9.4 The trapezium rule and limits on the estimate — Practise with help

Work each question closed book first. Write down the strip width, then list the heights in order before you substitute into the rule.

Warm-up: remember

Work out $\displaystyle\int_0^2 3x^2\, dx$ .
For $f(x) = x^2$ , work out $f(1)$ , $f(2)$ and the average $\dfrac{f(1) + f(2)}{2}$ .

Answers

$\left[x^3\right]_0^2 = 8 - 0 = 8$ .
$f(1) = 1$ and $f(2) = 4$ , so the average is $\dfrac{1 + 4}{2} = 2.5$ .

If the first part was hard, revise definite integration in Section 8.2. If the second was hard, revise substituting into a function in Section 2.9.

Practise with help

1. Use the trapezium rule with four strips to estimate $\displaystyle\int_0^2 x^2\, dx$ . Write down $h$ and the heights.

Hint

The strip width is $h = \dfrac{b - a}{n}$ . There are five heights for four strips.

Next step

$h = 0.5$ , and the heights are $0, 0.25, 1, 2.25, 4$ at $x = 0, 0.5, 1, 1.5, 2$ .

Full working

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

2. Find the exact value of $\displaystyle\int_0^2 x^2\, dx$ , and state whether the estimate in question 1 is too large or too small.

Hint

Integrate $x^2$ from Chapter 8, then compare it with $2.75$ .

Next step

The exact value is $\left[\dfrac{x^3}{3}\right]_0^2 = \dfrac{8}{3} = 2.6667$ .

Full working

The exact value is $\dfrac{8}{3} = 2.6667$ , and the estimate is $2.75$ . The estimate is too large. Also $f''(x) = 2 > 0$ , so the curve bends upwards and every chord lies above the curve.

3. Use four strips and rectangles to find a lower limit and an upper limit for $\displaystyle\int_0^2 x^2\, dx$ .

Hint

The curve increases across the interval, so the smaller height on each strip is on the left.

Next step

The left-hand heights are $0, 0.25, 1, 2.25$ , so their rectangle sum is $0.5(3.5)$ .

Full working

Lower sum, using the left-hand heights:

0.5(0 + 0.25 + 1 + 2.25) = 1.75.

Upper sum, using the right-hand heights:

0.5(0.25 + 1 + 2.25 + 4) = 3.75.

So $1.75 < \dfrac{8}{3} < 3.75$ . The exact value $2.6667$ lies between the two limits, as it must.

4. Use four strips to estimate $\displaystyle\int_0^1 \frac{1}{1 + x}\, dx$ . Decide whether the estimate is too large or too small.

Hint

With $a = 0$ , $b = 1$ and four strips, the edges are $0, 0.25, 0.5, 0.75$ and $1$ .

Next step

The heights are $1, 0.8, 0.6667, 0.5714, 0.5$ , and the interior heights add to $2.0381$ .

Full working

\frac{0.25}{2}\Bigl[1 + 0.5 + 2(0.8 + 0.6667 + 0.5714)\Bigr] = 0.125(1.5 + 4.0762) = 0.6970.

The exact value is $\ln 2 = 0.6931$ , so the estimate is too large. Also $f''(x) = \dfrac{2}{(1 + x)^3} > 0$ on the interval, so the curve bends upwards.

5. A car's speed is recorded every second. Estimate the distance travelled.

Time (s)	$0$	$1$	$2$	$3$	$4$
Speed (m/s)	$0$	$3$	$5$	$6$	$6$

Hint

The distance is the area under the speed–time graph. Here $h = 1$ .

Next step

The interior heights are $3, 5, 6$ , so they add to $14$ and twice that is $28$ .

Full working

s \approx \frac{1}{2}\Bigl[0 + 6 + 2(3 + 5 + 6)\Bigr] = \frac{1}{2}(6 + 28) = 17.

The car travels about $17$ metres.

6. Find lower and upper limits for the distance in question 5.

Hint

The speed increases throughout, so the smaller height on each strip is on the left.

Next step

The left-hand heights are $0, 3, 5, 6$ , so their rectangle sum is $1(14)$ .

Full working

Lower sum: $1(0 + 3 + 5 + 6) = 14$ metres.

Upper sum: $1(3 + 5 + 6 + 6) = 20$ metres.

So the distance lies between $14$ and $20$ metres. The trapezium estimate $17$ is the midpoint of the two limits.

7. One of these needs the trapezium rule, and one does not. Say which is which, and give a reason for each choice. (a) Find the root of $x^3 - 2x - 5 = 0$ . (b) Find the area under a speed–time graph built from a table of readings.

Hint

Ask what the unknown is in each part: a value of $x$ , or a number built from heights.

Next step

Part (a) asks for a value where the function is zero. Part (b) asks for an area under a set of readings.

Full working

Part (a) is a root-finding problem. Use change of sign to bracket the root, then Newton–Raphson or a simple iteration to narrow it.

Part (b) is an area problem and there is no formula to integrate. Use the trapezium rule, and state limits from the smaller and larger readings on each strip.