9.4 The trapezium rule and limits on the estimate — Learn

Prerequisites: substituting values into a function from Section 2.9; definite integration and the area under a curve from Section 8.2; second derivatives and convexity from Section 7.4.

By the end of this section you can: use the trapezium rule to estimate a definite integral; decide from the curve whether the estimate is too large or too small; and use rectangles to give limits that the true area must lie between.

1. Where the rule comes from

Chapter 8 measured an area by splitting it into rectangles. The rectangles leave gaps or overshoot, because their tops are flat. The trapezium rule joins consecutive points with straight lines instead, so the top of each strip follows the curve more closely.

Divide the interval from $a$ to $b$ into $n$ equal strips. Each strip has width

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

Let the heights of the curve at the strip edges be $y_0, y_1, y_2, \ldots, y_n$ , where $y_i = f(x_i)$ . Each strip is a trapezium with parallel sides $y_{i-1}$ and $y_i$ , and width $h$ . Its area is

\frac{h}{2}\left(y_{i-1} + y_i\right).

Add all the strip areas. Every interior height appears in two strips, so it is counted twice. The two end heights appear once each. The trapezium rule is

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

The extra factor of $2$ in front of the interior heights is the only trap in the formula. Check it by counting: with $n$ strips there are $n+1$ heights, and the interior ones are $y_1$ to $y_{n-1}$ .

2. A first estimate

Example 1. Use the trapezium rule with four strips to estimate $\displaystyle\int_0^2 x^2\, dx$ , and compare it with the exact value.

Here $a = 0$ , $b = 2$ and $n = 4$ , so the width of each strip is

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

The strip edges are $0, 0.5, 1, 1.5, 2$ . The heights are the squares of those values:

$x$	$0$	$0.5$	$1$	$1.5$	$2$
$y = x^2$	$0$	$0.25$	$1$	$2.25$	$4$

Substitute into the rule:

\int_0^2 x^2\, dx \approx \frac{0.5}{2}\Bigl[0 + 4 + 2\left(0.25 + 1 + 2.25\right)\Bigr] = 0.25\left[4 + 7\right] = 2.75.

The exact value from Chapter 8 is

\int_0^2 x^2\, dx = \left[\frac{x^3}{3}\right]_0^2 = \frac{8}{3} = 2.6667.

The estimate $2.75$ is larger than the exact value by about $0.08$ . The estimate is an overestimate.

3. Deciding whether the estimate is too large or too small

The shape of the curve decides the direction of the error. Each straight top of a trapezium is a chord, and the chord sits on one side of the curve.

If the curve bends upwards, a chord lies above the curve, so the trapezium rule gives an overestimate.
If the curve bends downwards, a chord lies below the curve, so the rule gives an underestimate.

The second derivative tells you which way the curve bends. If $f''(x) > 0$ on the whole interval, the curve bends upwards. If $f''(x) < 0$ , it bends downwards.

For Example 1, $f(x) = x^2$ , so $f''(x) = 2 > 0$ . The curve bends upwards on $[0, 2]$ , so every trapezium sits above the curve. This confirms the overestimate without drawing the graph.

Example 2. Use the trapezium rule with four strips to estimate $\displaystyle\int_0^1 \frac{1}{1 + x}\, dx$ , and decide whether it is an overestimate or an underestimate.

Here $h = 0.25$ , and the strip edges are $0, 0.25, 0.5, 0.75, 1$ . Work out the heights:

y_0 = 1, \qquad y_1 = \frac{1}{1.25} = 0.8, \qquad y_2 = \frac{1}{1.5} = 0.6667, \qquad y_3 = \frac{1}{1.75} = 0.5714, \qquad y_4 = 0.5.

Substitute into the rule:

\frac{0.25}{2}\Bigl[1 + 0.5 + 2\left(0.8 + 0.6667 + 0.5714\right)\Bigr] = 0.125\left[1.5 + 2(2.0381)\right] = 0.125(5.5762) = 0.6970.

The exact value is $\ln 2 = 0.6931$ . The estimate $0.6970$ is too large.

To see why, differentiate twice. Write $f(x) = (1 + x)^{-1}$ , so $f'(x) = -(1 + x)^{-2}$ and $f''(x) = 2(1 + x)^{-3}$ . On $[0, 1]$ the value $f''(x)$ is positive, so the curve bends upwards. The chords lie above the curve, and the rule overestimates.

4. Limits the true area must lie between

An estimate on its own does not say how far wrong it is. Rectangle sums give limits. On each strip, build two rectangles: one using the smaller height, and one using the larger height.

The lower sum adds the smaller rectangle on every strip. It is too small, so it is a lower limit for the area.
The upper sum adds the larger rectangle on every strip. It is too large, so it is an upper limit.

The true area lies between the two sums. For a curve that increases across the interval, the lower sum uses the left-hand height and the upper sum uses the right-hand height.

Example 3. Find limits for $\displaystyle\int_0^2 x^2\, dx$ using four strips, and check that the exact value lies between them.

The curve $y = x^2$ increases on $[0, 2]$ , so the smaller height on each strip is the left-hand one. The lower sum is

0.5\left(0 + 0.25 + 1 + 2.25\right) = 0.5(3.5) = 1.75.

The larger height on each strip is the right-hand one, so the upper sum is

0.5\left(0.25 + 1 + 2.25 + 4\right) = 0.5(7.5) = 3.75.

The exact area satisfies

1.75 < \frac{8}{3} < 3.75, \qquad \text{that is} \qquad 1.75 < 2.6667 < 3.75.

The limits hold. Notice that the trapezium estimate is the average of the two rectangle sums:

\frac{1.75 + 3.75}{2} = 2.75.

That is not a coincidence. Each trapezium is the average of the two rectangles built on its strip.

5. Estimating from a table

Sometimes the readings arrive as a table, with no formula at all. The trapezium rule still works, because it only needs the heights.

Example 4. A river survey records the water depth at points across a $12$ metre channel. Estimate the cross-sectional area of the water.

Distance across (m)	$0$	$2$	$4$	$6$	$8$	$10$	$12$
Depth (m)	$0$	$1.2$	$2.5$	$3.1$	$2.4$	$0.9$	$0$

The readings are equally spaced, with width $h = 2$ . The two end heights are $0$ and $0$ . Sum the interior heights:

1.2 + 2.5 + 3.1 + 2.4 + 0.9 = 10.1.

Substitute into the rule:

A \approx \frac{2}{2}\Bigl[0 + 0 + 2(10.1)\Bigr] = 20.2.

The cross-sectional area is about $20.2$ square metres.

Now find limits. On each strip, take the smaller of its two readings and the larger of its two readings.

Strip	$0$ – $2$	$2$ – $4$	$4$ – $6$	$6$ – $8$	$8$ – $10$	$10$ – $12$
Smaller (m)	$0$	$1.2$	$2.5$	$2.4$	$0.9$	$0$
Larger (m)	$1.2$	$2.5$	$3.1$	$3.1$	$2.4$	$0.9$

The lower sum is $2(0 + 1.2 + 2.5 + 2.4 + 0.9 + 0) = 2(7.0) = 14.0$ .

The upper sum is $2(1.2 + 2.5 + 3.1 + 3.1 + 2.4 + 0.9) = 2(13.2) = 26.4$ .

So the true area lies between $14.0$ and $26.4$ square metres. The estimate $20.2$ sits inside those limits, as it must.

The limits are wide because the channel is deep in the middle and shallow at the edges, and six readings cannot pin down the shape. The answer is an estimate, not an exact value, because the survey only records the depth at six points. Between the readings the depth is unknown.

6. More strips give a better estimate

A smaller strip width follows the curve more closely, so the estimate improves. For $\displaystyle\int_0^2 x^2\, dx$ :

Number of strips	$h$	Estimate	Error
$4$	$0.5$	$2.75$	$0.0833$
$8$	$0.25$	$2.6875$	$0.0208$
$16$	$0.125$	$2.671875$	$0.0052$

Each time the strip width halves, the error falls to about a quarter of its previous value. The estimates approach $\dfrac{8}{3} = 2.6667$ from above, because the curve bends upwards everywhere on the interval.

Watch out. Three mistakes are common.

Using the whole width instead of the strip width. The formula needs $h = \dfrac{b - a}{n}$ , not $b - a$ . Re-check: count the strip edges. With $n$ strips there are $n + 1$ of them, and the gaps between them all equal $h$ .

Forgetting to double the interior heights. Only $y_0$ and $y_n$ are counted once. Every other height is counted once from the strip on its left and once from the strip on its right. Re-check: the bracket must contain $n + 1$ heights in total, counting the doubled ones twice.

Guessing the direction of the error. An estimate is not always too large. Re-check: look at the second derivative, or at the bend of the curve, before you say which way the error goes.

7. Try it: more trapezia, smaller error

Predict first. The figure shows a curve with trapezia drawn under it. Before you change anything, predict whether the estimate on the screen is larger or smaller than the true area. Write your prediction and one reason.

Test. Choose each curve and move the slider to change the number of strips. Watch the estimate, the true area and the message about the direction of the error.

Explain. Write two sentences. The first explains why increasing the number of strips improves the estimate on any curve. The second explains why one curve overestimates and another underestimates.

Progress check

Return to the chapter problem. The river survey gives depths $0, 1.2, 2.5, 3.1, 2.4, 0.9, 0$ metres at distances $0, 2, 4, 6, 8, 10, 12$ metres.

Estimate the cross-sectional area with the trapezium rule.
Find a lower limit and an upper limit for the area.
The water flows at about $0.8$ metres per second. Estimate the flow rate in cubic metres per second.

You should now be able to

state the trapezium rule and explain where it comes from;
estimate a definite integral from a formula or from a table;
use the second derivative to decide whether the estimate is too large or too small;
use rectangle sums to give lower and upper limits for an area;
explain why more strips give a better estimate;
explain why an estimate from a table is not an exact value.