Trapezium Rule (Estimation)

\( A\approx \tfrac{h}{2}\,[y_0+y_n+2(y_1+y_2+\cdots+y_{n-1})] \)
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Question 4 of 20
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\( y=[1,9,25,49], h=2. \)

Hint (H)
Work carefully with trapezium rule.

Explanation

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Statement

The trapezium rule is a method to estimate the area under a curve by approximating the region with trapezia instead of rectangles. The formula is:

\[ A \approx \frac{h}{2} \Big[ y_0 + y_n + 2(y_1 + y_2 + \cdots + y_{n-1}) \Big] \]

Here, \(h\) is the interval width and \(y_0, y_1, \dots, y_n\) are the values of the function at equally spaced points.

Why it’s true

  • The area under a curve can be split into strips of equal width.
  • Each strip is approximated by a trapezium, not a rectangle, giving a better fit to the curve.
  • The sum of these trapezia areas gives the approximation formula above.
  • The end ordinates are counted once (\(y_0\) and \(y_n\)), and the interior ordinates are counted twice, because each interior ordinate is shared between two trapezia.

Recipe (how to use it)

  1. Divide the interval \([a,b]\) into \(n\) equal parts: \(h=(b-a)/n\).
  2. Find the ordinates: \(y_0=f(x_0), y_1=f(x_1), \dots, y_n=f(x_n)\).
  3. Substitute into the trapezium rule formula.
  4. Compute the sum carefully, remembering to double the interior ordinates.

Spotting it

This formula is used whenever you’re asked to estimate an area under a curve (integration) and exact integration is not required or possible. Exam questions often say “Use the trapezium rule with n strips”.

Common pairings

  • Simpson’s rule (an alternative estimation method).
  • Exact integration for comparison.

Mini examples

  1. Given: Curve values \(y_0=1, y_1=2, y_2=3\), interval \(h=0.5\). Answer: \(A\approx 0.5/2 [1+3+2(2)] = 0.25[1+3+4]=0.25×8=2\).
  2. Given: \(y_0=2, y_1=5, y_2=6, h=1\). Answer: \(A\approx 0.5[2+6+2(5)]=0.5[2+6+10]=0.5×18=9\).

Pitfalls

  • Forgetting to halve the sum (the \(\frac{h}{2}\) factor).
  • Not doubling the interior ordinates.
  • Using unequal intervals (not valid for trapezium rule).

Exam strategy

  • Write all ordinates in order and label which are doubled.
  • Perform the calculation step by step: inside brackets first, then multiply by \(h/2\).
  • Leave exact form if asked, or round to required decimals.

Summary

The trapezium rule is a practical way to estimate areas under curves. By averaging top and bottom of each strip and summing, the formula \(\frac{h}{2}[y_0+y_n+2(y_1+\dots+y_{n-1})]\) emerges. It’s widely used in GCSE and A-level mathematics for numerical integration.

Worked examples

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  1. \( Use trapezium rule with h=1, y=[2,5,6]. Find area. \)
    1. A≈0.5[2+6+2(5)]
    2. \( =0.5[18]=9 \)
    Answer: 9
  2. \( y=[1,2,3], h=0.5. Estimate area. \)
    1. A≈0.25[1+3+2(2)]
    2. \( =0.25×8=2 \)
    Answer: 2
  3. \( y=[4,6,8], h=2. Estimate area. \)
    1. A≈1[4+8+2(6)]
    2. \( =1[24]=24 \)
    Answer: 24
  4. \( y=[3,5,7,9], h=1. Estimate area. \)
    1. A≈0.5[3+9+2(5+7)]
    2. \( =0.5[3+9+24]=18 \)
    Answer: 18
  5. \( y=[0,1,4,9], h=1. Estimate area. \)
    1. A≈0.5[0+9+2(1+4)]
    2. \( =0.5[0+9+10]=9.5 \)
    Answer: 9.5
  6. \( y=[1,4,9,16], h=1. Estimate area. \)
    1. A≈0.5[1+16+2(4+9)]
    2. \( =0.5[17+26]=21.5 \)
    Answer: 21.5
  7. \( y=[2,8,18,32], h=2. Estimate area. \)
    1. A≈1[2+32+2(8+18)]
    2. \( =1[34+52]=86 \)
    Answer: 86
  8. \( y=[1,7,19,37,61], h=2. Find area. \)
    1. A≈1[1+61+2(7+19+37)]
    2. \( =1[62+126]=188 \)
    Answer: 188
  9. \( y=[2,5,10,17,26], h=1. Find area. \)
    1. A≈0.5[2+26+2(5+10+17)]
    2. \( =0.5[28+64]=46 \)
    Answer: 46
  10. \( y=[0,2,8,18,32], h=2. Estimate area. \)
    1. A≈1[0+32+2(2+8+18)]
    2. \( =1[32+56]=88 \)
    Answer: 88