The trapezium rule: numerical integration and area estimates

The trapezium rule estimates a definite integral by replacing a curve with straight line segments. The area below each segment is a trapezium, so the method is a form of numerical integration.

For $n$ equal strips of width

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

the rule is

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

where $y_r=f(a+rh)$. The first and last ordinates occur once; every interior ordinate occurs twice.

Before you begin

You should be able to:

• evaluate a function accurately with a calculator;
• interpret a definite integral as signed area;
• find the area of a trapezium;
• distinguish the number of strips from the number of ordinates.

If exact integration is available, it is usually preferable. The trapezium rule is needed when an antiderivative is unavailable, awkward, or when the function is known only through measured data.

From one trapezium to the formula

For one strip from $x_r$ to $x_{r+1}$, let the parallel vertical sides have lengths $y_r$ and $y_{r+1}$ and let the horizontal width be $h$. Its area is

$$A_r=\frac12(y_r+y_{r+1})h.$$

For four strips, adding the four trapezia gives

$$\begin{aligned} A&\approx\frac h2\big[(y_0+y_1)+(y_1+y_2)+(y_2+y_3)+(y_3+y_4)\big]\\ &=\frac h2\big[y_0+2y_1+2y_2+2y_3+y_4\big]. \end{aligned}$$

This explains the coefficient pattern

$$\boxed{1,2,2,\ldots,2,1}.$$

It is safer to understand this pattern than to memorise an unexplained formula.

A reliable method

1. Calculate the strip width $h=(b-a)/n$.
2. List all $n+1$ values of $x$, including both endpoints.
3. Calculate the ordinates $y_r=f(x_r)$, keeping extra digits.
4. Apply the coefficient pattern $1,2,2,\ldots,2,1$.
5. Multiply the completed bracket by $h/2$.
6. Round only the final answer, as instructed.

Worked example 1: a function over four strips

Use four strips to estimate

$$\int_0^2 x^2\,dx.$$

There are $n=4$ strips, so

$$h=\frac{2-0}{4}=0.5.$$

Four strips require five ordinates:

$r$	$0$	$1$	$2$	$3$	$4$
$x_r$	$0$	$0.5$	$1$	$1.5$	$2$
$y_r=x_r^2$	$0$	$0.25$	$1$	$2.25$	$4$

Substitute carefully:

$$\begin{aligned} \int_0^2x^2\,dx &\approx\frac{0.5}{2}\left[0+2(0.25+1+2.25)+4\right]\\ &=0.25[0+7+4]\\ &=\boxed{2.75}. \end{aligned}$$

The exact value is $8/3=2.666\ldots$, so this estimate is slightly too large. The reason is geometric, not accidental: $y=x^2$ curves upwards, and the joining chords lie above the curve.

Self-check 1

Use three strips to estimate

$$\int_1^4\sqrt{x}\,dx.$$

Give the answer to three decimal places.

Answer

Here $h=(4-1)/3=1$, and the ordinates are

$$1,\quad \sqrt2,\quad \sqrt3,\quad 2.$$

Therefore

$$\begin{aligned} \int_1^4\sqrt{x}\,dx &\approx\frac12\left[1+2(\sqrt2+\sqrt3)+2\right]\\ &=4.646\ldots\\ &=\boxed{4.646}. \end{aligned}$$

Three strips require four ordinates.

Using values from a table

Numerical integration is particularly valuable for experimental data because there may be no formula to integrate. First check that the $x$ values are equally spaced.

Worked example 2: tabulated values

The table gives values of $y=1/x$. Use the trapezium rule with four strips to estimate the area from $x=1$ to $x=2$.

$x$	$1.00$	$1.25$	$1.50$	$1.75$	$2.00$
$y$	$1.0000$	$0.8000$	$0.6667$	$0.5714$	$0.5000$

The common spacing is $h=0.25$. Hence

$$\begin{aligned} \int_1^2\frac1x\,dx &\approx\frac{0.25}{2} \left[1.0000+2(0.8000+0.6667+0.5714)+0.5000\right]\\ &=0.125(5.5762)\\ &=\boxed{0.697025}. \end{aligned}$$

Using the rounded table values limits the precision of the estimate. It would be misleading to claim many reliable decimal places.

Worked example 3: find a missing ordinate

The ordinates $3$, $5$, $p$, and $4$ are equally spaced by $2$ units. The trapezium rule gives an estimated area of $25$. Find $p$.

There are four ordinates, so there are three strips. Apply the coefficients $1,2,2,1$:

$$25=\frac22[3+2(5)+2p+4].$$

Thus

$$25=17+2p,$$

so

$$\boxed{p=4}.$$

Do not interpret four ordinates as four strips. The intervals between consecutive ordinates are the strips.

Unequal strip widths

The compact formula assumes one common width $h$. If the $x$ values are not equally spaced, calculate each trapezium separately:

$$\boxed{ \int_a^b f(x)\,dx \approx \sum_{r=0}^{n-1}\frac{x_{r+1}-x_r}{2}(y_r+y_{r+1}) }.$$

Worked example 4: unequal intervals

Estimate the integral represented by this table.

$x$	$0$	$1$	$3$	$4$
$y$	$2$	$5$	$4$	$8$

The widths are $1$, $2$, and $1$, so a single value of $h$ cannot be used:

$$\begin{aligned} I&\approx\frac12(2+5) +\frac22(5+4) +\frac12(4+8)\\ &=3.5+9+6\\ &=\boxed{18.5}. \end{aligned}$$

Overestimates and underestimates

The rule replaces each piece of curve by a chord joining its endpoints.

• If $f''(x)>0$, the graph is convex, or curves upwards. Chords lie above the graph, so the rule gives an overestimate.
• If $f''(x)<0$, the graph is concave, or curves downwards. Chords lie below the graph, so the rule gives an underestimate.
• If $f$ is linear, the chords are the graph itself and the rule is exact.
• If concavity changes within the interval, errors from different strips can have different signs. A single global conclusion may not be possible from $f''$ alone.

This discussion concerns the signed integral. If the graph crosses the $x$ axis, negative ordinates produce negative trapezium contributions. To estimate geometric area, split the interval at each crossing and make each contribution positive.

Worked example 5: predict the direction of error

Without finding the exact integral, decide whether the trapezium rule overestimates or underestimates

$$\int_0^1 e^{-x}\,dx.$$

For $f(x)=e^{-x}$,

$$f''(x)=e^{-x}>0.$$

The graph curves upwards throughout $[0,1]$, so every chord lies above the curve. The trapezium rule gives an

$$\boxed{\text{overestimate}.}$$

Notice that a decreasing function can still curve upwards. Increasing or decreasing behaviour does not determine the direction of trapezium rule error.

Self-check 2

State whether the trapezium rule gives an overestimate, an underestimate, or an exact value in each case.

1. $\displaystyle \int_0^2(3x+1)\,dx$
2. $\displaystyle \int_0^{\pi/2}\sin x\,dx$
3. $\displaystyle \int_1^3\ln x\,dx$

Answers

1. Exact. The graph is a straight line, so every trapezium matches the region exactly.
2. Underestimate. On this interval, $f''(x)=-\sin x\leq0$, so the graph curves downwards.
3. Underestimate. Since $f''(x)=-1/x^2<0$, the graph curves downwards.

Improving accuracy

For a smooth curve, using more equal strips makes each chord follow the curve more closely. Since

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

increasing $n$ decreases $h$. The trapezium rule converges to the exact integral as $h$ tends to zero under the usual smoothness conditions.

For sufficiently smooth $f$, the total error is proportional to $h^2$ in leading order. Consequently, halving $h$ usually reduces the error by a factor of about $4$. This is a guide, not a justification for inventing precision when the original data are rounded or noisy.

Worked example 6: compare refinements

Estimate $\int_0^2x^2\,dx$ with two strips and four strips.

With two strips, $h=1$:

$$T_2=\frac12[0+2(1)+4]=3.$$

With four strips, Example 1 gave

$$T_4=2.75.$$

The exact value is $8/3$, so the absolute errors are

$$\left|3-\frac83\right|=\frac13$$

and

$$\left|2.75-\frac83\right|=\frac1{12}.$$

Halving the width has divided the error by exactly $4$ in this quadratic example.

Common misconceptions

Confusing strips and ordinates

$n$ strips require $n+1$ ordinates. The endpoints are both included.

Doubling the endpoints

Only interior ordinates have coefficient $2$. The endpoint coefficients are $1$.

Forgetting the factor $h/2$

The bracket is a weighted sum of heights, not an area. Multiplication by $h/2$ supplies the horizontal scale.

Rounding each ordinate too early

Store full calculator values and round once at the end. If a question supplies a rounded table, use those given values.

Assuming more strips always means a better real world answer

More strips reduce the discretisation error for a smooth known function. They cannot remove measurement error or restore information absent from sparse data.

Calling every result an area

A definite integral is signed. Below the $x$ axis, the ordinate and the trapezium contribution are negative.

Final self-check

1. Use four strips to estimate $\displaystyle \int_0^1\frac{1}{1+x^2}\,dx$. Give your answer to four decimal places.
2. Explain, without evaluating the exact integral, whether your answer to question 1 is necessarily an overestimate or underestimate over the whole interval.
3. Five ordinates are used between $x=2$ and $x=5$. If they are equally spaced, find $n$ and $h$.

Answers

1. Here $h=0.25$ and

$$\begin{array}{c|ccccc} x&0&0.25&0.5&0.75&1\\ \hline \dfrac1{1+x^2}&1&0.941176\ldots&0.8&0.64&0.5 \end{array}$$

so

$$\begin{aligned} T_4 &=\frac{0.25}{2}[1+2(0.941176\ldots+0.8+0.64)+0.5]\\ &=0.782794\ldots\\ &=\boxed{0.7828}. \end{aligned}$$

2. For $f(x)=(1+x^2)^{-1}$,

$$f''(x)=\frac{6x^2-2}{(1+x^2)^3}.$$

This changes sign at $x=1/\sqrt3$, so the graph changes concavity within $[0,1]$. The simple concavity test does not determine one error direction for the whole interval.

3. Five ordinates form four strips, so $n=4$ and

$$h=\frac{5-2}{4}=\boxed{0.75}.$$

What to learn next

• Revisit definite integrals and areas to connect the numerical estimate with exact integration and signed area.
• Study integration as a limit to understand why increasingly fine sums approach a definite integral.
• Continue with fixed point iteration and the Newton-Raphson method for numerical solutions of equations.
• Learn when iteration fails to develop error awareness across numerical methods.