,\big] = \frac{15}{2}\big[\,12 + 2(17 + 21 + 20) + 14\,\big].

Evaluate the brackets. The interior sum is $17 + 21 + 20 = 58$ , so the bracket is $12 + 2(58) + 14 = 12 + 116 + 14 = 142$ .

Finish with units. $A \approx 7.5 \times 142 = 1065 \text{ m}^2$ . The area between the baseline and the creek is about $1065$ square metres.

Estimate a definite integral from a formula

Estimate $\displaystyle\int_1^4 \sqrt{x}\,\,dx$ using the trapezoidal rule with three strips, and state whether the estimate is too large or too small.

Work out $h$ and the ordinates. With $n = 3$ on $[1, 4]$ , $h = \frac{4 - 1}{3} = 1$ , so the $x$ -values are $1, 2, 3, 4$ and the ordinates are $y_0 = \sqrt{1} = 1$ , $y_1 = \sqrt{2} \approx 1.4142$ , $y_2 = \sqrt{3} \approx 1.7321$ , $y_3 = \sqrt{4} = 2$ .

Apply the rule.

\int_1^4 \sqrt{x}\,\,dx \approx \frac{1}{2}\big[\,1 + 2(1.4142 + 1.7321) + 2\,\big] = \frac{1}{2}\big[\,3 + 2(3.1463)\,\big] = \frac{1}{2}(9.2926) \approx 4.6463.

Decide over or under. The curve $y = \sqrt{x}$ is concave down ( $f''(x) = -\tfrac{1}{4}x^{-3/2} < 0$ ), so the chords lie below the curve and the estimate is an under-estimate. As a check, the exact value is $\left[\frac{2}{3}x^{3/2}\right]_1^4 = \frac{2}{3}(8 - 1) = \frac{14}{3} \approx 4.6667$ , and indeed $4.6463 < 4.6667$ .

Estimate the area under $f(x) = \ln(1 + x^2)$ (2024 HSC Q22)

For $f(x) = \ln(1 + x^2)$ , a table gives the values below (to 4 decimal places). Using the trapezoidal rule, estimate the shaded area under the curve from $x = 0$ to $x = 1$ , and say whether it is an over- or under-estimate.

$x$	$0$	$0.25$	$0.5$	$0.75$	$1$
$\ln(1 + x^2)$	$0$	$0.0606$	$0.2231$	$0.4463$	$0.6931$

Read off $h$ and the ordinates. The five $x$ -values are $0.25$ apart, so $h = 0.25$ and $n = 4$ : $y_0 = 0$ , $y_1 = 0.0606$ , $y_2 = 0.2231$ , $y_3 = 0.4463$ , $y_4 = 0.6931$ .

Apply the rule.

A \approx \frac{0.25}{2}\big[\,y_0 + 2(y_1 + y_2 + y_3) + y_4\,\big] = 0.125\big[\,0 + 2(0.0606 + 0.2231 + 0.4463) + 0.6931\,\big].

Evaluate. The interior sum is $0.0606 + 0.2231 + 0.4463 = 0.73$ , so $A \approx 0.125\,[\,0.6931 + 1.46\,] = 0.125 \times 2.1531 = 0.2691$ square units.

Over or under, with a reason. It is an over-estimate. On $0 < x < 1$ the curve is concave up: $f''(x) = \dfrac{2(1 - x^2)}{(1 + x^2)^2}$ , which is positive there because $1 - x^2 > 0$ . With the curve concave up the chords lie above it, so each trapezium adds a little extra and the total overestimates the true area. (The exact value is $\ln 2 - 2 + \frac{\pi}{2} \approx 0.2639$ , confirming $0.2691$ is slightly too big.)

Estimate the area under $y = \left(\tfrac{1}{2}\right)^x$ with two applications (2025 HSC Q27)

The region is bounded by $y = \left(\tfrac{1}{2}\right)^x$ , the coordinate axes and $x = 2$ . Use two applications of the trapezoidal rule to estimate its area.

Turn "two applications" into $n$ and $h$ . Two applications means $n = 2$ strips, so $h = \frac{2 - 0}{2} = 1$ , with ordinates at $x = 0, 1, 2$ .

Compute the ordinates. $y_0 = \left(\tfrac{1}{2}\right)^0 = 1$ , $y_1 = \left(\tfrac{1}{2}\right)^1 = 0.5$ , $y_2 = \left(\tfrac{1}{2}\right)^2 = 0.25$ .

Apply the rule (one interior ordinate, doubled).

A \approx \frac{h}{2}\big[\,y_0 + 2 y_1 + y_2\,\big] = \frac{1}{2}\big[\,1 + 2(0.5) + 0.25\,\big] = \frac{1}{2}(2.25) = 1.125 \text{ square units}.

Sense-check against the concavity. $y = \left(\tfrac{1}{2}\right)^x$ is an exponential decay curve, which is concave up, so this should be an over-estimate; the exact area is $\frac{3}{4\ln 2} \approx 1.0820$ , and indeed $1.125 > 1.0820$ , as expected.

Estimate distance travelled from a speed table

A car's speed is recorded every $2$ seconds during the first $10$ seconds of a drive, giving the table below (speed in metres per second). Estimate the distance travelled in the $10$ seconds.

$t$ (s)	$0$	$2$	$4$	$6$	$8$	$10$
$v$ (m/s)	$0$	$8$	$15$	$19$	$21$	$22$

Recognise distance as area under the speed-time graph. Distance is $\int_0^{10} v\,dt$ , the area under the $v$ - $t$ curve, so estimate that area with the trapezoidal rule. The readings are $2$ s apart, so $h = 2$ and $n = 5$ strips: $y_0 = 0, y_1 = 8, y_2 = 15, y_3 = 19, y_4 = 21, y_5 = 22$ .

Apply the rule.

\text{distance} \approx \frac{h}{2}\big[\,y_0 + 2(y_1 + y_2 + y_3 + y_4) + y_5\,\big] = \frac{2}{2}\big[\,0 + 2(8 + 15 + 19 + 21) + 22\,\big].

Evaluate. The interior sum is $8 + 15 + 19 + 21 = 63$ , so the bracket is $0 + 2(63) + 22 = 148$ , and with $\frac{h}{2} = 1$ the distance is $148$ metres.

State with units. The car travels approximately $148$ m in the first $10$ seconds.

Common traps

Doubling the wrong ordinates: Only the interior ordinates are doubled; the first and last are counted once. Doubling the ends (or forgetting to double the middles) is the most frequent error.
Confusing strips with ordinates: $n$ strips need $n + 1$ ordinates, and the width is $h = \frac{b - a}{n}$ . "Two applications" or "two strips" means $n = 2$ and three ordinates, not two.
Wrong strip width: $h = \frac{b - a}{n}$ , the spacing between adjacent $x$ -values. Reading $h$ off a table means checking the $x$ -values really are evenly spaced.
Naming concavity but giving no reason: "Over-estimate because concave up" earns the mark; "over-estimate" alone usually does not. Link the concavity to the chords lying above or below the curve.
Getting over and under back to front: Concave up overestimates; concave down underestimates. Picture the straight chord as a shortcut: over a valley it is too high, under a hill it is too low.
Forgetting units: Area answers are in square units (or m $^2$ ); a distance from a speed table is in metres. Leaving units off a real-context answer drops marks.

Exam technique

Lay the working out as a small table of $x$ against $y$ first, even when a formula is given, so each ordinate is paired with the right strip and nothing is doubled by mistake. Write the rule with the brackets before substituting, $\frac{h}{2}[\,y_0 + 2(y_1 + \cdots + y_{n-1}) + y_n\,]$ , because markers award a method mark for the correct structure even if the arithmetic slips. State $h$ explicitly and confirm the count of strips matches the wording ("two applications" is $n = 2$ ). For the over- or under-estimate part, give a one-line concavity reason: name concave up or down, say the chords lie above or below the curve, then state over or under, and quote $f''$ or the given graph if you have it. Keep the units of the context (square metres for a block of land, metres for a distance) and round only as the question asks.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2024 HSC Q223 marksFor

f(x) = \ln(1 + x^2)

, a table of values at

x = 0, 0.25, 0.5, 0.75, 1

is given (to 4 decimal places). Using the trapezoidal rule, estimate the shaded area under the curve from

x = 0

x = 1

, then state whether the estimate is an over- or under-estimate, with a reason.

Show worked answer →

The table gives $y_0 = 0$ , $y_1 = 0.0606$ , $y_2 = 0.2231$ , $y_3 = 0.4463$ , $y_4 = 0.6931$ , with strip width $h = \frac{1 - 0}{4} = 0.25$ .

Apply the multi-strip rule, adding the two end values once and the three interior values twice:

$A \approx \frac{h}{2}\left[ y_0 + 2(y_1 + y_2 + y_3) + y_4 \right] = \frac{0.25}{2}\left[ 0 + 2(0.0606 + 0.2231 + 0.4463) + 0.6931 \right]$ .

The interior sum is $0.0606 + 0.2231 + 0.4463 = 0.73$ , so $A \approx 0.125 \left[ 0.6931 + 1.46 \right] = 0.125 \times 2.1531 = 0.2691$ square units.

It is an overestimate. On $0 < x < 1$ the curve is concave up (it was proved in part (a) that $f''(x) > 0$ there), so each straight chord lies above the curve and every trapezium captures slightly more than the true area. Markers reward the correct $h$ , the rule with the interior values doubled, the value $0.2691$ , and the concave-up reason for the overestimate.

2025 HSC Q272 marksThe region is bounded by

y = \left(\tfrac{1}{2}\right)^x

, the coordinate axes and

x = 2

. Use two applications of the trapezoidal rule to estimate the area of the shaded region.

Show worked answer →

Two applications means two strips, so $n = 2$ and $h = \frac{2 - 0}{2} = 1$ , with ordinates at $x = 0, 1, 2$ .

The function values are $y_0 = \left(\tfrac{1}{2}\right)^0 = 1$ , $y_1 = \left(\tfrac{1}{2}\right)^1 = 0.5$ and $y_2 = \left(\tfrac{1}{2}\right)^2 = 0.25$ .

$A \approx \frac{h}{2}\left[ y_0 + 2 y_1 + y_2 \right] = \frac{1}{2}\left[ 1 + 2(0.5) + 0.25 \right] = \frac{1}{2}(2.25) = 1.125$ square units.

Equivalently, this is the two trapezia $\frac{1}{2}(1 + 0.5) = 0.75$ and $\frac{1}{2}(0.5 + 0.25) = 0.375$ , which sum to $1.125$ . The exact area is $\frac{3}{4\ln 2} \approx 1.0820$ , so the trapezoidal value is an overestimate, as expected for a concave-up curve. Markers reward $h = 1$ , the three correct ordinates, and the value $1.125$ .

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marksUse a single application of the trapezoidal rule to estimate

\int_2^6 f(x)\,dx

given that

f(2) = 5

and

f(6) = 13

Show worked solution →

Set up the single-application rule. One application uses the two end ordinates and the full width $b - a = 6 - 2 = 4$ :

\int_2^6 f(x)\,dx \approx \frac{b - a}{2}\big[\,f(a) + f(b)\,\big] = \frac{4}{2}\big[\,5 + 13\,\big].

Evaluate.

\frac{4}{2}(18) = 2 \times 18 = 36.

So the estimate is $\int_2^6 f(x)\,dx \approx 36$ .

Marker's note: one mark for the correct rule with $\frac{b-a}{2} = 2$ , one for the value $36$ . Using $\frac{b-a}{n}$ with the wrong width, or averaging the ordinates without multiplying by the width, is the usual slip.

foundation3 marksEstimate

\int_0^4 \sqrt{x + 1}\,dx

using the trapezoidal rule with two strips.

Show worked solution →

Work out $h$ and the ordinates. With $n = 2$ on $[0, 4]$ , $h = \frac{4 - 0}{2} = 2$ , so the $x$ -values are $0, 2, 4$ :

y_0 = \sqrt{0 + 1} = 1, \qquad y_1 = \sqrt{2 + 1} = \sqrt{3} \approx 1.7321, \qquad y_2 = \sqrt{4 + 1} = \sqrt{5} \approx 2.2361.

Apply the rule (one interior ordinate, doubled).

\int_0^4 \sqrt{x + 1}\,dx \approx \frac{h}{2}\big[\,y_0 + 2 y_1 + y_2\,\big] = \frac{2}{2}\big[\,1 + 2(1.7321) + 2.2361\,\big].

Evaluate.

1 \times \big[\,1 + 3.4642 + 2.2361\,\big] = 6.7003 \approx 6.70.

So the estimate is about $6.70$ square units.

Marker's note: one mark for $h = 2$ and the three ordinates, one for the rule with $y_1$ doubled, one for the value $6.70$ (to two decimal places). Doubling an end ordinate, or forgetting to double $y_1$ , is the standard error.

foundation2 marksThe curve

y = x^2 + 1

is concave up on

[1, 4]

. If the trapezoidal rule is used to estimate the area under this curve, state whether the estimate is an over-estimate or an under-estimate, and give a reason.

Show worked solution →

Link the concavity to the chords. The curve is concave up, so every chord joining two points on the curve lies above the curve between them.

Conclude. Each trapezium therefore captures a little extra area between the chord and the curve, so the total is an over-estimate.

Marker's note: one mark for stating over-estimate, one for the reason that concave up means the chords lie above the curve. "Over-estimate" with no reason usually earns nothing; naming the chord position is what secures the mark.

core4 marksA river channel is surveyed by measuring its depth at equal horizontal spacings across a straight

24

m cross-section. The depths in metres are recorded in the table: at

0

m the depth is

0

; at

6

m it is

2.4

; at

12

m it is

3.1

; at

18

m it is

2.0

; at

24

m it is

0

. (a) Use the trapezoidal rule to estimate the area of the cross-section. (b) Water flows through this cross-section at

0.5

m/s; estimate the volume of water passing per second.

Show worked solution →

Part (a): identify $h$ , $n$ and the ordinates. The depths are measured $6$ m apart, so $h = 6$ and there are five ordinates, hence $n = 4$ strips:

y_0 = 0, \quad y_1 = 2.4, \quad y_2 = 3.1, \quad y_3 = 2.0, \quad y_4 = 0.

Apply the multi-strip rule, ends once and middles twice.

A \approx \frac{h}{2}\big[\,y_0 + 2(y_1 + y_2 + y_3) + y_4\,\big] = \frac{6}{2}\big[\,0 + 2(2.4 + 3.1 + 2.0) + 0\,\big].

The interior sum is $2.4 + 3.1 + 2.0 = 7.5$ , so

A \approx 3\big[\,2(7.5)\,\big] = 3 \times 15 = 45 \text{ m}^2.

Part (b): multiply the cross-sectional area by the flow speed. Volume per second is area times speed:

45 \times 0.5 = 22.5 \text{ m}^3\text{/s}.

Marker's note: one mark for $h = 6$ with the five ordinates, one for the rule with the interior three doubled, one for $A = 45$ m $^2$ , one for the flow rate $22.5$ m $^3$ /s with units. Treating the two zero end depths as interior values (and doubling them) changes nothing here, but the habit fails elsewhere; keep the ends single.

core4 marksEstimate

\int_1^3 \ln x\,dx

using the trapezoidal rule with four strips, giving your answer to three decimal places. State, with a reason, whether the estimate is an over-estimate or an under-estimate. Take

\ln 1 = 0

\ln 1.5 = 0.4055

\ln 2 = 0.6931

\ln 2.5 = 0.9163

and

\ln 3 = 1.0986

Show worked solution →

Work out $h$ and list the ordinates. With $n = 4$ on $[1, 3]$ , $h = \frac{3 - 1}{4} = 0.5$ , so the $x$ -values are $1, 1.5, 2, 2.5, 3$ :

y_0 = 0, \quad y_1 = 0.4055, \quad y_2 = 0.6931, \quad y_3 = 0.9163, \quad y_4 = 1.0986.

Apply the rule.

\int_1^3 \ln x\,dx \approx \frac{h}{2}\big[\,y_0 + 2(y_1 + y_2 + y_3) + y_4\,\big] = \frac{0.5}{2}\big[\,0 + 2(0.4055 + 0.6931 + 0.9163) + 1.0986\,\big].

The interior sum is $0.4055 + 0.6931 + 0.9163 = 2.0149$ , so

\int_1^3 \ln x\,dx \approx 0.25\big[\,1.0986 + 2(2.0149)\,\big] = 0.25\big[\,1.0986 + 4.0298\,\big] = 0.25 \times 5.1284 = 1.2821.

To three decimal places the estimate is $1.282$ .

Decide over or under. The curve $y = \ln x$ has $f''(x) = -\dfrac{1}{x^2} < 0$ , so it is concave down. The chords lie below the curve, so each trapezium misses a sliver of area and the estimate is an under-estimate.

Marker's note: one mark for $h = 0.5$ and the ordinates, one for the correctly weighted rule, one for the value $1.282$ , one for the under-estimate justified by $f'' < 0$ (concave down). The exact value is $[x\ln x - x]_1^3 = 3\ln 3 - 2 \approx 1.2958$ , confirming $1.282$ is slightly low.

exam5 marksA particle moves in a straight line. Its velocity

v

in m/s is recorded every

3

seconds over a

12

second interval, giving the readings: at

t = 0

v = 4

; at

t = 3

v = 10

; at

t = 6

v = 18

; at

t = 9

v = 24

; at

t = 12

v = 27

. (a) Use the trapezoidal rule to estimate the distance travelled in the

12

seconds. (b) The velocity graph is concave down throughout; state whether your estimate is greater than or less than the true distance, with a reason. (c) Hence estimate the average speed of the particle over the

12

seconds.

Show worked solution →

Part (a): distance is the area under the velocity-time graph. The distance travelled is $\int_0^{12} v\,dt$ , so estimate that area. The readings are $3$ s apart, so $h = 3$ and $n = 4$ strips:

y_0 = 4, \quad y_1 = 10, \quad y_2 = 18, \quad y_3 = 24, \quad y_4 = 27.

Apply the multi-strip rule, ends once and interior three doubled:

\text{distance} \approx \frac{h}{2}\big[\,y_0 + 2(y_1 + y_2 + y_3) + y_4\,\big] = \frac{3}{2}\big[\,4 + 2(10 + 18 + 24) + 27\,\big].

The interior sum is $10 + 18 + 24 = 52$ , so the bracket is $4 + 2(52) + 27 = 4 + 104 + 27 = 135$ , giving

\text{distance} \approx 1.5 \times 135 = 202.5 \text{ m}.

Part (b): use the concavity. The graph is concave down, so each chord lies below the curve and every trapezium misses a sliver of area. The estimate is therefore less than the true distance (an under-estimate).

Part (c): average speed is total distance over total time.

\text{average speed} = \frac{\text{distance}}{\text{time}} = \frac{202.5}{12} = 16.875 \text{ m/s}.

Marker's note: one mark for recognising distance as the area under the $v$ - $t$ graph with $h = 3$ , one for the correctly weighted rule, one for the distance $202.5$ m; one for the under-estimate justified by concave-down chords lying below the curve; one for the average speed $16.875$ m/s from distance over time. Doubling the end readings $4$ and $27$ is the trap that inflates the bracket.

exam5 marksConsider

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

, which cannot be evaluated by elementary means. (a) Using the trapezoidal rule with four strips and the values

e^0 = 1

e^{-0.0625} = 0.9394

e^{-0.25} = 0.7788

e^{-0.5625} = 0.5698

and

e^{-1} = 0.3679

, show that

\int_0^1 e^{-x^2}\,dx \approx 0.7430

. (b) Given that

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

has

f'(x) = (4x^2 - 2)e^{-x^2}

, determine the sign of

f'(x)

on the interval

0 \le x \le 0.5

and hence state whether the trapezoidal estimate over the whole of

[0, 1]

can be trusted to be a strict over-estimate. Explain.

Show worked solution →

Part (a): set up the ordinates. With $n = 4$ on $[0, 1]$ , $h = \frac{1 - 0}{4} = 0.25$ , so the $x$ -values are $0, 0.25, 0.5, 0.75, 1$ . Each ordinate is $e^{-x^2}$ :

y_0 = e^{0} = 1, \quad y_1 = e^{-0.0625} = 0.9394, \quad y_2 = e^{-0.25} = 0.7788,

y_3 = e^{-0.5625} = 0.5698, \quad y_4 = e^{-1} = 0.3679.

Apply the rule with the interior three doubled:

\int_0^1 e^{-x^2}\,dx \approx \frac{0.25}{2}\big[\,y_0 + 2(y_1 + y_2 + y_3) + y_4\,\big] = 0.125\big[\,1 + 2(0.9394 + 0.7788 + 0.5698) + 0.3679\,\big].

The interior sum is $0.9394 + 0.7788 + 0.5698 = 2.288$ , so

\int_0^1 e^{-x^2}\,dx \approx 0.125\big[\,1 + 4.576 + 0.3679\,\big] = 0.125 \times 5.9439 = 0.7430,

as required.

Part (b): test the sign of $f''(x)$ . With $f''(x) = (4x^2 - 2)e^{-x^2}$ , the exponential factor is always positive, so the sign of $f''(x)$ is the sign of $4x^2 - 2$ . On $0 \le x \le 0.5$ ,

4x^2 - 2 \le 4(0.25) - 2 = 1 - 2 = -1 < 0,

so $f''(x) < 0$ there: the curve is concave down on $[0, 0.5]$ .

Interpret for the full interval. Concavity is not constant across $[0, 1]$ : $f''(x) = 0$ when $4x^2 = 2$ , that is $x = \frac{1}{\sqrt{2}} \approx 0.707$ , and for $x > 0.707$ we have $4x^2 - 2 > 0$ , so the curve is concave up on $(0.707, 1]$ . Because the curve is concave down over most of the interval but concave up near the right end, the chords lie below the curve on part of the range and above it on another part. The over- or under-estimate contributions therefore work in opposite directions, so the trapezoidal estimate cannot be guaranteed to be a strict over-estimate (nor a strict under-estimate); the simple concavity test only settles the sign of the error when the concavity keeps one sign across the whole interval.

Marker's note: one mark for $h = 0.25$ and the ordinates, one for the correctly weighted rule reaching $0.7430$ ; one for showing $4x^2 - 2 < 0$ on $[0, 0.5]$ (concave down there), one for locating the change of concavity at $x = \frac{1}{\sqrt{2}}$ , one for the reasoned conclusion that a mixed concavity means the over/under test is inconclusive over all of $[0, 1]$ . Asserting "concave down so under-estimate" for the whole interval, without checking the concavity stays one sign, is the trap.