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Euler's or Forward Rectangular Rule

Mathematics · FE Reference Handbook section

Mathematics
2 formulas
10 exam-style examples
~49 min
All Mathematics lectures

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
Newton's algorithm applied to a square root — Euler's or Forward Rectangular Rule

Use Newton's algorithm on f(x) = x² − 6 with a starting value x₀ = 2 to obtain two improved estimates of √6, then report the error after the second iteration.

Given

  • f(x)=x2−6f(x) = x^{2} - 6
  • x0=2x_{0} = 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

  1. Formula

    xk+1=xk−f(xk)/f′(xk)=½(xk+N/xk)x_{k+1} = x_k - f(x_k)/f'(x_k) = ½(x_k + N/x_k)
  2. Iteration 1

    x1=½(2+6/2)=2.50000x_{1} = ½(2 + 6/2) = 2.50000
  3. Iteration 2

    x2=½(2.50000+6/2.50000)=2.45000x_{2} = ½(2.50000 + 6/2.50000) = 2.45000
  4. Exact value — √6 = 2.44949

  5. Error — |x₂ − √6| = 0.000510

Answer:
x2=2.45000witherror0.000510x_{2} = 2.45000 with error 0.000510

Why the other options are there

  • 3.0000 (single division, no averaging)
  • 4.0000 (forgot to divide by f′)

Reference: FE Reference Handbook — Mathematics → Euler's or Forward Rectangular Rule

Example 2
Trapezoidal rule versus the exact integral — Euler's or Forward Rectangular Rule

Estimate ∫₀^3 (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+1f(x) = x^{2} + 1
  • a=0,b=3a = 0, b = 3
  • n=4n = 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

  1. Step size

    h=(b−a)/n=(3−0)/4=0.7500h = (b - a)/n = (3 - 0)/4 = 0.7500
  2. Formula — I ≈ (h/2)[f(a) + 2Σf(xᵢ) + f(b)]

  3. Ordinates — f(0) = 1.000, interior sum Σf(xᵢ) = 10.8750, f(3) = 10.000

  4. Substituting

    I≈(0.7500/2)[32.7500]=12.2813I \approx (0.7500/2)[32.7500] = 12.2813
  5. Exact

    ∫=b3/3+b=12.0000\int = b^{3}/3 + b = 12.0000
  6. Error

    ∣I−exact∣/exact=2.344|I - exact|/exact = 2.344%
Answer:

Trapezoidal I ≈ 12.281 versus exact 12.000 (2.34% high)

Why the other options are there

  • 24.563 (dropped the ½)
  • 4.125 (single trapezoid)

Reference: FE Reference Handbook — Mathematics → Euler's or Forward Rectangular Rule

Example 3
Newton's algorithm applied to a square root — Euler's or Forward Rectangular Rule (2)

Use Newton's algorithm on f(x) = x² − 5 with a starting value x₀ = 4 to obtain two improved estimates of √5, then report the error after the second iteration.

Given

  • f(x)=x2−5f(x) = x^{2} - 5
  • x0=4x_{0} = 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

  1. Formula

    xk+1=xk−f(xk)/f′(xk)=½(xk+N/xk)x_{k+1} = x_k - f(x_k)/f'(x_k) = ½(x_k + N/x_k)
  2. Iteration 1

    x1=½(4+5/4)=2.62500x_{1} = ½(4 + 5/4) = 2.62500
  3. Iteration 2

    x2=½(2.62500+5/2.62500)=2.26488x_{2} = ½(2.62500 + 5/2.62500) = 2.26488
  4. Exact value — √5 = 2.23607

  5. Error — |x₂ − √5| = 0.028813

Answer:
x2=2.26488witherror0.028813x_{2} = 2.26488 with error 0.028813

Why the other options are there

  • 1.2500 (single division, no averaging)
  • -7.0000 (forgot to divide by f′)

Reference: FE Reference Handbook — Mathematics → Euler's or Forward Rectangular Rule

Example 4
Trapezoidal rule versus the exact integral — Euler's or Forward Rectangular Rule (2)

Estimate ∫₀^2 (x² + 1) dx with the trapezoidal rule using n = 8 equal intervals, then compare with the exact value and report the percentage error.

Given

  • f(x)=x2+1f(x) = x^{2} + 1
  • a=0,b=2a = 0, b = 2
  • n=8n = 8

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

  1. Step size

    h=(b−a)/n=(2−0)/8=0.2500h = (b - a)/n = (2 - 0)/8 = 0.2500
  2. Formula — I ≈ (h/2)[f(a) + 2Σf(xᵢ) + f(b)]

  3. Ordinates — f(0) = 1.000, interior sum Σf(xᵢ) = 15.7500, f(2) = 5.000

  4. Substituting

    I≈(0.2500/2)[37.5000]=4.6875I \approx (0.2500/2)[37.5000] = 4.6875
  5. Exact

    ∫=b3/3+b=4.6667\int = b^{3}/3 + b = 4.6667
  6. Error

    ∣I−exact∣/exact=0.446|I - exact|/exact = 0.446%
Answer:

Trapezoidal I ≈ 4.688 versus exact 4.667 (0.45% high)

Why the other options are there

  • 9.375 (dropped the ½)
  • 0.750 (single trapezoid)

Reference: FE Reference Handbook — Mathematics → Euler's or Forward Rectangular Rule

Example 5
Newton's algorithm applied to a square root — Euler's or Forward Rectangular Rule (3)

Use Newton's algorithm on f(x) = x² − 5 with a starting value x₀ = 4 to obtain two improved estimates of √5, then report the error after the second iteration.

Given

  • f(x)=x2−5f(x) = x^{2} - 5
  • x0=4x_{0} = 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

  1. Formula

    xk+1=xk−f(xk)/f′(xk)=½(xk+N/xk)x_{k+1} = x_k - f(x_k)/f'(x_k) = ½(x_k + N/x_k)
  2. Iteration 1

    x1=½(4+5/4)=2.62500x_{1} = ½(4 + 5/4) = 2.62500
  3. Iteration 2

    x2=½(2.62500+5/2.62500)=2.26488x_{2} = ½(2.62500 + 5/2.62500) = 2.26488
  4. Exact value — √5 = 2.23607

  5. Error — |x₂ − √5| = 0.028813

Answer:
x2=2.26488witherror0.028813x_{2} = 2.26488 with error 0.028813

Why the other options are there

  • 1.2500 (single division, no averaging)
  • -7.0000 (forgot to divide by f′)

Reference: FE Reference Handbook — Mathematics → Euler's or Forward Rectangular Rule

Example 6
Trapezoidal rule versus the exact integral — Euler's or Forward Rectangular Rule (3)

Estimate ∫₀^4 (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+1f(x) = x^{2} + 1
  • a=0,b=4a = 0, b = 4
  • n=4n = 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

  1. Step size

    h=(b−a)/n=(4−0)/4=1.0000h = (b - a)/n = (4 - 0)/4 = 1.0000
  2. Formula — I ≈ (h/2)[f(a) + 2Σf(xᵢ) + f(b)]

  3. Ordinates — f(0) = 1.000, interior sum Σf(xᵢ) = 17.0000, f(4) = 17.000

  4. Substituting

    I≈(1.0000/2)[52.0000]=26.0000I \approx (1.0000/2)[52.0000] = 26.0000
  5. Exact

    ∫=b3/3+b=25.3333\int = b^{3}/3 + b = 25.3333
  6. Error

    ∣I−exact∣/exact=2.632|I - exact|/exact = 2.632%
Answer:

Trapezoidal I ≈ 26.000 versus exact 25.333 (2.63% high)

Why the other options are there

  • 52.000 (dropped the ½)
  • 9.000 (single trapezoid)

Reference: FE Reference Handbook — Mathematics → Euler's or Forward Rectangular Rule

Example 7
Newton's algorithm applied to a square root — Euler's or Forward Rectangular Rule (4)

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

Given

  • f(x)=x2−4f(x) = x^{2} - 4
  • x0=4x_{0} = 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

  1. Formula

    xk+1=xk−f(xk)/f′(xk)=½(xk+N/xk)x_{k+1} = x_k - f(x_k)/f'(x_k) = ½(x_k + N/x_k)
  2. Iteration 1

    x1=½(4+4/4)=2.50000x_{1} = ½(4 + 4/4) = 2.50000
  3. Iteration 2

    x2=½(2.50000+4/2.50000)=2.05000x_{2} = ½(2.50000 + 4/2.50000) = 2.05000
  4. Exact value — √4 = 2.00000

  5. Error — |x₂ − √4| = 0.050000

Answer:
x2=2.05000witherror0.050000x_{2} = 2.05000 with error 0.050000

Why the other options are there

  • 1.0000 (single division, no averaging)
  • -8.0000 (forgot to divide by f′)

Reference: FE Reference Handbook — Mathematics → Euler's or Forward Rectangular Rule

Example 8
Trapezoidal rule versus the exact integral — Euler's or Forward Rectangular Rule (4)

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

Given

  • f(x)=x2+1f(x) = x^{2} + 1
  • a=0,b=4a = 0, b = 4
  • n=6n = 6

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

  1. Step size

    h=(b−a)/n=(4−0)/6=0.6667h = (b - a)/n = (4 - 0)/6 = 0.6667
  2. Formula — I ≈ (h/2)[f(a) + 2Σf(xᵢ) + f(b)]

  3. Ordinates — f(0) = 1.000, interior sum Σf(xᵢ) = 29.4444, f(4) = 17.000

  4. Substituting

    I≈(0.6667/2)[76.8889]=25.6296I \approx (0.6667/2)[76.8889] = 25.6296
  5. Exact

    ∫=b3/3+b=25.3333\int = b^{3}/3 + b = 25.3333
  6. Error

    ∣I−exact∣/exact=1.170|I - exact|/exact = 1.170%
Answer:

Trapezoidal I ≈ 25.630 versus exact 25.333 (1.17% high)

Why the other options are there

  • 51.259 (dropped the ½)
  • 6.000 (single trapezoid)

Reference: FE Reference Handbook — Mathematics → Euler's or Forward Rectangular Rule

Example 9
Newton's algorithm applied to a square root — Euler's or Forward Rectangular Rule (5)

Use Newton's algorithm on f(x) = x² − 6 with a starting value x₀ = 5 to obtain two improved estimates of √6, then report the error after the second iteration.

Given

  • f(x)=x2−6f(x) = x^{2} - 6
  • x0=5x_{0} = 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

  1. Formula

    xk+1=xk−f(xk)/f′(xk)=½(xk+N/xk)x_{k+1} = x_k - f(x_k)/f'(x_k) = ½(x_k + N/x_k)
  2. Iteration 1

    x1=½(5+6/5)=3.10000x_{1} = ½(5 + 6/5) = 3.10000
  3. Iteration 2

    x2=½(3.10000+6/3.10000)=2.51774x_{2} = ½(3.10000 + 6/3.10000) = 2.51774
  4. Exact value — √6 = 2.44949

  5. Error — |x₂ − √6| = 0.068252

Answer:
x2=2.51774witherror0.068252x_{2} = 2.51774 with error 0.068252

Why the other options are there

  • 1.2000 (single division, no averaging)
  • -14.0000 (forgot to divide by f′)

Reference: FE Reference Handbook — Mathematics → Euler's or Forward Rectangular Rule

Example 10
Trapezoidal rule versus the exact integral — Euler's or Forward Rectangular Rule (5)

Estimate ∫₀^8 (x² + 1) dx with the trapezoidal rule using n = 6 equal intervals, then compare with the exact value and report the percentage error.

Given

  • f(x)=x2+1f(x) = x^{2} + 1
  • a=0,b=8a = 0, b = 8
  • n=6n = 6

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

  1. Step size

    h=(b−a)/n=(8−0)/6=1.3333h = (b - a)/n = (8 - 0)/6 = 1.3333
  2. Formula — I ≈ (h/2)[f(a) + 2Σf(xᵢ) + f(b)]

  3. Ordinates — f(0) = 1.000, interior sum Σf(xᵢ) = 102.8, f(8) = 65.000

  4. Substituting

    I≈(1.3333/2)[271.6]=181.0I \approx (1.3333/2)[271.6] = 181.0
  5. Exact

    ∫=b3/3+b=178.7\int = b^{3}/3 + b = 178.7
  6. Error

    ∣I−exact∣/exact=1.327|I - exact|/exact = 1.327%
Answer:

Trapezoidal I ≈ 181.0 versus exact 178.7 (1.33% high)

Why the other options are there

  • 362.1 (dropped the ½)
  • 44.000 (single trapezoid)

Reference: FE Reference Handbook — Mathematics → Euler's or Forward Rectangular Rule

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