Mathematics 9709 · Pure Mathematics 2

The trapezium rule

When a function won't integrate by hand, estimate the area instead: slice it into vertical strips and cap each with a straight line, making a trapezium. Add the strip areas and you have a good approximation, no integration needed. It is the one tool examined only in Paper 2, and it is worth easy marks if you are tidy.

Heights at the strip boundaries, then $\tfrac{h}{2}\big[\text{first}+\text{last}+2(\text{all the middles})\big]$ . Ends once, middles twice.

The rule: ends once, middles twice

Split $[a,b]$ into $n$ strips of equal width $h=\dfrac{b-a}{n}$ . That gives $n+1$ heights $y_0,y_1,\dots,y_n$ (the textbook calls them ordinates), one at each boundary. Then:

$\int_a^b y\,dx\approx \tfrac{h}{2}\big[\,y_0+y_n+2(y_1+y_2+\cdots+y_{n-1})\,\big]$

The two ends count once; every middle height counts twice.

Each strip is capped by a straight chord (a trapezium); add their areas. Here the curve bends down, so the chords sit below it and the estimate falls a little short.

Notice that $n$ strips need $n+1$ heights: the number of ordinates is always one more than the number of strips. The width $h$ is the gap along the $x$ -axis, not a height.

Answer format: 2 decimal places

Almost every trapezium question ends with “give your answer correct to 2 decimal places.” Carry the heights at full accuracy through the working and only round the final number to 2 d.p. Round the heights early and you can lose the accuracy mark even with a perfect method.

Common mistake

This is an estimate, so the question almost always asks for 2 d.p. Do not leave a long decimal, and do not round to 3 s.f. by habit: read the instruction and give exactly 2 d.p.

Worked example: 4 strips

Worked example

Estimate ∫₀¹ 2ˣ dx using 4 strips

$h=\tfrac{1-0}{4}=0.25$ ; heights at $x=0,0.25,0.5,0.75,1$ are $1,\ 1.189,\ 1.414,\ 1.682,\ 2$ .

$\approx\tfrac{0.25}{2}\big[\,1+2+2(1.189+1.414+1.682)\,\big]=0.125\,[\,3+8.570\,]$

Answer

\approx 1.45

Ends $y_0=1$ and $y_4=2$ once; the three middles doubled. $2^x$ bends up ( $\cup$ ), so the chords sit above it and this slightly over-estimates.

Second example: watch the zero height

A real trap: a height can be 0, and a zero still uses up an ordinate slot. The curve below is $y=2\sqrt{4x-x^2}$ on $[0,4]$ , which starts and ends on the $x$ -axis.

Worked example

Estimate ∫₀⁴ 2√(4x − x²) dx using 4 strips

$h=\tfrac{4-0}{4}=1$ ; heights at $x=0,1,2,3,4$ are $0,\ 3.464,\ 4,\ 3.464,\ 0$ .

$\approx\tfrac{1}{2}\big[\,0+0+2(3.464+4+3.464)\,\big]=\tfrac12\,[\,0+21.856\,]$

Answer

\approx 10.93

The ends are $y_0=0$ and $y_4=0$ : they still go in the bracket, they just add nothing. This curve bends down ( $\cap$ ), so the chords lie below it and the answer under-estimates.

More strips, more accurate

The strips only miss the thin slivers between each chord and the curve. Halve the strip width (double $n$ ) and those slivers shrink fast:

Double the number of strips and the error drops to about a quarter of what it was. So 4 strips is roughly four times better than 2.

You can only use as many strips as the question gives you, but this tells you which way to lean: a question that wants accuracy will hand you more strips.

Over- or under-estimate? Look at the chords

The textbook's labels “convex” and “concave” are easy to confuse, so do not depend on the words. Use the picture instead. Sketch the curve and ask one thing: where do the chords sit?

Curve bends up ( $\cup$ ): the chords arch above the curve, so each strip includes a little too much. That is an over-estimate.
Curve bends down ( $\cap$ ): the chords sit below the curve, so each strip misses a sliver at the top. That is an under-estimate.

Common mistake

Don't guess from the formula. Sketch the curve, draw one chord, and see whether it is above or below. Chord above the curve, you overshoot; chord below, you fall short.

Where the marks go

Common mistake

Don't forget the ×2 on the middle heights (ends once, middles twice), and remember

n

strips need

n+1

heights.

Common mistake

A height of 0 still counts as an ordinate. Write it in its slot. Skip it and every later height shifts one place, so an end gets doubled or a middle gets counted once. Keep the slots lined up.

Common mistake

Only use this when asked. If a question can be integrated exactly, integrate it. The trapezium rule is for areas you cannot find algebraically, or when the paper explicitly says to use it.

Now you try

Estimate $\int_0^4 \sqrt{x}\,dx$ using 4 strips, give your answer to 2 decimal places, and say whether it over- or under-estimates. (9709-style)

Your turn— tap to reveal the worked answer (9709-style)

$h=1$ ; heights $0,\ 1,\ \sqrt2,\ \sqrt3,\ 2$ : $\tfrac12\big[\,0+2+2(1+1.414+1.732)\,\big]=\tfrac12[\,2+8.293\,]$ .

$\sqrt{x}$ bends down ( $\cap$ ), so the chords sit below the curve, which is an under-estimate. The first height is 0, but it still fills the $y_0$ slot.

Answer

\approx 5.15\ \text{(an under-estimate)}

PreviousDefinite integrals and the area under a curve NextLocating a Root: Sign Change and Sketching a Pair of Graphs

The trapezium rule

Heights at the strip boundaries, then $\tfrac{h}{2}\big[\text{first}+\text{last}+2(\text{all the middles})\big]$ . Ends once, middles twice.

The rule: ends once, middles twice

Split $[a,b]$ into $n$ strips of equal width $h=\dfrac{b-a}{n}$ . That gives $n+1$ heights $y_0,y_1,\dots,y_n$ (the textbook calls them ordinates), one at each boundary. Then:

$\int_a^b y\,dx\approx \tfrac{h}{2}\big[\,y_0+y_n+2(y_1+y_2+\cdots+y_{n-1})\,\big]$

The two ends count once; every middle height counts twice.

Each strip is capped by a straight chord (a trapezium); add their areas. Here the curve bends down, so the chords sit below it and the estimate falls a little short.

Notice that $n$ strips need $n+1$ heights: the number of ordinates is always one more than the number of strips. The width $h$ is the gap along the $x$ -axis, not a height.

Answer format: 2 decimal places

Common mistake

This is an estimate, so the question almost always asks for 2 d.p. Do not leave a long decimal, and do not round to 3 s.f. by habit: read the instruction and give exactly 2 d.p.

Worked example: 4 strips

Worked example

Estimate ∫₀¹ 2ˣ dx using 4 strips

$h=\tfrac{1-0}{4}=0.25$ ; heights at $x=0,0.25,0.5,0.75,1$ are $1,\ 1.189,\ 1.414,\ 1.682,\ 2$ .

$\approx\tfrac{0.25}{2}\big[\,1+2+2(1.189+1.414+1.682)\,\big]=0.125\,[\,3+8.570\,]$

Answer

\approx 1.45

Ends $y_0=1$ and $y_4=2$ once; the three middles doubled. $2^x$ bends up ( $\cup$ ), so the chords sit above it and this slightly over-estimates.

Second example: watch the zero height

A real trap: a height can be 0, and a zero still uses up an ordinate slot. The curve below is $y=2\sqrt{4x-x^2}$ on $[0,4]$ , which starts and ends on the $x$ -axis.

Worked example

Estimate ∫₀⁴ 2√(4x − x²) dx using 4 strips

$h=\tfrac{4-0}{4}=1$ ; heights at $x=0,1,2,3,4$ are $0,\ 3.464,\ 4,\ 3.464,\ 0$ .

$\approx\tfrac{1}{2}\big[\,0+0+2(3.464+4+3.464)\,\big]=\tfrac12\,[\,0+21.856\,]$

Answer

\approx 10.93

The ends are $y_0=0$ and $y_4=0$ : they still go in the bracket, they just add nothing. This curve bends down ( $\cap$ ), so the chords lie below it and the answer under-estimates.

More strips, more accurate

The strips only miss the thin slivers between each chord and the curve. Halve the strip width (double $n$ ) and those slivers shrink fast:

Double the number of strips and the error drops to about a quarter of what it was. So 4 strips is roughly four times better than 2.

You can only use as many strips as the question gives you, but this tells you which way to lean: a question that wants accuracy will hand you more strips.

Over- or under-estimate? Look at the chords

The textbook's labels “convex” and “concave” are easy to confuse, so do not depend on the words. Use the picture instead. Sketch the curve and ask one thing: where do the chords sit?

Curve bends up ( $\cup$ ): the chords arch above the curve, so each strip includes a little too much. That is an over-estimate.
Curve bends down ( $\cap$ ): the chords sit below the curve, so each strip misses a sliver at the top. That is an under-estimate.

Common mistake

Don't guess from the formula. Sketch the curve, draw one chord, and see whether it is above or below. Chord above the curve, you overshoot; chord below, you fall short.

Where the marks go

Common mistake

Don't forget the ×2 on the middle heights (ends once, middles twice), and remember

n

strips need

n+1

heights.

Common mistake

A height of 0 still counts as an ordinate. Write it in its slot. Skip it and every later height shifts one place, so an end gets doubled or a middle gets counted once. Keep the slots lined up.

Common mistake

Only use this when asked. If a question can be integrated exactly, integrate it. The trapezium rule is for areas you cannot find algebraically, or when the paper explicitly says to use it.

Now you try

Estimate $\int_0^4 \sqrt{x}\,dx$ using 4 strips, give your answer to 2 decimal places, and say whether it over- or under-estimates. (9709-style)

Your turn— tap to reveal the worked answer (9709-style)

$h=1$ ; heights $0,\ 1,\ \sqrt2,\ \sqrt3,\ 2$ : $\tfrac12\big[\,0+2+2(1+1.414+1.732)\,\big]=\tfrac12[\,2+8.293\,]$ .

$\sqrt{x}$ bends down ( $\cap$ ), so the chords sit below the curve, which is an under-estimate. The first height is 0, but it still fills the $y_0$ slot.

Answer

\approx 5.15\ \text{(an under-estimate)}