Trapezoidal rule (composite; error)

At a Glance

• Frequency: 3 sub-parts across 3 of 13 years (2014, 2024, 2025)
• Priority tier: T3
• Marks (count): 10 (1), 7.5 (1), 15 (1)
• Average solve time: ~9 min
• Difficulty mix: easy 2, medium 1
• Section: B | Dominant type: computation

Why This Chapter Matters

The trapezoidal rule is tested in two modes: straightforward application (2014, 2024) and theoretical derivation of the correction constant and error term (2025). The application questions are among the fastest 10-mark problems in the paper — tabulate $f(x_i)$, apply the formula, done. The 2025 derivation question is harder (15 marks) but follows a single pattern: expand both sides in Taylor series about $x_0$, match powers of $h$, and the surviving term gives the error. Both modes are mechanical once the template is known.

Minimum Theory

Composite trapezoidal rule. For $\int_a^b f(x)\,dx$ with $n$ equal subintervals of width $h=(b-a)/n$ and nodes $x_i=a+ih$:

$\int_a^b f(x)\,dx \approx \frac{h}{2}\Big[f(x_0) + 2f(x_1) + 2f(x_2) + \cdots + 2f(x_{n-1}) + f(x_n)\Big]$

Weights: 1 for the two endpoints; 2 for each interior node. The total number of function evaluations is $n+1$.

Truncation error. The composite trapezoidal rule has global error $O(h^2)$: specifically $E_T = -\dfrac{(b-a)h^2}{12}f''(\xi)$ for some $\xi\in(a,b)$. Halving $h$ reduces the error by a factor of 4.

Corrected trapezoidal rule (2025). Adding derivative correction terms $\tfrac{h^2}{12}(f'(a)-f'(b))$ raises accuracy to $O(h^4)$. This is derived by choosing the constant $p$ in $\tfrac{h}{2}(f_0+f_1)+ph^2(f_0'-f_1')$ to kill the $h^3$ Taylor residual, giving $p=\tfrac{1}{12}$.

Question Archetypes

trapezoidal-rule (2 question(s); 2014, 2024)

Recognition Cues

• “Use $n$ subintervals to integrate … using the trapezoidal rule.”
• “Integrate $f(x)$ from $x=a$ to $x=b$ using the trapezoidal rule with width $h=\ldots$”
• A specific $h$ or $n$ is given; the function is evaluated at the grid points.

Solution Template

1. Compute $h$ and list nodes. $h=(b-a)/n$; nodes $x_0,x_1,\ldots,x_n$.
2. Tabulate $f(x_i)$. Compute each value, keeping 6 decimal places.
3. Apply the formula. Sum = $f_0 + f_n + 2(f_1+f_2+\cdots+f_{n-1})$; multiply by $h/2$.
4. State the result. Optionally compare with the exact value to show $O(h^2)$ accuracy.

Worked Example

2014 Paper 2, 2014-P2-Q5c (10 marks)

Use five subintervals to integrate $\displaystyle\int_0^1\dfrac{dx}{1+x^2}$ using the trapezoidal rule.

$h=(1-0)/5=0.2$. Nodes: $0,\,0.2,\,0.4,\,0.6,\,0.8,\,1.0$.

$T = \frac{0.2}{2}\Big[1.000000 + 0.500000 + 2(0.961538+0.862069+0.735294+0.609756)\Big]$

$= 0.1\cdot[1.500000 + 2\times 3.168657] = 0.1\times 7.837314$

$\boxed{\int_0^1\frac{dx}{1+x^2}\approx 0.7837}$

(Exact: $\pi/4\approx0.7854$; error $\approx0.0017$, consistent with $O(h^2)=O(0.04)$.)

2024 Paper 2, 2024-P2-Q7bii (7.5 marks)

Integrate $f(x)=5x^3-3x^2+2x+1$ from $x=-2$ to $x=4$ using the trapezoidal rule with $h=1$.

Nodes: $-2,-1,0,1,2,3,4$ ($n=6$). Function values: $-55,-9,1,5,33,115,281$.

$T = \frac{1}{2}\Big[-55+281 + 2(-9+1+5+33+115)\Big] = \frac{1}{2}[226 + 2(145)] = \frac{516}{2}$

$\boxed{\int_{-2}^4 f(x)\,dx \approx 258}$

(Exact value 246; error 12, expected for a cubic with $h=1$.)

quadrature-derivation (1 question(s); 2025)

Recognition Cues

• “Find the constant $p$ such that $\int_{x_0}^{x_1}f\,dx = \tfrac{h}{2}(f_0+f_1)+ph^2(f_0'-f_1')$ achieves maximum accuracy.”
• “Find the error term. Deduce the composite rule.”
• Taylor expansion matching; the formula has a derivative correction term.

Solution Template

1. Expand both sides in powers of $h$ about $x_0$ using $f_1=f+f'h+\tfrac{f''}{2}h^2+\cdots$ and $f_0'=f'$, $f_1'=f'+f''h+\cdots$.
2. Compute the error $E =$ LHS $-$ RHS term by term.
3. Set the $h^3$ coefficient to zero to find $p$.
4. Identify the leading surviving error term (the first non-vanishing power of $h$).
5. Compose over $[a,b]$: sum $N$ panels; derivative correction telescopes to $f'(a)-f'(b)$.

Worked Example

2025 Paper 2, 2025-P2-Q8b (15 marks)

Find the constant $p$ and error term for $\int_{x_0}^{x_1}f\,dx = \tfrac{h}{2}(f_0+f_1)+ph^2(f_0'-f_1')$. Deduce the composite rule on $[a,b]$.

Taylor expansions about $x_0$ (writing $f^{(k)}=f^{(k)}(x_0)$):

$\text{LHS} = fh + \frac{f'h^2}{2} + \frac{f''h^3}{6} + \frac{f'''h^4}{24} + \frac{f^{(4)}h^5}{120}+\cdots$

$\frac{h}{2}(f_0+f_1) = fh + \frac{f'h^2}{2} + \frac{f''h^3}{4} + \frac{f'''h^4}{12} + \frac{f^{(4)}h^5}{48}+\cdots$

$ph^2(f_0'-f_1') = -pf''h^3 - \frac{p}{2}f'''h^4 - \frac{p}{6}f^{(4)}h^5 - \cdots$

Error $E=$ LHS $-$ RHS:

$E = \left(\frac{1}{6}-\frac{1}{4}+p\right)f''h^3 + \left(\frac{1}{24}-\frac{1}{12}+\frac{p}{2}\right)f'''h^4 + \cdots$

Set $h^3$ coefficient to zero: $\tfrac{1}{6}-\tfrac{1}{4}+p=0 \Rightarrow$

$\boxed{p = \frac{1}{12}}$

With $p=\tfrac{1}{12}$, the $h^4$ coefficient also vanishes. The first surviving term is:

$E = \frac{h^5}{720}f^{(4)}(\xi),\quad \xi\in(x_0,x_1)$

Composite rule on $[a,b]$ with $N$ panels ($h=(b-a)/N$):

$\int_a^b f\,dx = h\left[\frac{f_0}{2}+f_1+f_2+\cdots+f_{N-1}+\frac{f_N}{2}\right] + \frac{h^2}{12}\big(f'(a)-f'(b)\big) + \frac{(b-a)h^4}{720}f^{(4)}(\eta)$

Common Traps

• $n$ subintervals vs. $n+1$ points. With 5 subintervals there are 6 grid points ($x_0$ through $x_5$). Counting 5 points gives the wrong answer.
• Interior vs. endpoint weights. The two endpoints have weight 1; all interior points have weight 2. Adding an interior point with weight 1 (or vice versa) is a common slip.
• Taylor sign of $f_0'-f_1'$. The correction term is $ph^2(f_0'-f_1')$, not $(f_1'-f_0')$. The sign difference swaps the role of $p$.
• Telescoping in the composite formula. The derivative correction $\sum(f_i'-f_{i+1}')$ telescopes to $f'(a)-f'(b)$ — not $N$ times the single-panel correction.

Marks-Aware Writing

A 10-mark trapezoidal application answer must show: the step size $h$ computed, a table of $f(x_i)$ values, the formula with interior/endpoint weights identified, and the numerical result. A 15-mark derivation answer must show: the Taylor expansion of each side to at least $h^5$, the coefficient-matching step, the value of $p$ with clear algebra, the error term with its exponent identified, and the composite rule written out.

Practice Set

• 2014-P2-Q5c (10 m) — — Hint: $h=0.2$, 6 grid points; interior sum $\approx3.169$; exact value is $\pi/4$.
• 2024-P2-Q7b-ii (7.5 m) — — Hint: cubic function so trapezoidal is not exact; error $= 258-246=12$, consistent with $O(h^2)$ error bound.
• 2025-P2-Q8b (15 m) — — Hint: match $h^3$ coefficient to get $p=1/12$; the $h^4$ term also vanishes automatically; composite rule has a derivative end-correction term.