The Trapezium Rule and Justifying an Over- or Underestimate

A curve is tabulated at nine equally spaced x-values, from $x = 2$ to $x = 6$, for use with the trapezium rule. What is $h$?

• $h = 0.5$

  Correct. The interval has length $6 - 2 = 4$, and nine points bound eight equal strips, so $h = \frac{4}{8} = 0.5$.

• B$h \approx 0.4444$, from dividing $4$ by the $9$ given x-values

  This divides by the number of points, not the number of gaps between them — the same error a real examiner report records candidates making on this exact step, with a different pair of numbers.

• C$h = 4$

  This is the whole interval length with no division at all — the width of the entire region, not of a single strip within it.

• D$h = 9$, the number of ordinates

  The number of ordinates plays no direct role in the value of $h$ at all beyond determining, once one is subtracted, how many strips there are to divide the interval length by.

Traps tested: Strip width from point count not gap count · Strip width not divided by strip count · Strip count confused with ordinate count

The trapezium rule is used with $n = 6$ strips, giving seven ordinates $y_0$ to $y_6$. Which ordinates are doubled inside the bracket of the compressed formula?

• $y_1, y_2, y_3, y_4, y_5$ — the five ordinates strictly between the two ends

  Correct. Six strips share five internal boundaries, and each of those five ordinates belongs to two neighbouring strips at once, so each is counted twice when the individual trapezium areas are added together.

• B$y_0$ and $y_6$ only

  These are the two endpoints, each belonging to a single strip — the reverse of which ordinates are actually doubled.

• CAll seven ordinates, $y_0$ through $y_6$

  Doubling the endpoints as well overcounts the two outer strips, treating $y_0$ and $y_6$ as if they too were shared between two strips.

• DOnly $y_3$, the ordinate exactly in the middle

  Every one of the five interior ordinates is shared between two adjacent strips, not just the one nearest the centre of the interval.

Traps tested: Endpoints and interior ordinates swapped · All ordinates doubled including endpoints · Only central ordinates doubled

Four ordinates $y_0 = 2$, $y_1 = 5$, $y_2 = 4$, $y_3 = 6$, $y_4 = 3$ are used with the trapezium rule, $h = 0.5$. What is the correct estimate?

• $8.75$

  Correct. $\frac{1}{2}(0.5)\{(2+3) + 2(5+4+6)\} = 0.25 \times \{5 + 30\} = 0.25 \times 35 = 8.75$.

• B$31.25$, from $\frac{1}{2}(0.5)(2+3) + 2(5+4+6)$

  This closes the bracket after the endpoints and adds the doubled interior sum afterwards, so the $\frac{1}{2}h$ scales only the endpoint pair and never touches the interior terms at all — exactly the bracket-closed-early error a real examiner report records as gaining no credit.

• C$6.25$, from doubling the endpoints instead of the interior ordinates

  $\frac{1}{2}(0.5)\{2(2+3) + (5+4+6)\} = 0.25 \times \{10+15\} = 6.25$ — the endpoints and interior ordinates have swapped roles: $y_0$ and $y_4$ are each used once, $y_1, y_2, y_3$ are each used twice, not the other way round.

• D$10.0$, from doubling every ordinate including the endpoints

  $\frac{1}{2}(0.5) \times 2(2+5+4+6+3) = 0.25 \times 40 = 10.0$ — this treats every ordinate, including the two endpoints, as shared between two strips, which only the interior ordinates actually are.

Traps tested: Bracket closed before scaling · Endpoints and interior ordinates swapped · All ordinates doubled including endpoints

The trapezium rule is used to approximate $\int_0^2 x^2\,dx$ with 4 strips. Is the result an over- or underestimate, and why?

• An overestimate, because $y = x^2$ bends away from the x-axis on $[0,2]$, so each chord lies above the curve and every trapezium covers more area than the true strip beneath the curve

  Correct. $y = x^2$ curves upward, away from the x-axis, across the whole interval, so the straight chords sit above the curve throughout: every trapezium area exceeds the true area of its strip. (Checked directly: the trapezium estimate here is $2.75$, and the true value is $\int_0^2 x^2\,dx = \frac{8}{3} \approx 2.6667$ — the estimate is indeed larger.)

• BAn underestimate, because straight-line approximations always fall short of the curve they are approximating

  There is no such general rule — whether the chord falls short of or exceeds the curve depends entirely on which way the curve bends across the strip. For a curve bending the other way, exactly as $y = \sqrt{2x+1}$ does in the worked chain earlier in this lesson, the same reasoning gives an underestimate instead.

• CAn overestimate, because $y = x^2$ is an increasing function on $[0,2]$

  Whether a function is increasing or decreasing over an interval is a separate question from which way it bends, and it is the bending — not the increasing or decreasing — that decides whether the chords sit above or below the curve. A decreasing function can bend either way, exactly as an increasing one can.

• DIt cannot be determined without first calculating $\int_0^2 x^2\,dx$ exactly and comparing it to the trapezium-rule value

  The direction can be read off the shape of the curve alone, before any exact value is calculated: $y = x^2$ bends away from the x-axis throughout the interval, which is enough on its own to predict that the chords sit above the curve and the estimate overestimates.

Traps tested: Straight line approximation assumed always under · Monotonicity confused with concavity · Over under direction assumed to require exact value

A trapezium-rule estimate for an area comes out as $1.40$; the true value, found separately, is $1.4353$. Which explanation would earn full credit for "state, with a reason, whether the estimate is an over- or underestimate"?

• Underestimate — the curve bends towards the x-axis over the interval used, so the chords joining consecutive ordinates lie below the curve, making the sum of trapezium areas less than the true area under the curve

  Correct: this names the mechanism (the direction the curve bends, and what that does to the chords) rather than only restating that the numbers differ in a particular direction — exactly the distinction a real examiner report draws for this mark.

• BUnderestimate, because $1.40$ is less than $1.4353$

  This restates the definition of underestimate using the two given numbers; it gives no reason why the trapezium sum came out smaller than the true area. An examiner report records this exact gap as making the reasoning mark the hardest single mark on the whole paper to earn.

• CUnderestimate, because straight lines are always shorter than the curve they replace

  Straight-line chords are not "always" shorter or below a curve — it depends on which way the curve bends, and the opposite bend produces chords above the curve and an overestimate instead.

• DCannot be justified without seeing the original curve, since the numbers alone do not show which is bigger

  The numbers given, $1.40 < 1.4353$, already show the estimate is smaller than the true value — what is missing for full credit is not more numerical information but the geometric reason (how the curve bends) that explains why.

Traps tested: Conclusion restated instead of reasoned · Straight line approximation assumed always under · Reasoning mark mistaken for missing data

The specification allows a question to ask for the same integral approximated with an increasing number of strips. What generally happens to the trapezium-rule estimate as $n$ increases (strips get narrower)?

• It generally gets closer to the true value, because each chord spans a shorter piece of the curve and a shorter piece of a smooth curve is straighter

  Correct, and it is exactly the effect the spec's own guidance names when it says increasing the number of trapezia may be required to improve accuracy: a chord across a narrower strip has less curve left to depart from, so the gap between the chord and the curve — and hence the gap between the trapezium sum and the true integral — generally shrinks as $n$ grows.

• BIt generally gets less accurate, because more strips means more individual roundings and more chances for arithmetic error to accumulate

  This confuses an increase in the arithmetic WORK (more terms to add, more chances for a slip in an individual calculation) with the accuracy of the underlying method, which improves with more, narrower strips regardless of how much arithmetic that takes.

• CIt stays exactly the same, since the total interval being approximated has not changed

  The total interval length is unchanged, but the SHAPE of the approximation is not — narrower strips mean each individual chord departs less from the curve it replaces, which does change the total error even though the interval itself is fixed.

• DIt becomes the exact value once $n$ reaches a fixed number such as 10, regardless of the curve

  There is no universal strip count at which the trapezium rule becomes exact for every curve — the estimate approaches the true value as $n$ grows, but for a genuinely curved function it remains an approximation, however small the remaining gap, for any finite $n$.

Traps tested: More strips assumed less accurate · Strip count increase assumed no effect · Exact at fixed strip threshold