Numerical Integration

Mathematics · FE Reference Handbook section

Mathematics

1 formulas

10 exam-style examples

~47 min

All Mathematics lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers Numerical Integration within Mathematics. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what numerical integration describes physically and when it applies.
State every one of the 1 relation the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: radians vs degrees — set the calculator before the first trig entry.

Lecture

Why this section exists. Numerical Integration is the part of Mathematics that lets you connect an algebraic or calculus expression that must be evaluated exactly to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as short symbolic manipulations with one numeric evaluation at the end. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. radians vs degrees — set the calculator before the first trig entry. Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Three engineers in hard hats and safety vests reviewing drawings on a truck tailgate. — Photo 1. Where this shows up in practice: numerical integration.
Capstone Studio instructional photograph

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes an algebraic or calculus expression that must be evaluated exactly. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 1 relation on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

Three of the more common numerical integration algorithms used to evaluate the integral
are:

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Earthwork volume by the trapezoidal rule

Cross-section areas at 20 m stations are 0, 12, 18, 15 and 6 m². Estimate the earthwork volume by the trapezoidal rule.

Given

Δx = 20 m
A = 0, 12, 18, 15, 6 m²

Find

Volume V (m³)

Start with the thinking

End areas get half weight; interior areas get full weight.
Do not double-count the first and last stations.

Step-by-step solution

Trapezoidal rule — V = Δx[(A₀ + Aₙ)/2 + ΣA_interior]
End terms
$(0 + 6) /2 = 3.00 m^{2}$
Interior sum
$12 + 18 + 15 = 45.0 m^{2}$
Substitute
$V = 20 (3.00 + 45.0)$
Evaluate
$V = 20 (48.0) = 960 m^{3}$

Answer: V ≈ 960 m³

Why the other options are there

1,020 m³ (end areas taken at full weight)
48 m³ (Δx omitted)

Reference: FE Reference Handbook — Mathematics — Numerical integration

Example 2

Definite integral of a distributed load

A beam carries w(x) = 5x kN/m over 0 ≤ x ≤ 4 m. What is the total load and where does its resultant act?

Given

w(x) = 5x kN/m
L = 4 m

Find

Resultant W and its location x̄

Start with the thinking

Total load is the area under the loading diagram.
The resultant acts at the centroid of that area.

Step-by-step solution

Total load
$W = \int_{0}^{4} 5 x d x$
Antiderivative
$\int_{5} x d x = 2.5 x^{2}$
Evaluate
$W = 2.5 (16) - 0 = 40.0 k N$
First moment — ∫₀⁴ x(5x) dx = (5/3)x³ evaluated 0 to 4 = 106.7 kN·m
Centroid
$x ˉ = 106.7/40.0 = 2.67 m f r o m t h eor i g in$

Answer: W = 40.0 kN acting 2.67 m from the small end

Why the other options are there

20 kN (average intensity times half length)
x̄ = 2.00 m (uniform load assumed)

Reference: FE Reference Handbook — Mathematics — Integral calculus

Example 3

Definite integral — Numerical Integration

Evaluate ∫₀^5 (5x² + 1x) dx.

Given

Integrand 5x² + 1x
Limits 0 to 5

Find

The definite integral

Start with the thinking

Antidifferentiate term by term, then apply the limits.
The lower limit of zero kills the second evaluation.

Step-by-step solution

Antiderivative
$F (x) = (5/3) x^{3} + (1/2) x^{2}$
Upper limit
$F (5) = (5/3) (125) + (1/2) (25)$
Evaluate
$F (5) = 208.3 + 12.500 = 220.8$
Lower limit
$F (0) = 0$
Result
$\int = 220.8$

Answer: 220.8

Why the other options are there

130.0 (integrand evaluated instead of integrated)
44.17 (average value reported)

Reference: FE Reference Handbook — Mathematics → Numerical Integration

Example 4

Newton's algorithm applied to a square root — Numerical Integration

Use Newton's algorithm on f(x) = x² − 8 with a starting value x₀ = 2 to obtain two improved estimates of √8, then report the error after the second iteration.

Given

f(x) = x² − 8
x₀ = 2

Find

x₁, x₂ and the absolute error |x₂ − √N|

Start with the thinking

Newton's algorithm is x_{k+1} = x_k − f(x_k)/f′(x_k); for f = x² − N it collapses to the averaging form.
Convergence is quadratic, so two iterations already give several correct digits.

Step-by-step solution

Formula
$x_{k + 1} = x_{k} - f (x_{k}) / f^{'} (x_{k}) = ½ (x_{k} + N / x_{k})$
Iteration 1
$x_{1} = ½ (2 + 8/2) = 3.00000$
Iteration 2
$x_{2} = ½ (3.00000 + 8/3.00000) = 2.83333$
Exact value — √8 = 2.82843
Error — |x₂ − √8| = 0.004906

Answer: x₂ = 2.83333 with error 0.004906

Why the other options are there

4.0000 (single division, no averaging)
6.0000 (forgot to divide by f′)

Reference: FE Reference Handbook — Mathematics → Numerical Integration

Example 5

Trapezoidal rule versus the exact integral — Numerical Integration

Estimate ∫₀^6 (x² + 1) dx with the trapezoidal rule using n = 4 equal intervals, then compare with the exact value and report the percentage error.

Given

f(x) = x² + 1
a = 0, b = 6
n = 4

Find

Trapezoidal estimate, exact value, and percent error

Start with the thinking

The trapezoidal rule weights interior ordinates by 2 and the two end ordinates by 1.
For a convex function the trapezoidal rule over-estimates the true area.

Step-by-step solution

Step size
$h = (b - a) / n = (6 - 0) /4 = 1.5000$
Formula — I ≈ (h/2)[f(a) + 2Σf(xᵢ) + f(b)]
Ordinates — f(0) = 1.000, interior sum Σf(xᵢ) = 34.5000, f(6) = 37.000
Substituting
$I \approx (1.5000/2) [107.0] = 80.2500$
Exact
$\int = b^{3} /3 + b = 78.0000$
Error
$∣ I - e x a c t ∣/ e x a c t = 2.885$

Answer: Trapezoidal I ≈ 80.250 versus exact 78.000 (2.88% high)

Why the other options are there

160.5 (dropped the ½)
28.500 (single trapezoid)

Reference: FE Reference Handbook — Mathematics → Numerical Integration

Example 6

Definite integral — Numerical Integration (2)

Evaluate ∫₀^2 (6x² + 2x) dx.

Given

Integrand 6x² + 2x
Limits 0 to 2

Find

The definite integral

Start with the thinking

Antidifferentiate term by term, then apply the limits.
The lower limit of zero kills the second evaluation.

Step-by-step solution

Antiderivative
$F (x) = (6/3) x^{3} + (2/2) x^{2}$
Upper limit
$F (2) = (6/3) (8) + (2/2) (4)$
Evaluate
$F (2) = 16.000 + 4.000 = 20.000$
Lower limit
$F (0) = 0$
Result
$\int = 20.000$

Answer: 20.000

Why the other options are there

28.00 (integrand evaluated instead of integrated)
10.00 (average value reported)

Reference: FE Reference Handbook — Mathematics → Numerical Integration

Example 7

Newton's algorithm applied to a square root — Numerical Integration (2)

Use Newton's algorithm on f(x) = x² − 8 with a starting value x₀ = 5 to obtain two improved estimates of √8, then report the error after the second iteration.

Given

f(x) = x² − 8
x₀ = 5

Find

x₁, x₂ and the absolute error |x₂ − √N|

Start with the thinking

Newton's algorithm is x_{k+1} = x_k − f(x_k)/f′(x_k); for f = x² − N it collapses to the averaging form.
Convergence is quadratic, so two iterations already give several correct digits.

Step-by-step solution

Formula
$x_{k + 1} = x_{k} - f (x_{k}) / f^{'} (x_{k}) = ½ (x_{k} + N / x_{k})$
Iteration 1
$x_{1} = ½ (5 + 8/5) = 3.30000$
Iteration 2
$x_{2} = ½ (3.30000 + 8/3.30000) = 2.86212$
Exact value — √8 = 2.82843
Error — |x₂ − √8| = 0.033694

Answer: x₂ = 2.86212 with error 0.033694

Why the other options are there

1.6000 (single division, no averaging)
-12.0000 (forgot to divide by f′)

Reference: FE Reference Handbook — Mathematics → Numerical Integration

Example 8

Trapezoidal rule versus the exact integral — Numerical Integration (2)

Estimate ∫₀^7 (x² + 1) dx with the trapezoidal rule using n = 4 equal intervals, then compare with the exact value and report the percentage error.

Given

f(x) = x² + 1
a = 0, b = 7
n = 4

Find

Trapezoidal estimate, exact value, and percent error

Start with the thinking

The trapezoidal rule weights interior ordinates by 2 and the two end ordinates by 1.
For a convex function the trapezoidal rule over-estimates the true area.

Step-by-step solution

Step size
$h = (b - a) / n = (7 - 0) /4 = 1.7500$
Formula — I ≈ (h/2)[f(a) + 2Σf(xᵢ) + f(b)]
Ordinates — f(0) = 1.000, interior sum Σf(xᵢ) = 45.8750, f(7) = 50.000
Substituting
$I \approx (1.7500/2) [142.8] = 124.9$
Exact
$\int = b^{3} /3 + b = 121.3$
Error
$∣ I - e x a c t ∣/ e x a c t = 2.945$

Answer: Trapezoidal I ≈ 124.9 versus exact 121.3 (2.94% high)

Why the other options are there

249.8 (dropped the ½)
44.625 (single trapezoid)

Reference: FE Reference Handbook — Mathematics → Numerical Integration

Example 9

Definite integral — Numerical Integration (3)

Evaluate ∫₀^5 (6x² + 5x) dx.

Given

Integrand 6x² + 5x
Limits 0 to 5

Find

The definite integral

Start with the thinking

Antidifferentiate term by term, then apply the limits.
The lower limit of zero kills the second evaluation.

Step-by-step solution

Antiderivative
$F (x) = (6/3) x^{3} + (5/2) x^{2}$
Upper limit
$F (5) = (6/3) (125) + (5/2) (25)$
Evaluate
$F (5) = 250.0 + 62.500 = 312.5$
Lower limit
$F (0) = 0$
Result
$\int = 312.5$

Answer: 312.5

Why the other options are there

175.0 (integrand evaluated instead of integrated)
62.50 (average value reported)

Reference: FE Reference Handbook — Mathematics → Numerical Integration

Example 10

Newton's algorithm applied to a square root — Numerical Integration (3)

Use Newton's algorithm on f(x) = x² − 6 with a starting value x₀ = 4 to obtain two improved estimates of √6, then report the error after the second iteration.

Given

f(x) = x² − 6
x₀ = 4

Find

x₁, x₂ and the absolute error |x₂ − √N|

Start with the thinking

Newton's algorithm is x_{k+1} = x_k − f(x_k)/f′(x_k); for f = x² − N it collapses to the averaging form.
Convergence is quadratic, so two iterations already give several correct digits.

Step-by-step solution

Formula
$x_{k + 1} = x_{k} - f (x_{k}) / f^{'} (x_{k}) = ½ (x_{k} + N / x_{k})$
Iteration 1
$x_{1} = ½ (4 + 6/4) = 2.75000$
Iteration 2
$x_{2} = ½ (2.75000 + 6/2.75000) = 2.46591$
Exact value — √6 = 2.44949
Error — |x₂ − √6| = 0.016419

Answer: x₂ = 2.46591 with error 0.016419

Why the other options are there

1.5000 (single division, no averaging)
-6.0000 (forgot to divide by f′)

Reference: FE Reference Handbook — Mathematics → Numerical Integration

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given an algebraic or calculus expression that must be evaluated exactly, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

Numerical Integration contains 1 relation; you must be able to find this page in under 15 seconds.
Exam style: short symbolic manipulations with one numeric evaluation at the end.
Unit rule: radians vs degrees — set the calculator before the first trig entry.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

radians vs degrees — set the calculator before the first trig entry
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.