The trapezium rule | CAIE 9709 Pure Mathematics 2 Notes

Chapters

Pure Mathematics 2

Mathematics9709

Build the formula from equal strips

Five strips need six heights

×1

×2

×1

h/2 [ends + 2(interior heights)]

Five strips need six heights: the ends count once and the middles count twice.

A strip is one narrow interval of width $h$ . Its vertical edge is an ordinate, so $y_i=f(x_i)$ is the height at $x_i$ . The area of one trapezium is $\frac h2(\text{left height}+\text{right height})$ .

h=\frac{b-a}{n}

I=\int_a^b f(x)\,dx

\begin{aligned}I&\approx\frac h2\left[y_0+y_n\right.\\&\left.\quad+2(y_1+\cdots+y_{n-1})\right]\end{aligned}

Here $I$ is the definite integral being estimated.

$n$ strips require $n+1$ heights.
The first and last heights count once; every middle height counts twice.
The heights must be equally spaced in $x$ .

Key idea

Write the x-values and heights in a table before using the formula. This exposes a missing endpoint or a wrongly doubled height.

Connect the approximation to a current paper

A calculator is the quickest reliable way to generate the heights. Work out $h$ , list the equally spaced x-values, then use the calculator table or function mode for $f(x)$ . Keep the stored values unrounded and round only the final answer.

Worked example

Two strips, then an exact comparison

In 9709/22/O/N/24 Q6, region A lies under $y=(5x^2+7)^{1/3}$ from $x=0$ to $x=2$ . Two strips give $h=1$ and the heights $\sqrt[3]{7},\sqrt[3]{12},3$ .

A\approx\frac12\left[y_0+y_2+2y_1\right]

=\frac12\left[\sqrt[3]{7}+3+2\sqrt[3]{12}\right]

=4.75\quad(3\text{ s.f.})

The exact total of regions A and B comes from the other curve:

\int_0^2\frac{27}{2x+5}\,dx=\frac{27}{2}\ln\frac95

Hence $B\approx\frac{27}{2}\ln(9/5)-4.75=3.19$ . (9709/22/O/N/24 Q6)

Examiner note

Store unrounded heights in the calculator and round only the requested final answer. The paper may ask for significant figures rather than decimal places.

Common mistake

Two strips need three ordinates. The words strips and ordinates do not name the same count.

Judge over or under from the chord

One chord shows the error direction

chord above

over-estimate

chord below

under-estimate

If the bend changes, one verdict may not be possible.

Judge the estimate by where each straight chord sits relative to the curve.

Sketch the curve over the required interval and draw one straight chord between neighbouring points.
A chord above the curve adds too much area: over-estimate.
A chord below the curve misses area: under-estimate.

Common mistake

Do not rely on the words concave and convex if you may reverse them. The chord position gives the reason the mark scheme wants.

If the curve changes its bending direction inside the interval, some chords may lie above and others below. A sketch alone may then be insufficient to decide the overall direction. The rule is exact for a straight-line graph because every chord is the graph itself.

Number of equal strips2

Strip width, h: 0.5
Trapezium estimate: 1.75393
Error: +0.03565

The chords stay above this upward-bending curve, so every shown value is an over-estimate. More strips make the chords follow the curve more closely.

Move the slider. The exact value is e − 1 ≈ 1.71828; the estimate approaches it from above.

Your turnIndependent check [5 marks]

Use the trapezium rule with two strips to estimate $\displaystyle\int_0^1 e^x\,dx$ . Give 3 significant figures and state whether the result is an over- or under-estimate.

Show worked answer

$h=0.5$ and the heights are $1,\sqrt e,e$ .

T\approx\frac{0.5}{2}\left[1+e+2\sqrt e\right]

T=1.75\quad(3\text{ s.f.})

The graph of $e^x$ bends upward, so its chords lie above the curve.

Answer

1.75

over-estimate

Build the formula from equal strips

Five strips need six heights

×1

×2

×1

h/2 [ends + 2(interior heights)]

Five strips need six heights: the ends count once and the middles count twice.

h=\frac{b-a}{n}

I=\int_a^b f(x)\,dx

\begin{aligned}I&\approx\frac h2\left[y_0+y_n\right.\\&\left.\quad+2(y_1+\cdots+y_{n-1})\right]\end{aligned}

Here $I$ is the definite integral being estimated.

$n$ strips require $n+1$ heights.
The first and last heights count once; every middle height counts twice.
The heights must be equally spaced in $x$ .

Key idea

Write the x-values and heights in a table before using the formula. This exposes a missing endpoint or a wrongly doubled height.

Connect the approximation to a current paper

Worked example

Two strips, then an exact comparison

In 9709/22/O/N/24 Q6, region A lies under $y=(5x^2+7)^{1/3}$ from $x=0$ to $x=2$ . Two strips give $h=1$ and the heights $\sqrt[3]{7},\sqrt[3]{12},3$ .

A\approx\frac12\left[y_0+y_2+2y_1\right]

=\frac12\left[\sqrt[3]{7}+3+2\sqrt[3]{12}\right]

=4.75\quad(3\text{ s.f.})

The exact total of regions A and B comes from the other curve:

\int_0^2\frac{27}{2x+5}\,dx=\frac{27}{2}\ln\frac95

Hence $B\approx\frac{27}{2}\ln(9/5)-4.75=3.19$ . (9709/22/O/N/24 Q6)

Examiner note

Store unrounded heights in the calculator and round only the requested final answer. The paper may ask for significant figures rather than decimal places.

Common mistake

Two strips need three ordinates. The words strips and ordinates do not name the same count.

Judge over or under from the chord

One chord shows the error direction

chord above

over-estimate

chord below

under-estimate

If the bend changes, one verdict may not be possible.

Judge the estimate by where each straight chord sits relative to the curve.

Sketch the curve over the required interval and draw one straight chord between neighbouring points.
A chord above the curve adds too much area: over-estimate.
A chord below the curve misses area: under-estimate.

Common mistake

Do not rely on the words concave and convex if you may reverse them. The chord position gives the reason the mark scheme wants.

Number of equal strips2

Strip width, h: 0.5
Trapezium estimate: 1.75393
Error: +0.03565

The chords stay above this upward-bending curve, so every shown value is an over-estimate. More strips make the chords follow the curve more closely.

Move the slider. The exact value is e − 1 ≈ 1.71828; the estimate approaches it from above.

Your turnIndependent check [5 marks]

Use the trapezium rule with two strips to estimate $\displaystyle\int_0^1 e^x\,dx$ . Give 3 significant figures and state whether the result is an over- or under-estimate.

Show worked answer

$h=0.5$ and the heights are $1,\sqrt e,e$ .

T\approx\frac{0.5}{2}\left[1+e+2\sqrt e\right]

T=1.75\quad(3\text{ s.f.})

The graph of $e^x$ bends upward, so its chords lie above the curve.

Answer

1.75

over-estimate