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

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

1. Trapezoidal rule — V = Δx[(A₀ + Aₙ)/2 + ΣA_interior]

2. End terms

   $$(0+6)/2=3.00m^2$$

3. Interior sum

   $$12+18+15=45.0m^2$$

4. Substitute

   $$V=20(3.00+45.0)$$

5. Evaluate

   $$V=20(48.0)=960m^3$$

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

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

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

1. Total load

   $$W={\int }_0^{4}5xdx$$

2. Antiderivative

   $${\int }_{5}xdx=2.5x^2$$

3. Evaluate

   $$W=2.5(16)-0=40.0kN$$

4. First moment — ∫₀⁴ x(5x) dx = (5/3)x³ evaluated 0 to 4 = 106.7 kN·m

5. Centroid

   $$\bar{x}=106.7/40.0=2.67mfromtheorigin$$

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

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.

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

   $$x_{k+1}=x_k-f(x_k)/f^{\prime }(x_k)=½(x_k+N/x_k)$$

2. Iteration 1

   $$x_1=½(4+4/4)=2.50000$$

3. Iteration 2

   $$x_2=½(2.50000+4/2.50000)=2.05000$$

4. Exact value — √4 = 2.00000

5. 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 → Trapezoidal 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.

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

   $$h=(b-a)/n=(3-0)/4=0.7500$$

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

   $$I\approx (0.7500/2)[32.7500]=12.2813$$

5. Exact

   $$\int =b^3/3+b=12.0000$$

6. Error

   $$|I-exact|/exact=2.344$$

Why the other options are there

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

Reference: FE Reference Handbook — Mathematics → Trapezoidal Rule

Use Newton's algorithm on f(x) = x² − 5 with a starting value x₀ = 3 to obtain two improved estimates of √5, then report the error after the second iteration.

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

   $$x_{k+1}=x_k-f(x_k)/f^{\prime }(x_k)=½(x_k+N/x_k)$$

2. Iteration 1

   $$x_1=½(3+5/3)=2.33333$$

3. Iteration 2

   $$x_2=½(2.33333+5/2.33333)=2.23810$$

4. Exact value — √5 = 2.23607

5. Error — |x₂ − √5| = 0.002027

Why the other options are there

• 1.6667 (single division, no averaging)
• -1.0000 (forgot to divide by f′)

Reference: FE Reference Handbook — Mathematics → Trapezoidal Rule

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.

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

   $$h=(b-a)/n=(7-0)/4=1.7500$$

2. Formula — I ≈ (h/2)[f(a) + 2Σf(xᵢ) + f(b)]

3. Ordinates — f(0) = 1.000, interior sum Σf(xᵢ) = 45.8750, f(7) = 50.000

4. Substituting

   $$I\approx (1.7500/2)[142.8]=124.9$$

5. Exact

   $$\int =b^3/3+b=121.3$$

6. Error

   $$|I-exact|/exact=2.945$$

Why the other options are there

• 249.8 (dropped the ½)
• 44.625 (single trapezoid)

Reference: FE Reference Handbook — Mathematics → Trapezoidal Rule

Use Newton's algorithm on f(x) = x² − 4 with a starting value x₀ = 2 to obtain two improved estimates of √4, then report the error after the second iteration.

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

   $$x_{k+1}=x_k-f(x_k)/f^{\prime }(x_k)=½(x_k+N/x_k)$$

2. Iteration 1

   $$x_1=½(2+4/2)=2.00000$$

3. Iteration 2

   $$x_2=½(2.00000+4/2.00000)=2.00000$$

4. Exact value — √4 = 2.00000

5. Error — |x₂ − √4| = 0.000000

Why the other options are there

• 2.0000 (single division, no averaging)
• 2.0000 (forgot to divide by f′)

Reference: FE Reference Handbook — Mathematics → Trapezoidal Rule

Estimate ∫₀^4 (x² + 1) dx with the trapezoidal rule using n = 8 equal intervals, then compare with the exact value and report the percentage 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

   $$h=(b-a)/n=(4-0)/8=0.5000$$

2. Formula — I ≈ (h/2)[f(a) + 2Σf(xᵢ) + f(b)]

3. Ordinates — f(0) = 1.000, interior sum Σf(xᵢ) = 42.0000, f(4) = 17.000

4. Substituting

   $$I\approx (0.5000/2)[102.0]=25.5000$$

5. Exact

   $$\int =b^3/3+b=25.3333$$

6. Error

   $$|I-exact|/exact=0.658$$

Why the other options are there

• 51.000 (dropped the ½)
• 4.500 (single trapezoid)

Reference: FE Reference Handbook — Mathematics → Trapezoidal Rule

Use Newton's algorithm on f(x) = x² − 8 with a starting value x₀ = 3 to obtain two improved estimates of √8, then report the error after the second iteration.

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

   $$x_{k+1}=x_k-f(x_k)/f^{\prime }(x_k)=½(x_k+N/x_k)$$

2. Iteration 1

   $$x_1=½(3+8/3)=2.83333$$

3. Iteration 2

   $$x_2=½(2.83333+8/2.83333)=2.82843$$

4. Exact value — √8 = 2.82843

5. Error — |x₂ − √8| = 0.000004

Why the other options are there

• 2.6667 (single division, no averaging)
• 2.0000 (forgot to divide by f′)

Reference: FE Reference Handbook — Mathematics → Trapezoidal Rule

Estimate ∫₀^7 (x² + 1) dx with the trapezoidal rule using n = 8 equal intervals, then compare with the exact value and report the percentage 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

   $$h=(b-a)/n=(7-0)/8=0.8750$$

2. Formula — I ≈ (h/2)[f(a) + 2Σf(xᵢ) + f(b)]

3. Ordinates — f(0) = 1.000, interior sum Σf(xᵢ) = 114.2, f(7) = 50.000

4. Substituting

   $$I\approx (0.8750/2)[279.4]=122.2$$

5. Exact

   $$\int =b^3/3+b=121.3$$

6. Error

   $$|I-exact|/exact=0.736$$

Why the other options are there

• 244.5 (dropped the ½)
• 22.313 (single trapezoid)

Reference: FE Reference Handbook — Mathematics → Trapezoidal Rule