Euler's or Forward Rectangular Rule
Mathematics · FE Reference Handbook section
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.
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
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
Iteration 1
Iteration 2
Exact value — √6 = 2.44949
Error — |x₂ − √6| = 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
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
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
Formula — I ≈ (h/2)[f(a) + 2Σf(xᵢ) + f(b)]
Ordinates — f(0) = 1.000, interior sum Σf(xᵢ) = 10.8750, f(3) = 10.000
Substituting
Exact
Error
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
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
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
Iteration 1
Iteration 2
Exact value — √5 = 2.23607
Error — |x₂ − √5| = 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
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
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
Formula — I ≈ (h/2)[f(a) + 2Σf(xᵢ) + f(b)]
Ordinates — f(0) = 1.000, interior sum Σf(xᵢ) = 15.7500, f(2) = 5.000
Substituting
Exact
Error
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
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
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
Iteration 1
Iteration 2
Exact value — √5 = 2.23607
Error — |x₂ − √5| = 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
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
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
Formula — I ≈ (h/2)[f(a) + 2Σf(xᵢ) + f(b)]
Ordinates — f(0) = 1.000, interior sum Σf(xᵢ) = 17.0000, f(4) = 17.000
Substituting
Exact
Error
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
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
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
Iteration 1
Iteration 2
Exact value — √4 = 2.00000
Error — |x₂ − √4| = 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
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
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
Formula — I ≈ (h/2)[f(a) + 2Σf(xᵢ) + f(b)]
Ordinates — f(0) = 1.000, interior sum Σf(xᵢ) = 29.4444, f(4) = 17.000
Substituting
Exact
Error
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
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
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
Iteration 1
Iteration 2
Exact value — √6 = 2.44949
Error — |x₂ − √6| = 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
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
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
Formula — I ≈ (h/2)[f(a) + 2Σf(xᵢ) + f(b)]
Ordinates — f(0) = 1.000, interior sum Σf(xᵢ) = 102.8, f(8) = 65.000
Substituting
Exact
Error
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