Numerical Integration — Trapezoid Rule

§ 01When Exact Integration Fails

Not every integrand has a nice antiderivative. Functions like e^−x², sin(x²), and √(1 + cos² x) cannot be expressed in terms of elementary functions, yet their definite integrals have perfectly well-defined numerical values. Numerical integration methods compute those values to any desired precision without ever finding an antiderivative.

Even when an exact antiderivative exists, numerical methods are often used in engineering and science because the integrand is given as a table of measured values rather than a formula. In those cases, no symbolic method applies at all.

Both methods on this page share the same setup: divide the interval [a, b] into n equal subintervals of width h = (b−a)/n, label the endpoints x₀ = a, x₁, x₂, …, x_n = b, and evaluate f at each node.

Common Setup for Both Methods

h = (b − a)/n x_i = a + ih i = 0, 1, 2, …, n

n must be even for Simpson's Rule. It can be any positive integer for the Trapezoid Rule.

Relationship to Riemann Sums Riemann sums use rectangles. Numerical integration methods use better geometric shapes — trapezoids and parabolic arcs — to approximate the same area with far fewer subintervals. The Trapezoid Rule is exact for linear functions; Simpson's Rule is exact for polynomials of degree ≤ 3.

§ 02The Trapezoid Rule — Derivation and Formula

On each subinterval [x_i−1, x_i], replace the curve with the straight line connecting (x_i−1, f(x_i−1)) to (x_i, f(x_i)). The area under that line segment is the area of a trapezoid:

Area of One Trapezoid

A_i = (h/2)[f(x_i−1) + f(x_i)]

Summing all n trapezoids, the endpoint values f(x₀) and f(x_n) appear once each; every interior value f(x_i) appears in two adjacent trapezoids and therefore carries coefficient 2.

Trapezoid Rule — T_n

T_n = (h/2)[ f(x₀) + 2f(x₁) + 2f(x₂) + … + 2f(x_n−1) + f(x_n) ]

Coefficient pattern: 1, 2, 2, …, 2, 1 — multiply by h/2

A compact shorthand: T_n = (h/2)[f(x₀) + f(x_n)] + h∑_i=1ⁿ⁻¹ f(x_i). In plain English: half the step-size times (endpoints once, interior points twice).

Example 1 — Trapezoid Rule with n = 4

Example 01Approximate ∫₁³ (1/x) dx using T₄

Setup

a=1, b=3, n=4 → h = (3−1)/4 = 0.5x₀=1, x₁=1.5, x₂=2, x₃=2.5, x₄=3

Values table

i	x_i	f(x_i) = 1/x_i	Coefficient	Contribution
0	1.000	1.0000	1	1.0000
1	1.500	0.6667	2	1.3333
2	2.000	0.5000	2	1.0000
3	2.500	0.4000	2	0.8000
4	3.000	0.3333	1	0.3333
Sum				4.4667

Apply formula

T₄ = (0.5/2) × 4.4667 = 0.25 × 4.4667

Result

T₄ ≈ 1.1167 (exact: ln 3 ≈ 1.0986)

Example 2 — Trapezoid Rule with n = 6 on a Tabulated Function

Example 02Use T₆ to estimate ∫₀³ f(x) dx given the table of values

Given data

x	0	0.5	1.0	1.5	2.0	2.5	3.0
f(x)	1.0	1.8	2.3	2.1	1.6	1.2	0.9

h = 0.5, n = 6

Weighted sum

1(1.0) + 2(1.8) + 2(2.3) + 2(2.1) + 2(1.6) + 2(1.2) + 1(0.9)= 1.0 + 3.6 + 4.6 + 4.2 + 3.2 + 2.4 + 0.9 = 19.9

Result

T₆ = (0.5/2) × 19.9 = 0.25 × 19.9 ≈ 4.975

§ 03Simpson’s Rule — Derivation and Formula

The Trapezoid Rule fits straight lines between nodes. Simpson's Rule does better: it fits a quadratic (parabolic arc) through each pair of consecutive subintervals. Since a parabola matches the curve at three points instead of two, the approximation is significantly more accurate for the same n.

On the subinterval [x₀, x₂], the unique parabola through (x₀, f₀), (x₁, f₁), (x₂, f₂) integrates to give area (h/3)[f₀ + 4f₁ + f₂]. Applying this to every consecutive pair of subintervals and summing:

Simpson's Rule — S_n (n must be even)

S_n = (h/3)[ f(x₀) + 4f(x₁) + 2f(x₂) + 4f(x₃) + 2f(x₄) + … + 4f(x_n−1) + f(x_n) ]

Coefficient pattern: 1, 4, 2, 4, 2, …, 4, 2, 4, 1 — multiply by h/3

The 1-4-2-4-2-…-4-1 pattern is the signature of Simpson's Rule. Interior odd-indexed points always carry coefficient 4; interior even-indexed points always carry coefficient 2; the two endpoints carry coefficient 1.

Simpson's Rule Requires Even n Each parabolic arc spans two subintervals. If n is odd, the intervals cannot be paired completely and the formula breaks down. Always choose an even value of n when applying Simpson's Rule — n = 4, 6, 8, … are all valid.

Example 3 — Simpson's Rule with n = 4

Example 03Approximate ∫₁³ (1/x) dx using S₄

Setup

Same nodes as Example 1: h=0.5, x₀=1, x₁=1.5, x₂=2, x₃=2.5, x₄=3

Values table

i	x_i	f(x_i)	Coeff (1,4,2,4,1)	Contribution
0	1.000	1.0000	1	1.0000
1	1.500	0.6667	4	2.6667
2	2.000	0.5000	2	1.0000
3	2.500	0.4000	4	1.6000
4	3.000	0.3333	1	0.3333
Sum				6.6000

Apply formula

S₄ = (0.5/3) × 6.6000 = (1/6) × 6.6000

Result

S₄ ≈ 1.1000 (exact: ln 3 ≈ 1.0986 — error < 0.002)

Example 4 — Simpson's Rule with n = 6

Example 04Approximate ∫₀³ e^−x² dx using S₆

Setup

h = (3−0)/6 = 0.5 x_i = 0, 0.5, 1.0, 1.5, 2.0, 2.5, 3.0

This integrand has no elementary antiderivative: numerical integration is the only practical approach.

Values table

i	x_i	f(x_i) = e^−x_i²	Coeff	Contribution
0	0.0	1.000000	1	1.000000
1	0.5	0.778801	4	3.115203
2	1.0	0.367879	2	0.735759
3	1.5	0.105399	4	0.421598
4	2.0	0.018316	2	0.036631
5	2.5	0.001930	4	0.007722
6	3.0	0.000123	1	0.000123
Sum				5.317036

Apply formula

S₆ = (0.5/3) × 5.317036 ≈ 0.8862

Result

∫₀³ e^−x² dx ≈ 0.8862 (true value ≈ 0.8862 — excellent accuracy)

♦ ♦ ♦

§ 04Error Bounds — Guaranteed Accuracy

Numerical methods give approximations, but mathematics also provides hard upper bounds on how wrong those approximations can be. The error bounds below guarantee that the true error cannot exceed the stated value: regardless of the specific function values.

Trapezoid Rule Error Bound — |E_T|

|E_T| ≤ M₂(b−a)³ / (12n²)

where M₂ = max|f′′(x)| on [a, b]. The error shrinks as 1/n², doubling n quarters the error.

Simpson’s Rule Error Bound — |E_S|

|E_S| ≤ M₄(b−a)⁵ / (180n⁴)

where M₄ = max|f⁽⁴⁾(x)| on [a, b]. The error shrinks as 1/n⁴, doubling n reduces error by a factor of 16. This is why Simpson's Rule is dramatically more efficient than the Trapezoid Rule for smooth functions.

Interpreting the Error Bounds Both bounds give a worst-case guarantee. The actual error is usually much smaller. Also note: if f′′ = 0 on [a,b] (f is linear), the Trapezoid Rule is exact with zero error. If f⁽⁴⁾ = 0 on [a,b] (f is a cubic polynomial), Simpson's Rule is exact. This explains why the error bounds involve the second and fourth derivatives respectively.

Example 5 — Trapezoid Error Bound

Example 05Find n so that T_n approximates ∫₁³ (1/x) dx with error < 0.001

Find M₂

f(x) = 1/x → f′(x) = −1/x² → f′′(x) = 2/x³

On [1, 3], f′′ is decreasing, so max is at x = 1: M₂ = 2/1³ = 2.

Set up inequality

|E_T| ≤ 2 × (3−1)³ / (12n²) = 2 × 8 / (12n²) = 16/(12n²) = 4/(3n²)Require: 4/(3n²) < 0.001 → n² > 4/(0.003) ≈ 1333 → n > 36.5

Result

Use n ≥ 37 (or round up to n = 40 for a round number). T₃₇ guarantees error < 0.001.

Example 6 — Simpson's Error Bound

Example 06Find n so that S_n approximates ∫₀¹ e^x dx with error < 0.00001

Find M₄

f(x) = e^x → f⁽⁴⁾(x) = e^x

On [0,1], max|f⁽⁴⁾| = e¹ = e ≈ 2.718. So M₄ = e.

Set up inequality

|E_S| ≤ e(1−0)⁵ / (180n⁴) = e/(180n⁴)Require: e/(180n⁴) < 0.00001 → n⁴ > e/0.0018 ≈ 1510 → n > 6.23

Result

Use n = 8 (next even integer ≥ 7). S₈ guarantees error < 0.00001.

§ 05Comparison, Accuracy and When to Use Each Method

Property	Trapezoid Rule	Simpson's Rule
Geometric shape used	Straight lines (trapezoids)	Parabolic arcs
Coefficient pattern	1, 2, 2, …, 2, 1	1, 4, 2, 4, …, 4, 1
Multiplier	h/2	h/3
n restriction	Any positive integer	Must be even
Exact for	Linear functions (degree ≤ 1)	Polynomials of degree ≤ 3
Error order	O(h²) — error ∝ 1/n²	O(h⁴) — error ∝ 1/n⁴
Derivative needed for error bound	f′′ (second derivative)	f⁽⁴⁾ (fourth derivative)
Best suited for	Tabulated data with any spacing; quick estimates; linear-ish integrands	Smooth functions; when high accuracy is needed with fewer intervals

For a smooth function, Simpson's Rule with n = 8 will typically outperform the Trapezoid Rule with n = 100. When the integrand is given only as a table of values with no formula, both rules apply, but the Trapezoid Rule is simpler and has no parity requirement on n.

Example 7 — Comparing Both Methods on the Same Integral

Example 07Approximate ∫₀^π/2 cos x dx using T₄ and S₄; compare with exact value

Setup

h = (π/2)/4 = π/8 ≈ 0.3927 Exact value: [sin x]₀^π/2 = 1

Node values

i	x_i	cos(x_i)
0	0	1.000000
1	π/8 ≈ 0.3927	0.923880
2	π/4 ≈ 0.7854	0.707107
3	3π/8 ≈ 1.1781	0.382683
4	π/2 ≈ 1.5708	0.000000

T₄

= (π/8)/2 × [1(1) + 2(0.9239) + 2(0.7071) + 2(0.3827) + 1(0)] = (π/16) × 5.0274 ≈ 0.9871

S₄

= (π/8)/3 × [1(1) + 4(0.9239) + 2(0.7071) + 4(0.3827) + 1(0)] = (π/24) × 7.6284 ≈ 0.9998

Comparison

T₄ ≈ 0.9871 (error 0.0129) S₄ ≈ 0.9998 (error 0.0002) Exact: 1

Example 8 — Finding the Minimum n for a Given Accuracy

Example 08How many intervals does Simpson's Rule need to approximate ∫₀² x⁴ dx to within 0.0001?

Find M₄

f(x) = x⁴ → f⁽⁴⁾(x) = 24 → M₄ = 24 (constant)

Error bound

|E_S| ≤ 24 × (2−0)⁵ / (180n⁴) = 24 × 32 / (180n⁴) = 768/(180n⁴) = 4.267/n⁴Require 4.267/n⁴ < 0.0001 → n⁴ > 42667 → n > 14.4

Note

Since f⁽⁴⁾(x) = 24 (a polynomial of degree exactly 4), and the Simpson error bound involves f⁽⁴⁾, you might expect exact results for lower degree polynomials. Indeed, for f = x³, f⁽⁴⁾ = 0 and Simpson's Rule is exact with any even n.

Result

Use n = 16 (next even integer ≥ 15). S₁₆ guarantees error < 0.0001.

§ 0610-Question Quiz

Test your understanding of both numerical methods, coefficient patterns, and error bounds.

§ 07Continue the Integration Series

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Improper Integrals

Volumes of Revolution