The Trapezoidal Rule
Not all integrals can be found exactly. The trapezoidal rule approximates the area under a curve by dividing it into trapezoids — like estimating the area of an irregular field by splitting it into straight-edged strips.
Practise this lesson
Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.
Before we start, what do you already know about approximating areas under curves?
There are only two moves in this lesson. Lock them in and the rest is arithmetic.
Move 1: Calculate $h$ and all $y$-values. Find the subinterval width $h = \frac{b-a}{n}$ and evaluate the function at each equally spaced point.
Move 2: Apply the formula. The first and last $y$-values get coefficient 1; all interior values get coefficient 2. Organise into a table first.
Key facts
- The trapezoidal rule formula and what each symbol means
- That $h = \frac{b-a}{n}$ and $n$ strips require $n+1$ function values
- When the rule overestimates vs underestimates
Concepts
- Why trapezoids approximate the area under a curve
- How more strips improve accuracy
- The link between concavity and the direction of error
Skills
- Apply the trapezoidal rule to approximate definite integrals
- Calculate function values at equally spaced points
- Assess whether the trapezoidal rule gives an overestimate or underestimate
The trapezoidal rule approximates the area under a curve by replacing the curve with straight line segments between points, forming trapezoids. The area of each trapezoid is the average of the two parallel sides times the width: $\frac{h}{2}(y_i + y_{i+1})$. Summing all trapezoid areas gives the approximation.
where $h = \dfrac{b-a}{n}$ and $y_i = f(x_i)$
More strips mean greater accuracy because the straight-line segments better approximate the curve. When the curve is concave up, the straight-line tops of the trapezoids lie above the curve — giving an overestimate. When concave down, the opposite occurs.
Trapezoidal rule: $\int_a^b f(x) \, dx \approx \frac{h}{2}[y_0 + 2(y_1 + \cdots + y_{n-1}) + y_n]$; $h = \frac{b-a}{n}$; $n$ strips require $n+1$ function values
Pause — copy the trapezoidal rule $\int_a^b f(x)\,dx \approx \dfrac{h}{2}[y_0 + 2(y_1 + \cdots + y_{n-1}) + y_n]$ where $h = \dfrac{b-a}{n}$ into your book.
Quick check: True or false — if you use 4 strips with the trapezoidal rule, you need exactly 4 function values.
Worked examples · 3 in a row, reveal as you go
Use the trapezoidal rule with 4 strips to approximate $\int_0^2 x^2 \, dx$.
Use the trapezoidal rule with 5 function values to approximate $\int_1^3 \frac{1}{x} \, dx$.
A curve has the following values. Approximate $\int_0^{1.2} y \, dx$ using the trapezoidal rule.
| $x$ | 0 | 0.3 | 0.6 | 0.9 | 1.2 |
|---|---|---|---|---|---|
| $y$ | 1.00 | 1.09 | 1.36 | 1.81 | 2.44 |
Quick check: Using 3 strips to approximate $\int_0^3 f(x) \, dx$, the value of $h$ is:
Common errors · the 3 traps that cost marks
Odd one out: Three of the following are true about the trapezoidal rule. Which one is FALSE?
Quick-fire practice · 5 problems
Use the trapezoidal rule with 4 strips to approximate $\int_0^4 x \, dx$.
Use 3 strips to approximate $\int_1^4 x^2 \, dx$.
A curve has values: $(0, 1)$, $(1, 3)$, $(2, 5)$, $(3, 7)$. Approximate $\int_0^3 y \, dx$.
Use 4 strips to approximate $\int_0^2 e^x \, dx$ (use $e^0 = 1$, $e^{0.5} \approx 1.65$, $e^1 \approx 2.72$, $e^{1.5} \approx 4.48$, $e^2 \approx 7.39$).
Explain whether the trapezoidal rule overestimates or underestimates $\int_0^1 x^3 \, dx$.
Fill the blanks: drag each token into the matching blank.
The subinterval width is ___ $= \frac{b-a}{n}$. Using $n$ strips requires ___ function values. Interior values have coefficient ___. A concave-up curve gives an ___.
Teach it back: Explain in your own words why interior $y$-values in the trapezoidal rule formula have a coefficient of 2 while the endpoint values have a coefficient of 1.
Earlier you were asked about approximating areas. The trapezoidal rule is a practical method when exact integration is difficult or impossible. It replaces curves with straight lines, trading exactness for computability. More strips give a better approximation because the straight-line segments more closely follow the actual curve.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Use the trapezoidal rule with 4 strips to approximate $\int_0^2 x^2 \, dx$. Give your answer to 2 decimal places. Show all working. 3 MARKS
Q2. The following table gives values of $y = f(x)$:
| $x$ | 0 | 0.5 | 1.0 | 1.5 | 2.0 |
|---|---|---|---|---|---|
| $y$ | 1.00 | 1.41 | 2.00 | 2.83 | 4.00 |
Use the trapezoidal rule to approximate $\int_0^2 f(x) \, dx$. 3 MARKS
Q3. Use the trapezoidal rule with 2 strips to approximate $\int_0^2 (x^3 - 3x) \, dx$. Determine whether your answer is an overestimate or underestimate, giving a reason. 4 MARKS
📖 Comprehensive answers (click to reveal)
Q1 (3 marks): $h = 0.5$ [0.5]. $y$-values: 0, 0.25, 1, 2.25, 4 [0.5]. $\approx \frac{0.5}{2}[0 + 2(0.25 + 1 + 2.25) + 4] = 0.25 \times 11 = 2.75$ [2].
Q2 (3 marks): $h = 0.5$ (4 strips from 5 values) [0.5]. $\approx \frac{0.5}{2}[1.00 + 2(1.41 + 2.00 + 2.83) + 4.00]$ [0.5]. $= 0.25[1 + 12.48 + 4] = 0.25 \times 17.48 = 4.37$ [2].
Q3 (4 marks): $h = 1$ [0.5]. Values at $x = 0, 1, 2$: $0, -2, 2$ [0.5]. $\approx \frac{1}{2}[0 + 2(-2) + 2] = \frac{1}{2}(-2) = -1$ [1]. $f''(x) = 6x > 0$ for $x > 0$ on $(0,2)$, so the curve is predominantly concave up [1]. When concave up, trapezoids lie above the curve, so the trapezoidal rule overestimates [1].
Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering trapezoidal rule questions. Lighter alternative to the boss.
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