The trapezium rule

The idea

With $n$ strips of width $h = \frac{b - a}{n}$ , the trapezium rule estimates $\int_a^b y\,dx \approx \frac{h}{2}[y_0 + y_n + 2(y_1 + \dots + y_{n-1})]$ It overestimates when the curve bends upwards (convex) and underestimates when it bends downwards.

Work out $h$ from the number of strips, not the number of $y$ -values.
Make a table of $y$ -values at each $x$ .
Ends once, middles twice, times $\frac{h}{2}$ .

Watch out $n$ strips need $n + 1$ $y$ -values.

See it worked

Worked example

Use the trapezium rule with $4$ strips to estimate $\int_0^2 x^2\,dx$ .

$h = 0.5$ ; $y$ : $0, 0.25, 1, 2.25, 4$ .
$\frac{0.5}{2}[0 + 4 + 2(0.25 + 1 + 2.25)] = 0.25 \times 11$ .
$= 2.75$ , an overestimate because $y = x^2$ is convex (the exact value is $2.67$ ).

Answer $2.75$

Your turn

Question 1Calculator allowed3 marks

Use the trapezium rule with $2$ strips to estimate $\int_1^3 \frac{1}{x}\,dx$ . Give your answer to 4 decimal places.

Question 2Calculator allowed3 marks

Use the trapezium rule with $4$ strips to estimate $\int_0^2 e^x\,dx$ . Give your answer to 3 decimal places.

Question 3Non-calculator3 marks

Is the estimate of $\int_0^2 e^x\,dx$ from the trapezium rule an overestimate or an underestimate? Explain.

Write your answer, then show the working

$e^x$ curves upwards: its second derivative $e^x$ is positive.
Each straight top of a trapezium lies above the curve.
So the rule overestimates (exact value $e^2 - 1 = 6.389$ ).

Answer An overestimate: $e^x$ is convex, so each trapezium lies above the curve

In past papers

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