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Numerical integration — the trapezium rule

Covers AS91579

When you need it

Some functions cannot be integrated with the methods in this standard — $e^{x^2}$ and $\sqrt{1 + x^3}$ among them.
Sometimes you have data points rather than a formula at all.
In both cases the area can still be estimated numerically, and the standard names "the rectangle or trapezium rule".

The idea

Divide the interval into equal strips.
Approximate each strip by a shape whose area is easy — a rectangle or a trapezium.
Add the areas.

The trapezium rule

$\int_a^b f(x)\,dx \approx \frac{h}{2}\left[y_0 + y_n + 2\left(y_1 + y_2 + \cdots + y_{n-1}\right)\right]$

Each part:
- $n$ — the number of strips
- $h = \dfrac{b-a}{n}$ — the width of each strip
- $y_0, y_1, \ldots, y_n$ — the ordinates, the function values at the strip boundaries
There are $n + 1$ ordinates for $n$ strips. Counting them wrong is the most common error.
The first and last ordinates are counted once; every interior one is doubled.

The method

Work out $h$ using $h = \frac{b-a}{n}$ .
List the $x$ -values: $a$ , $a+h$ , $a+2h$ , ..., up to $b$ .
Calculate each $y$ -value by substituting into the function. Keep full accuracy.
Substitute into the formula, keeping the first/last and the doubled interior values separate.
Evaluate and round at the end.
The report lists "did not recognise the value of $h$ in the Trapezium rule" among Not Achieved behaviours, so getting $h$ right is worth the care.

A tabulated layout

Setting the work out in a table makes it checkable and shows the marker your method:

$x$	$y$	Weight
$x_0 = a$	$y_0$	1
$x_1$	$y_1$	2
$x_2$	$y_2$	2
...	...	2
$x_n = b$	$y_n$	1

Whether the estimate is too big or too small

The trapezium rule replaces the curve with straight chords, so:
- Concave up ( $f'' > 0$ ) — the chords lie above the curve, so the estimate is an overestimate
- Concave down ( $f'' < 0$ ) — the chords lie below, so it is an underestimate
Note that this is a genuinely answerable exam question, and the answer comes from the second derivative or from a sketch — not from guessing.

Improving the estimate

More strips means a better estimate, because each chord hugs the curve more closely.
Doubling $n$ roughly quarters the error for the trapezium rule.

The rectangle rule

A cruder alternative, using rectangles instead of trapeziums:
- Left-endpoint rule: rectangle heights taken from the left of each strip
- Right-endpoint rule: heights taken from the right
- Midpoint rule: heights taken from the middle, which is usually the most accurate of the three
$\text{Area} \approx h\left(y_0 + y_1 + \cdots + y_{n-1}\right)$ for the left-endpoint rule.
The trapezium estimate is exactly the average of the left and right rectangle estimates.

Tips and Tricks

Count the ordinates, not the strips. Five strips means six $y$ -values. Writing the formula with the wrong number of terms is the classic error, and it is easy to catch by listing the $x$ -values first.

$h$ is the strip width, not the number of strips. $h = \frac{b-a}{n}$ . The report names failure to recognise $h$ as a Not Achieved behaviour.

Only the interior ordinates are doubled. The first and last appear once. Doubling all of them, or none, both give wrong answers of about the right size — which makes the error hard to spot without care.

Keep full accuracy in the $y$ -values and round only the final answer. Rounding each ordinate to 2 dp and then adding compounds the error across every strip.

If asked whether the estimate is an over- or underestimate, answer from the concavity — check the sign of $f''$ , or sketch the curve and one chord. Do not guess.

Radian mode for any trigonometric ordinates, and take care substituting: the report specifically warns about calculator errors "whilst evaluating and substituting radian values into trigonometric functions".

Worked ExampleEstimating an integral that cannot be found exactly

Use the trapezium rule with four strips to estimate $\displaystyle\int_0^2 e^{-x^2}dx$ , giving your answer to 4 decimal places. State, with a reason, whether your estimate is an overestimate or an underestimate.

Step 1 — Note why a numerical method is needed

$e^{-x^2}$ has no elementary anti-derivative — the reverse chain rule fails because the multiplier $-2x$ is not present and cannot be supplied. A numerical estimate is the only route.

Step 2 — Find the strip width

$h = \frac{b - a}{n} = \frac{2 - 0}{4} = 0.5$

Step 3 — List the $x$ -values

Four strips means five ordinates:

$x = 0,\; 0.5,\; 1,\; 1.5,\; 2$

Step 4 — Calculate each ordinate

Substituting into $y = e^{-x^2}$ , keeping full accuracy:

$x$	$x^2$	$y = e^{-x^2}$	Weight
$0$	$0$	$1.000000$	1
$0.5$	$0.25$	$0.778801$	2
$1$	$1$	$0.367879$	2
$1.5$	$2.25$	$0.105399$	2
$2$	$4$	$0.018316$	1

Step 5 — Apply the formula

$\int_0^2 e^{-x^2}dx \approx \frac{h}{2}\left[y_0 + y_4 + 2\left(y_1 + y_2 + y_3\right)\right]$

The ends, counted once:

$y_0 + y_4 = 1.000000 + 0.018316 = 1.018316$

The interior values, summed then doubled:

$y_1 + y_2 + y_3 = 0.778801 + 0.367879 + 0.105399 = 1.252079$

$2 \times 1.252079 = 2.504158$

Step 6 — Combine

$\approx \frac{0.5}{2}\left[1.018316 + 2.504158\right]$

$= 0.25 \times 3.522474$

$= 0.880619$

$\approx 0.8806$

Step 7 — Decide over- or underestimate

Find the second derivative to determine concavity. With $y = e^{-x^2}$ :

$\frac{dy}{dx} = -2xe^{-x^2}$

$\frac{d^2y}{dx^2} = -2e^{-x^2} + \left(-2x\right)\left(-2xe^{-x^2}\right) = e^{-x^2}\left(4x^2 - 2\right)$

Since $e^{-x^2} > 0$ always, the sign is that of $4x^2 - 2$ , which is zero at $x = \frac{1}{\sqrt{2}} \approx 0.707$ :

Concave down ( $f'' < 0$ ) for $0 < x < 0.707$ — the chords lie below the curve there
Concave up ( $f'' > 0$ ) for $0.707 < x < 2$ — the chords lie above the curve there

The concavity is not constant across the interval, so the two effects work in opposite directions.

Comparing with the true value, $\displaystyle\int_0^2 e^{-x^2}dx = 0.882081...$ , the estimate 0.8806 is a slight underestimate — the concave-down region near the origin, where the function values are largest, dominates.

Numerical integration — the trapezium rule

Covers AS91579

When you need it

Some functions cannot be integrated with the methods in this standard — $e^{x^2}$ and $\sqrt{1 + x^3}$ among them.
Sometimes you have data points rather than a formula at all.
In both cases the area can still be estimated numerically, and the standard names "the rectangle or trapezium rule".

The idea

Divide the interval into equal strips.
Approximate each strip by a shape whose area is easy — a rectangle or a trapezium.
Add the areas.

The trapezium rule

$\int_a^b f(x)\,dx \approx \frac{h}{2}\left[y_0 + y_n + 2\left(y_1 + y_2 + \cdots + y_{n-1}\right)\right]$

Each part:
- $n$ — the number of strips
- $h = \dfrac{b-a}{n}$ — the width of each strip
- $y_0, y_1, \ldots, y_n$ — the ordinates, the function values at the strip boundaries
There are $n + 1$ ordinates for $n$ strips. Counting them wrong is the most common error.
The first and last ordinates are counted once; every interior one is doubled.

The method

Work out $h$ using $h = \frac{b-a}{n}$ .
List the $x$ -values: $a$ , $a+h$ , $a+2h$ , ..., up to $b$ .
Calculate each $y$ -value by substituting into the function. Keep full accuracy.
Substitute into the formula, keeping the first/last and the doubled interior values separate.
Evaluate and round at the end.
The report lists "did not recognise the value of $h$ in the Trapezium rule" among Not Achieved behaviours, so getting $h$ right is worth the care.

A tabulated layout

Setting the work out in a table makes it checkable and shows the marker your method:

$x$	$y$	Weight
$x_0 = a$	$y_0$	1
$x_1$	$y_1$	2
$x_2$	$y_2$	2
...	...	2
$x_n = b$	$y_n$	1

Whether the estimate is too big or too small

The trapezium rule replaces the curve with straight chords, so:
- Concave up ( $f'' > 0$ ) — the chords lie above the curve, so the estimate is an overestimate
- Concave down ( $f'' < 0$ ) — the chords lie below, so it is an underestimate
Note that this is a genuinely answerable exam question, and the answer comes from the second derivative or from a sketch — not from guessing.

Improving the estimate

More strips means a better estimate, because each chord hugs the curve more closely.
Doubling $n$ roughly quarters the error for the trapezium rule.

The rectangle rule

A cruder alternative, using rectangles instead of trapeziums:
- Left-endpoint rule: rectangle heights taken from the left of each strip
- Right-endpoint rule: heights taken from the right
- Midpoint rule: heights taken from the middle, which is usually the most accurate of the three
$\text{Area} \approx h\left(y_0 + y_1 + \cdots + y_{n-1}\right)$ for the left-endpoint rule.
The trapezium estimate is exactly the average of the left and right rectangle estimates.

Tips and Tricks

$h$ is the strip width, not the number of strips. $h = \frac{b-a}{n}$ . The report names failure to recognise $h$ as a Not Achieved behaviour.

Keep full accuracy in the $y$ -values and round only the final answer. Rounding each ordinate to 2 dp and then adding compounds the error across every strip.

If asked whether the estimate is an over- or underestimate, answer from the concavity — check the sign of $f''$ , or sketch the curve and one chord. Do not guess.

Worked ExampleEstimating an integral that cannot be found exactly

Step 1 — Note why a numerical method is needed

$e^{-x^2}$ has no elementary anti-derivative — the reverse chain rule fails because the multiplier $-2x$ is not present and cannot be supplied. A numerical estimate is the only route.

Step 2 — Find the strip width

$h = \frac{b - a}{n} = \frac{2 - 0}{4} = 0.5$

Step 3 — List the $x$ -values

Four strips means five ordinates:

$x = 0,\; 0.5,\; 1,\; 1.5,\; 2$

Step 4 — Calculate each ordinate

Substituting into $y = e^{-x^2}$ , keeping full accuracy:

$x$	$x^2$	$y = e^{-x^2}$	Weight
$0$	$0$	$1.000000$	1
$0.5$	$0.25$	$0.778801$	2
$1$	$1$	$0.367879$	2
$1.5$	$2.25$	$0.105399$	2
$2$	$4$	$0.018316$	1

Step 5 — Apply the formula

$\int_0^2 e^{-x^2}dx \approx \frac{h}{2}\left[y_0 + y_4 + 2\left(y_1 + y_2 + y_3\right)\right]$

The ends, counted once:

$y_0 + y_4 = 1.000000 + 0.018316 = 1.018316$

The interior values, summed then doubled:

$y_1 + y_2 + y_3 = 0.778801 + 0.367879 + 0.105399 = 1.252079$

$2 \times 1.252079 = 2.504158$

Step 6 — Combine

$\approx \frac{0.5}{2}\left[1.018316 + 2.504158\right]$

$= 0.25 \times 3.522474$

$= 0.880619$

$\approx 0.8806$

Step 7 — Decide over- or underestimate

Find the second derivative to determine concavity. With $y = e^{-x^2}$ :

$\frac{dy}{dx} = -2xe^{-x^2}$

$\frac{d^2y}{dx^2} = -2e^{-x^2} + \left(-2x\right)\left(-2xe^{-x^2}\right) = e^{-x^2}\left(4x^2 - 2\right)$

Since $e^{-x^2} > 0$ always, the sign is that of $4x^2 - 2$ , which is zero at $x = \frac{1}{\sqrt{2}} \approx 0.707$ :

Concave down ( $f'' < 0$ ) for $0 < x < 0.707$ — the chords lie below the curve there
Concave up ( $f'' > 0$ ) for $0.707 < x < 2$ — the chords lie above the curve there

The concavity is not constant across the interval, so the two effects work in opposite directions.

Numerical integration — the trapezium rule

When you need it

The idea

The trapezium rule

The method

A tabulated layout

Whether the estimate is too big or too small

Improving the estimate

The rectangle rule

Step 1 — Note why a numerical method is needed

Step 2 — Find the strip width

Step 3 — List the xxx-values

Step 4 — Calculate each ordinate

Step 5 — Apply the formula

Step 6 — Combine

Step 7 — Decide over- or underestimate

Numerical integration — the trapezium rule

When you need it

The idea

The trapezium rule

The method

A tabulated layout

Whether the estimate is too big or too small

Improving the estimate

The rectangle rule

Step 1 — Note why a numerical method is needed

Step 2 — Find the strip width

Step 3 — List the xxx-values

Step 4 — Calculate each ordinate

Step 5 — Apply the formula

Step 6 — Combine

Step 7 — Decide over- or underestimate

Step 3 — List the $x$ -values

Step 3 — List the $x$ -values