Numerical integration using the trapezium rule, estimating accuracy.
Answer at least 3 of 3 correctly to complete this section.
Not every function can be integrated exactly. Try finding ∫ex2dx or ∫xsinxdx — there is no closed-form answer. Yet these integrals arise in real problems (the first appears in probability, the second in signal processing).
The trapezium rule gives you a way to approximate any definite integral numerically. Instead of finding the exact area under a curve, you approximate it using trapeziums — shapes whose areas are easy to calculate.
This is one of the simplest numerical methods, and it appears regularly on A-Level papers. You need to know the formula, how to apply it, and when it gives an overestimate or underestimate.
Divide the area under the curve into n equal-width vertical strips. Approximate each strip as a trapezium (a shape with one pair of parallel sides). The area of each trapezium is:
Area of trapezium=21(a+b)×h
where a and b are the parallel sides (the y-values at each end) and h is the width.
Adding up all n trapeziums gives the trapezium rule:
∫abydx≈2h[y0+yn+2(y1+y2+⋯+yn−1)]
where:
The first and last y-values (y0 and yn) each appear in only one trapezium, so they are counted once. Every intermediate y-value (y1 through yn−1) is shared between two adjacent trapeziums, so it is counted twice.
Think of it as:
2h[first+last+2×(all the middle ones)]
Use the trapezium rule with 4 strips to estimate ∫02x2dx.
Step 1: Find h: h=42−0=0.5
Step 2: Set up the table of values:
| x | 0 | 0.5 | 1 | 1.5 | 2 |
|---|---|---|---|---|---|
| y=x2 | 0 | 0.25 | 1 | 2.25 | 4 |
Step 3: Apply the formula:
∫02x2dx≈20.5[0+4+2(0.25+1+2.25)]
=0.25[4+2(3.5)]=0.25[4+7]=0.25×11=2.75
The exact answer is ∫02x2dx=[3x3]02=38≈2.667. So the trapezium rule gives 2.75, which is an overestimate (we will see why shortly).
Use the trapezium rule with 5 ordinates to estimate ∫13x1dx, giving your answer to 4 decimal places.
5 ordinates means n+1=5, so n=4 strips.
h=43−1=0.5
| x | 1 | 1.5 | 2 | 2.5 | 3 |
|---|---|---|---|---|---|
| y=x1 | 1 | 0.66 | 0.5 | 0.4 | 0.33 |
≈20.5[1+0.3333+2(0.6667+0.5+0.4)]
=0.25[1.3333+2(1.5667)]=0.25[1.3333+3.1333]
=0.25×4.4667=1.1167
The exact answer is ln3≈1.0986, so the trapezium rule overestimates here too.
Common error: Students confuse “4 strips” with “4 ordinates”. 4 strips = 5 ordinates. Always check which the question specifies.
Answer at least 3 of 3 correctly to complete this section.
Whether the trapezium rule overestimates or underestimates depends on the concavity of the curve.
When the curve is concave up, the straight line connecting two points on the curve lies above the curve (the trapezium top overshoots). When the curve is concave down, the straight line lies below the curve (the trapezium top undershoots).
The trapezium rule is used to estimate ∫02x2dx. State whether the estimate is an overestimate or underestimate, and explain why.
y=x2, so dxdy=2x and dx2d2y=2>0 for all x.
The curve is concave up throughout the interval, so the trapeziums lie above the curve. The trapezium rule gives an overestimate.
State whether the trapezium rule overestimates or underestimates ∫14xdx.
y=x1/2, so dxdy=21x−1/2 and dx2d2y=−41x−3/2<0 for x>0.
The curve is concave down, so the trapeziums lie below the curve. The trapezium rule gives an underestimate.
Exam tip: If the question asks “is this an overestimate or underestimate?”, always justify with the second derivative (concavity). Sketching the curve with trapeziums can help your reasoning, but the second derivative argument is what earns the marks.
If the curve has a point of inflection in the interval (concavity changes from up to down or vice versa), the overestimate on one part may partially cancel the underestimate on the other. In this case, you cannot make a simple statement — the exam is unlikely to ask this.
Answer at least 3 of 3 correctly to complete this section.
(a) Use the trapezium rule with 4 strips to estimate ∫01ex2dx, giving your answer to 4 significant figures.
(b) State, with a reason, whether your answer is an overestimate or underestimate.
(a) h=41−0=0.25
| x | 0 | 0.25 | 0.5 | 0.75 | 1 |
|---|---|---|---|---|---|
| x2 | 0 | 0.0625 | 0.25 | 0.5625 | 1 |
| y=ex2 | 1 | 1.0645 | 1.2840 | 1.7551 | 2.7183 |
≈20.25[1+2.7183+2(1.0645+1.2840+1.7551)]
=0.125[3.7183+2(4.1036)]=0.125[3.7183+8.2072]
=0.125×11.9255=1.491 (4 s.f.)
(b) Let y=ex2. Using the chain rule:
dxdy=2xex2
dx2d2y=2ex2+4x2ex2=2ex2(1+2x2)>0 for all x
Since dx2d2y>0 on [0,1], the curve is concave up, so the trapezium rule gives an overestimate.
(a) Use the trapezium rule with 3 strips to estimate ∫14(2x−1)dx.
(b) Find the exact value and comment on your answer.
(a) h=34−1=1
| x | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| y=2x−1 | 1 | 3 | 5 | 7 |
≈21[1+7+2(3+5)]=21[8+16]=21×24=12
(b) Exact: ∫14(2x−1)dx=[x2−x]14=(16−4)−(1−1)=12.
The trapezium rule gives the exact answer because y=2x−1 is a straight line. Trapeziums fit a straight line perfectly — there is no curvature to cause error.
Key insight: The trapezium rule is exact for linear functions. For any other function, it is an approximation. The error depends on how curved the function is.
There are two ways to improve the accuracy of the trapezium rule:
Doubling the number of strips roughly quarters the error.
Answer at least 3 of 3 correctly to complete this section.
Lock in what you've learned with exam-style questions and spaced repetition.