The Trapezium Rule

Estimate the area under a curve with the trapezium rule, say whether the estimate is too big or too small, and improve it with more strips.

In the exam for Edexcel International AS Maths · 17 real past-paper questions below

Area from straight lines

Some areas you cannot integrate. Try $\int_{0}^{2} x^{3} + 1 d x$ : none of your rules work on it. And an engineer measuring a river only has a list of depths, not an equation.

The fix: cut the area into strips of equal width. Join the tops with straight lines. Each strip becomes a trapezium, and you know how to find the area of a trapezium.

Example

One strip, width $h$ , with heights $y_{0}$ and $y_{1}$ at its two sides:

\frac{1}{2} (y_{0} + y_{1}) h

Add the strips: $\frac{1}{2} h {(y_{0} + y_{1}) + (y_{1} + y_{2}) + (y_{2} + y_{3}) + \dots}$ . Every middle height belongs to two strips, so it is counted twice. The first and last heights belong to one strip each.

Formula

\int_{a}^{b} y d x \approx \frac{1}{2} h {(y_{0} + y_{n}) + 2 (y_{1} + y_{2} + \dots + y_{n - 1})}

$h$ — the width of one strip: $h = \frac{b - a}{n}$ , where $n$ is the number of strips.
$y_{0}, y_{n}$ — the first and last y-values. Use them once.
$y_{1}, \dots, y_{n - 1}$ — all the y-values in the middle. Use them twice.

Memory trick

Ends once, middles twice, then times half a strip.

Rule

The rule is printed in the Pearson formula booklet (Mathematical Formulae and Statistical Tables), under Pure Mathematics P2, with $h = \frac{b - a}{n}$ . You still need to know how to use it fast.

Key idea

Babylonian astronomers used a similar idea over 2,000 years ago. Clay tablets show them finding how far Jupiter moved across the sky as the area of a trapezium under a speed–time graph.

Key idea

Straight lines in place of the curve. Strips become trapezia. Add them up.

Find h, then the x-values

Before any adding, set up the table. Two quick numbers do it.

Steps

1
1. Count the strips, $n$ . The question tells you, or you count gaps in the table.
2. Find $h = \frac{b - a}{n}$ .
3. List the x-values: start at $a$ and add $h$ each time until you reach $b$ .
4. Work out each y-value. Keep the accuracy the table uses.

Worked example

1
Complete a table for $y = x^{3} + 1$ from $x = 0$ to $x = 2$ , using 4 strips. Give y-values to 3 decimal places.
2
$h = \frac{2 - 0}{4} = 0.5$ — Width of one strip.
3
$x = 0, 0.5, 1, 1.5, 2$ — Start at 0, add 0.5 each time. 4 strips need 5 x-values.
4
$y_{1} = 0. 5^{3} + 1 = 1.125 = 1.061$ — Put each x into the equation.
5
$y = 1, 1.061, 1.414, 2.092, 3$ — The full row: $1$ , $1.125$ , $2$ , $4.375$ , $9$ .

Common mistake

Mixing up strips and values. 5 y-values make 4 strips, so $n = 4$ , not 5. There is always one more value than strips.

Common mistake

Putting a trig function in with the calculator in degrees. If the x-values are radians, switch to radians first.

Key idea

$h$ first, then the x-values, then the y-values. The rest is just adding.

Apply the rule

Now put the table into the formula. Write the whole bracket out: the examiner needs to see it.

Worked example

1
Use the trapezium rule with the table above to estimate $\int_{0}^{2} x^{3} + 1 d x$ , to 2 decimal places.
2
$\frac{1}{2} (0.5) {(1 + 3) + 2 (1.061 + 1.414 + 2.092)}$ — Ends once, middles twice. $h = 0.5$ .
3
$= 0.25 {4 + 2 (4.567)}$ — Add the ends, then add the middles.
4
$= 0.25 \times 13.134$ — $4 + 9.134 = 13.134$ .
5
$= 3.2835 \approx 3.28$ — To 2 decimal places.

Marks: B1 for $h$ , M1 for the right bracket shape with every y-value in it (none missing, none extra), and A1 for the answer. Adding four separate trapezia also earns the method mark.

Common mistake

Losing the big bracket: $\frac{1}{2} (0.5) \times 1 + 3 + 2 (\dots)$ multiplies only the 1. Write the curly bracket round everything.

Common mistake

Using the x-values in the bracket, or counting an end value twice. Only y-values go in, and each end appears once.

Some questions give y-values as logs and say "show that". Keep them exact and use the log laws.

Worked example

1
Using 3 strips, show that the trapezium rule gives $\int_{1}^{4} lo g_{10} x d x \approx lo g_{10} 12$ .
2
$h = 1, y = lo g 1, lo g 2, lo g 3, lo g 4$ — $h = \frac{4 - 1}{3} = 1$ . All logs are base 10.
3
$\frac{1}{2} (1) {lo g 1 + lo g 4 + 2 (lo g 2 + lo g 3)}$ — Ends once, middles twice.
4
$= \frac{1}{2} {lo g 4 + 2 lo g 6} = \frac{1}{2} lo g 144$ — $lo g 1 = 0$ . $lo g 2 + lo g 3 = lo g 6$ , and $lo g 4 + lo g 36 = lo g 144$ .
5
$= lo g 14 4^{\frac{1}{2}} = lo g_{10} 12$ — Half a log is the log of a square root.

Rule

"You must show all stages of your working" means a calculator answer alone scores nothing. In a "show that", write every line.

Key idea

Write $\frac{1}{2} h {\dots}$ with every y-value in place, then work it out.

Use your answer again

The next part often says "using your answer" to estimate a new integral. Do not use the trapezium rule again: it scores no marks. Instead, rewrite the new integral in terms of the old one.

Formula

\int_{a}^{b} (k y + g (x)) d x = k \int_{a}^{b} y d x + \int_{a}^{b} g (x) d x

$k$ — a number multiplying the curve. It multiplies your estimate.
$g (x)$ — an extra term you can integrate, such as a constant or $2 x$ . Integrate it exactly and add.

Worked example

1
Given $\int_{0}^{2} x^{3} + 1 d x \approx 3.28$ , estimate (i) $\int_{0}^{2} 3 x^{3} + 1 d x$ , (ii) $\int_{0}^{2} (x^{3} + 1 + 2 x) d x$ , (iii) $\int_{0}^{2} (x^{3} + 1 - 1) d x$ .
2
$3 \times 3.28 = 9.84$ — (i) Every height is 3 times bigger, so the area is too.
3
$3.28 + [x^{2}]_{0}^{2} = 3.28 + 4 = 7.28$ — (ii) Integrate $2 x$ exactly, then add.
4
$3.28 - 1 \times 2 = 1.28$ — (iii) Taking 1 off every height removes a rectangle 2 wide and 1 high.

Log questions use the log laws first, to bring back the original log.

Worked example

1
Given $\int_{1}^{4} lo g_{10} x d x \approx lo g_{10} 12$ , estimate $\int_{1}^{4} lo g_{10} (100 x^{3}) d x$ .
2
$lo g (100 x^{3}) = lo g 100 + 3 lo g x = 2 + 3 lo g x$ — Split with the log laws. $lo g_{10} 100 = 2$ .
3
$\int_{1}^{4} 2 d x = 2 \times 3 = 6$ — A rectangle: height 2, width 3.
4
$6 + 3 lo g_{10} 12 \approx 9.24$ — Add 3 times your estimate.

Example

Slides and mirrors keep the area. If $\int_{0}^{2} f (x) d x \approx 3.28$ , then $\int_{2}^{4} f (x - 2) d x \approx 3.28$ too: the same curve moved 2 to the right. A mirror in the y-axis, $\int_{- 2}^{0} f (- x) d x$ , is also 3.28.

Common mistake

Using the rule again on the new function. The mark scheme gives M0. The question wants your earlier answer used.

Key idea

Numbers multiply the estimate. Extra terms are integrated exactly and added. Never redo the rule.

Too big or too small?

The straight tops of the trapezia do not follow the curve exactly. Look at which way the curve bends.

Curve bends up, like a bowl

The straight tops sit above the curve. Each trapezium has a bit too much area.

Overestimate.

Curve bends down, like a hill

The straight tops sit below the curve. Each trapezium misses a sliver.

Underestimate.

$y = x^{3} + 1$ bends up between 0 and 2. So 3.28 should be too big, and it is: the true value is 3.241…

Rule

The mark needs a reason about the trapezia, such as "the tops of the trapezia lie above the curve". The words concave or convex alone, or just "it is bigger", do not earn it.

To improve the estimate, use more strips (narrower strips). The straight tops then hug the curve more closely.

Example

For $\int_{0}^{2} (x^{2} + 1) d x = \frac{14}{3} \approx 4.667$ : 2 strips give 5, 4 strips give 4.75, 8 strips give 4.6875. Percentage error with 4 strips: $\frac{4.75 - 4.667}{4.667} \times 100 \approx 1.8%$ .

Worked example

1
A river is 12 m wide. Its depth is 0, 1.4, 2.2, 1.9 and 0 m at 0, 3, 6, 9 and 12 m from one bank. Water flows at 0.8 m/s. Estimate the volume passing each minute.
2
$\frac{1}{2} (3) {(0 + 0) + 2 (1.4 + 2.2 + 1.9)} = 16.5 m^{2}$ — Cross-section area. $h = 3$ .
3
$16.5 \times 0.8 \times 60 = 792 m^{3}$ — Area times speed gives volume per second. Times 60 for a minute.
4
$Underestimate$ — Likely too small: the depths rise then fall like a hill, so the straight lines cut inside the riverbed and miss slivers of water.

Key idea

Bowl: too big. Hill: too small. More strips: closer to the truth.

Which method? Read the question

The wording tells you what to do.

You see	You do
"Complete the table"	Put each x into the equation; match the decimal places given
"Use the trapezium rule with all the values"	$h = \frac{b - a}{n}$ , then $\frac{1}{2} h {ends + 2 (middles)}$
"Show that … ≈ log k"	Keep the logs exact; combine with the log laws
"Using your answer", with $k y$	Multiply your estimate by $k$
"Using your answer", with an extra term	Integrate the extra term exactly; add or subtract it
Area between the curve and another graph	Your estimate minus the exact area under the other graph
"Overestimate or underestimate?"	Bends up: over. Bends down: under. Give the reason about the trapezia
"How could the estimate be improved?"	Use more strips of smaller width
A real context (river, energy)	Area first, then use the units the question gives

Common mistake

Rounding the y-values more than the table does. Use the values as given, and round only the final answer.

Key idea

Find $h$ , ends once, middles twice. Then reuse that answer, never the rule.

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