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Trapezium Rule

Practise using the trapezium rule to find an approximate value for the area under a curve.
Menu Level 1 Level 2 Level 3 Exam-Style Help Calculus

Answers with more than three significant figures should be rounded to three significant figures.

Use the trapezium rule with three intervals to find an approximation of the area above the x axis and under the graph of \(y = 0.56^x \) between \(x=0.2\) and \(x = 0.8\).
Graph

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Use the trapezium rule with six intervals to improve the estimate found in the previous question.

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Use the trapezium rule with nine ordinates to estimate
$$ \int^{2}_{-2} x^4 \; \text{dx} $$

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Find the exact value of the area referred to in the previous question by evaluating the definite integral.

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What was the percentage error encountered by using the trapezium rule?

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Use the trapezium rule with six intervals to estimate the area enclosed by the graph \( y= \cos(2x-90) \), the x axis, the line \(x=240\) and the line \(x=300\).

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This is Trapezium Rule level 3. You can also try:
Level 1 Level 2

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Description of Levels

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Level 1 - A structured single question with many parts

Level 2 - Five practice questions

Level 3 - Questions requiring a little more thought

Exam-Style - Have a go at an IB/A-level exam-style questions (worked solutions are available for Transum subscribers).

Calculus a collection of lesson Starters, visual aids, investigations and self-marking exercises.

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Example

The video above comes from Revision Village

Sometimes the area under a curve cannot be found by integration. In these cases a method to approximate the area under the curve called the trapezium rule can be used.

The rule divides the area under a curve into trapeziums, calculates their areas, then sums these areas to get an approsimation of the total area.

Area of a trapezium

Formula for the area of a trapezium

Trapezium Rule

If the width of each interval is \(h\) and the y values (ordinates) are denoted as \(y_0, y_1, y_2, y_3 ...\) then the formula for finding the sum of the areas of these trapeziums is

$$ \frac12 h ((y_0 + y_n) + 2(y_1 + y_2 + ... + y_{n-1})) $$

where n is the number of intervals.

Note that the number or ordinates is always one more than the number of intervals.

If the lower bound of the required area is \(p\) and the upper bound is \(q\) then

$$ h= \frac{q-p}{n} $$

The more trapeziums the area is divided into the more accurate the estimate.

When the gradient of the graph is increasing over the given interval the area given by the trapezium rule will be an overestimate of the actual area.

When the gradient of the graph is decreasing over the given interval the area given by the trapezium rule will be an underestimate of the actual area.

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