Year 12 Maths Advanced (2027) Integral calculus

The Trapezoidal Rule

20 practice questions 0 video lessons Theory + worked examples

NSW · Year 12 Mathematics Advanced · Integral calculus

The trapezoidal rule estimates a definite integral by adding trapezium areas: \(\displaystyle\int_a^b f(x)\,dx\approx\dfrac{h}{2}\big(y_0+2y_1+\dots+2y_{n-1}+y_n\big)\).

Part of the NSW Year 12 Mathematics Advanced course, in the Calculus area of study (Integral calculus focus area) of the 2024 syllabus. Work through practice questions with fully worked solutions and video lessons, or scroll down for the theory summary and worked examples.

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Practice questions

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Theory

The trapezoidal rule estimates a definite integral using trapezoids. This Year 12 Mathematics Advanced topic (MAV-12-05) approximates areas when exact integration is hard or only data is given.

With \(n\) strips of width \(h=\dfrac{b-a}{n}\): \(\displaystyle\int_a^b f\,dx\approx\dfrac{h}{2}\big[y_0+2(y_1+\cdots+y_{n-1})+y_n\big]\).

The two end heights count once; every interior height counts twice. A single trapezoid is \(\dfrac{h}{2}(y_0+y_1)\).

Two trapezoids estimate the area under \(y=x^{2}\) on \([0,2]\).

\[\int_a^b f\,dx\approx\dfrac{h}{2}\big[y_0+2(y_1+\cdots+y_{n-1})+y_n\big],\ h=\dfrac{b-a}{n}\]

trapezoidal rule: h over 2 times y nought plus twice the interior values plus y n

Method

Find \(h=\dfrac{b-a}{n}\) and the heights \(y_0,\dots,y_n\).
Double the interior heights; add the two ends once.
Multiply the total by \(\dfrac{h}{2}\).

Example 1 — Two strips

Estimate \(\displaystyle\int_0^2 x^{2}\,dx\) with \(n=2\) (\(h=1\)).

Solution

\(\ \)	\(\approx\)	\(\tfrac12[0+2(1)+4]\)
	\(=\)	\(3\)

Example 2 — One trapezoid

Estimate \(\displaystyle\int_1^3 f\,dx\) with \(f(1)=2,\ f(3)=8\).

Solution

\(\ \)

\(\approx\)

\(\tfrac22[2+8]=10\)

Example 3 — From a table

Estimate \(\displaystyle\int_0^4 f\,dx\), \(h=2\), \(f(0)=1,f(2)=3,f(4)=5\).

Solution

\(\ \)

\(\approx\)

\(\tfrac22[1+2(3)+5]=12\)

Example 4 — Compare

The estimate for \(\displaystyle\int_0^2 x^{2}dx\) was \(3\). How does it compare?

Solution

\(\text{exact}\)	\(=\)	\(\tfrac83\approx2.67\)
\(\ \)		\(3>\tfrac83\ (\text{over-estimate})\)

Common pitfalls

Double the interior heights only. Ends count once.

Find \(h\) first: \(h=\dfrac{b-a}{n}\).

It is an estimate — upward-bending curves are over-estimated.

Frequently asked questions

What is the trapezoidal rule?

It estimates a definite integral by replacing the area with trapezoids: h over 2 times the first height, plus twice the interior heights, plus the last height.

When do you use the trapezoidal rule?

When a function is hard or impossible to integrate exactly, or when you only have a table of measured values.

Why are the interior values multiplied by 2?

Each interior height is shared by two adjacent trapezoids, so it is counted twice in the formula.

Does the trapezoidal rule over- or under-estimate?

For a curve that bends upward (concave up) it slightly over-estimates the true area.

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