CELE Engineering Mathematics — Integral CalculusExam Answer Templates

Examiner tip

For 1-mark VSA on basic integration, the constant C and correct denominator are the two critical elements. Both must appear for full credit.

Model answer

∫ x⁴ dx = x⁵/5 + C

Answer structure

• Line 1: Apply the power rule and write the antiderivative with constant C [1 mark]

Scoring breakdown

Common mark deductions

• Omitting the constant of integration C — automatic 0 for this 1-mark item
• Writing x⁵ without dividing by 5
• Incorrect exponent: writing x⁵/4 instead of x⁵/5

Key phrases to include

• x^(n+1)/(n+1)
• + C
• x⁵/5

Examiner tip

The absolute value in ln|x| is non-negotiable in formal answers. Boards frequently test whether reviewees know the distinction from ln(x).

Model answer

∫ (1/x) dx = ln|x| + C

Answer structure

• Line 1: Recognize the special case n = −1 and write the natural logarithm antiderivative with absolute value bars and constant C [1 mark]

Scoring breakdown

Common mark deductions

• Writing ln(x) without absolute value bars — partial or zero credit
• Writing 1/x² / (−1) — applying the power rule incorrectly to the singular case
• Omitting + C

Key phrases to include

• ln|x|
• absolute value
• + C

Examiner tip

Always show the unsimplified integration step before simplifying. This middle line earns its own mark and protects you if you make an arithmetic slip in simplification.

Model answer

∫ (3x² + 2x − 5) dx
= 3·(x³/3) + 2·(x²/2) − 5x + C
= x³ + x² − 5x + C

Answer structure

• Line 1: Write the integral expression [setup — no marks alone]
• Line 2: Apply power rule term-by-term, showing coefficients divided by new exponent [1 mark]
• Line 3: Simplify each term and write final answer with + C [1 mark]

Scoring breakdown

Common mark deductions

• Not simplifying 3x³/3 to x³ — partial credit at best
• Sign error on the constant term (writing +5x instead of −5x)
• Omitting C in the final answer

Key phrases to include

• term-by-term integration
• x³ + x² − 5x + C
• power rule

Examiner tip

Always write the bracket notation [F(x)]ₐᵇ explicitly. It signals to the examiner that you understand the Fundamental Theorem of Calculus and earns the method mark.

Model answer

∫₁³ (2x + 1) dx
= [x² + x]₁³
= (3² + 3) − (1² + 1)
= (9 + 3) − (1 + 1)
= 12 − 2
= 10

Answer structure

• Line 1: Write the antiderivative in bracket notation with limits [1 mark]
• Line 2: Substitute upper limit b = 3, then subtract lower limit a = 1 [1 mark]
• Line 3: Simplify and state the numerical answer

Scoring breakdown

Common mark deductions

• Computing F(lower) − F(upper) — reversed limits give negative answer
• Forgetting to subtract F(a); only substituting the upper limit
• Arithmetic errors in squaring the limits

Key phrases to include

• F(b) − F(a)
• [x² + x]₁³
• substitute upper then lower limit

Examiner tip

Simplify the antiderivative before substituting limits. It reduces arithmetic errors and shows mathematical fluency.

Model answer

∫₀² 3x² dx
= [x³]₀²
= (2)³ − (0)³
= 8 − 0
= 8

Answer structure

• Line 1: Integrate to get antiderivative [x³] with limits shown [1 mark]
• Line 2: Substitute limits and evaluate F(2) − F(0) [1 mark]
• Line 3: State final numerical answer

Scoring breakdown

Common mark deductions

• Not simplifying 3x³/3 — leaving as 3x³/3 without simplifying before evaluating
• Arithmetic error: computing 2³ as 6 instead of 8

Key phrases to include

• [x³]₀²
• F(2) − F(0)
• = 8

Examiner tip

Always verify which curve is upper by testing a value inside the interval (e.g., x = 0.5: y = 0.5 vs y = 0.25). This one-line verification earns the concept mark and prevents setup errors.

Model answer

Given: Upper curve: y = x; Lower curve: y = x²; Limits: x = 0 to x = 1

Note: For 0 ≤ x ≤ 1, x ≥ x², so y = x is the upper curve.

Area = ∫₀¹ (x − x²) dx
= [x²/2 − x³/3]₀¹
= (1/2 − 1/3) − (0 − 0)
= 3/6 − 2/6
= 1/6

∴ Area = 1/6 square units ≈ 0.167 sq. units

Answer structure

• Line 1: Identify upper and lower curves and verify with a test value [1 mark]
• Line 2: Set up the integral A = ∫(upper − lower) dx with correct limits [1 mark]
• Line 3: Integrate, evaluate, and state the answer with units [1 mark]

Scoring breakdown

Common mark deductions

• Setting up ∫(x² − x)dx instead of ∫(x − x²)dx — gives negative area, showing conceptual error
• Not stating square units in the final answer
• Forgetting to subtract the lower limit evaluation (treating F(0) = 0 as automatic without showing it)

Key phrases to include

• upper minus lower
• ∫(f(x) − g(x))dx
• 1/6 square units
• verify which curve is upper

Examiner tip

When the region is symmetric about the y-axis, you may write A = 2∫₀²(4−x²)dx — but you MUST state 'by symmetry' to earn the method mark.

Model answer

Given: y = 4 − x², x-axis (y = 0)

Find limits: Set y = 0: 4 − x² = 0 → x = ±2

Note: y = 4 − x² ≥ 0 for −2 ≤ x ≤ 2 (parabola opens downward).

Area = ∫₋₂² (4 − x²) dx
= [4x − x³/3]₋₂²
= (8 − 8/3) − (−8 + 8/3)
= (16/3) − (−16/3)
= 32/3

∴ Area = 32/3 square units ≈ 10.667 sq. units

Answer structure

• Line 1: Find x-intercepts by setting y = 0 to determine limits of integration [1 mark]
• Line 2: Set up and integrate ∫₋₂²(4 − x²)dx, showing antiderivative [1 mark]
• Line 3: Evaluate at both limits, subtract, and state answer with units [1 mark]

Scoring breakdown

Common mark deductions

• Using limits 0 to 2 only and doubling — risky unless you explicitly state and justify the symmetry step
• Error in substituting the negative lower limit: sign errors with (−2)³ = −8
• Not finding limits first — setting up ∫₀²(4−x²)dx without justification

Key phrases to include

• x-intercepts
• limits x = −2 to x = 2
• 32/3 square units
• parabola opens downward

Examiner tip

The LIATE rule (Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential) guides u selection. Choose u as the function that appears earlier in LIATE. Here, x (Algebraic) comes before eˣ (Exponential), so u = x.

Model answer

Using Integration by Parts: ∫ u dv = uv − ∫ v du

Let: u = x → du = dx
dv = eˣ dx → v = eˣ

∫ x·eˣ dx = x·eˣ − ∫ eˣ dx
= x·eˣ − eˣ + C
= eˣ(x − 1) + C

Answer structure

• Line 1: State the IBP formula ∫u dv = uv − ∫v du [1 mark]
• Line 2: Identify u, dv, du, and v explicitly [1 mark]
• Line 3: Substitute into formula, integrate ∫eˣdx, simplify, and add C [1 mark]

Scoring breakdown

Common mark deductions

• Choosing u = eˣ and dv = x dx — this is not wrong but leads to a more complex integral; show it simplifies
• Forgetting to integrate v after substitution — writing x·eˣ − eˣ·x instead of x·eˣ − eˣ
• Omitting constant C

Key phrases to include

• ∫u dv = uv − ∫v du
• u = x, dv = eˣdx
• du = dx, v = eˣ
• eˣ(x − 1) + C

Examiner tip

The most common error in disk/washer problems is forgetting to square the radius. Emphasize this by writing [R(x)]² explicitly before substituting — never shortcut this step.

Model answer

Given: y = x², revolved about the x-axis; limits x = 0 to x = 1
Method: Disk Method — V = π∫ₐᵇ [R(x)]² dx

Here R(x) = y = x²

V = π∫₀¹ (x²)² dx
= π∫₀¹ x⁴ dx
= π [x⁵/5]₀¹
= π [(1/5) − 0]
= π/5

∴ V = π/5 cubic units ≈ 0.628 cubic units

Answer structure

• Line 1: State disk method formula V = π∫[R(x)]²dx and identify R(x) = x² [1 mark]
• Line 2: Square the radius and integrate: π∫₀¹ x⁴ dx = π[x⁵/5]₀¹ [1 mark]
• Line 3: Evaluate and express as π/5 with units [1 mark]

Scoring breakdown

Common mark deductions

• Writing V = π∫ x² dx instead of π∫(x²)² dx — forgetting to square the radius
• Not including π in the final answer
• Not stating cubic units

Key phrases to include

• disk method
• V = π∫[R(x)]²dx
• R(x) = x²
• (x²)² = x⁴
• π/5 cubic units

Examiner tip

Always simplify [R(x)]² before integrating. Here (√x)² = x turns a root integral into a simple power integral. This algebraic simplification is worth showing explicitly.

Model answer

Given: y = √x = x^(1/2), revolved about the x-axis; x from 0 to 4
Method: Disk Method — V = π∫₀⁴ [R(x)]² dx

R(x) = √x, [R(x)]² = (√x)² = x

V = π∫₀⁴ x dx
= π [x²/2]₀⁴
= π [(16/2) − 0]
= π · 8
= 8π

∴ V = 8π cubic units ≈ 25.13 cubic units

Answer structure

• Line 1: Identify R(x) = √x and square it: [R(x)]² = x [1 mark]
• Line 2: Set up and integrate π∫₀⁴ x dx = π[x²/2]₀⁴ [1 mark]
• Line 3: Evaluate to get 8π cubic units [1 mark]

Scoring breakdown

Common mark deductions

• Not simplifying (√x)² to x — integrating π∫(√x)²dx as π∫x^(1/2)dx — treating the square as unresolved
• Evaluating [x²/2]₀⁴ as 4²/2 = 8 but forgetting to multiply by π

Key phrases to include

• (√x)² = x
• π∫₀⁴ x dx
• [x²/2]₀⁴
• 8π cubic units

Examiner tip

For a 5-mark long answer on volumes, show all five steps and consider adding a brief verification using the alternate method. A verified answer signals mathematical confidence and can earn the final mark even if the primary working has a minor error.

Model answer

Given: y = x², x = 0 to x = 2; Axis of revolution: y-axis
Method: Cylindrical Shell Method

Formula: V = 2π ∫ₐᵇ x · f(x) dx

Here f(x) = x², a = 0, b = 2

V = 2π ∫₀² x · x² dx
= 2π ∫₀² x³ dx
= 2π [x⁴/4]₀²
= 2π [(2⁴/4) − (0⁴/4)]
= 2π [(16/4) − 0]
= 2π · 4
= 8π

∴ V = 8π cubic units ≈ 25.13 cubic units

Verification (Disk method about y-axis):
Express x in terms of y: x = √y, limits y = 0 to y = 4
V = π∫₀⁴ (√y)² dy = π∫₀⁴ y dy = π[y²/2]₀⁴ = π·8 = 8π ✓

Answer structure

• Line 1: State method (cylindrical shells) and write the formula V = 2π∫x·f(x)dx [1 mark]
• Line 2: Identify f(x) = x², limits a = 0 to b = 2, and set up the integral [1 mark]
• Line 3: Multiply x·x² = x³ inside the integral [1 mark]
• Line 4: Integrate and evaluate [x⁴/4]₀² [1 mark]
• Line 5: State final answer 8π cubic units, with optional verification [1 mark]

Scoring breakdown

Common mark deductions

• Using disk method formula (π∫R²dx) when shell method is explicitly required
• Forgetting the 2π factor in the shell formula
• Using limits y = 0 to y = 4 in a shell integral (shell method integrates in x when revolving about y-axis)
• Not simplifying x·x² to x³ — integrating x·x² directly without algebraic preparation

Key phrases to include

• shell method
• V = 2π∫x·f(x)dx
• axis of revolution: y-axis
• 2π∫₀² x³ dx
• 8π cubic units

Examiner tip

The y/2 for the vertical strip centroid is the #1 source of error in centroid problems. Write 'y_el = y/2 for vertical strip' explicitly as a labeled note in your solution — it earns the concept mark and reminds you not to use y alone.

Model answer

Given: y = x², x from 0 to 2; Axis: x-axis (lower boundary)

Step 1 — Compute Total Area A:
A = ∫₀² y dx = ∫₀² x² dx = [x³/3]₀² = 8/3

Step 2 — Compute Moment about y-axis (for x̄):
Mᵧ = ∫₀² x · y dx = ∫₀² x · x² dx = ∫₀² x³ dx = [x⁴/4]₀² = 4

x̄ = Mᵧ/A = 4 ÷ (8/3) = 4 × (3/8) = 3/2 = 1.5

Step 3 — Compute Moment about x-axis (for ȳ):
For vertical strips, centroid of strip is at y_el = y/2
Mₓ = ∫₀² (y/2) · y dx = ∫₀² y²/2 dx = (1/2)∫₀² x⁴ dx = (1/2)[x⁵/5]₀² = (1/2)(32/5) = 16/5

ȳ = Mₓ/A = (16/5) ÷ (8/3) = (16/5) × (3/8) = 48/40 = 6/5 = 1.2

∴ Centroid: (x̄, ȳ) = (3/2, 6/5) = (1.5, 1.2)

Answer structure

• Step 1: Compute total area A = ∫₀² x² dx = 8/3 [1 mark]
• Step 2: Compute Mᵧ = ∫₀² x·x² dx = 4, then x̄ = Mᵧ/A = 3/2 [1 mark]
• Step 3: State that strip centroid is at y/2 for ȳ calculation [1 mark]
• Step 4: Compute Mₓ = ½∫₀² x⁴ dx = 16/5, then ȳ = Mₓ/A = 6/5 [1 mark]
• Step 5: State final centroid (x̄, ȳ) = (1.5, 1.2) [1 mark]

Scoring breakdown

Common mark deductions

• Using y instead of y/2 for the elemental centroid height in the ȳ calculation — the single most common centroid error in board exams
• Forgetting to divide moment by area to get centroid coordinate
• Setting up Mᵧ = ∫y·y dx instead of ∫x·y dx

Key phrases to include

• x̄ = ∫x dA / ∫dA
• ȳ = ∫y_el dA / ∫dA
• strip centroid at y/2
• Mᵧ = ∫x·y dx
• Mₓ = ∫(y/2)·y dx

Examiner tip

Pay careful attention to the double negative. When v = −cos(x), the IBP formula gives uv − ∫v du = −x cos x − ∫(−cos x)dx = −x cos x + ∫cos x dx. Write each sign change on its own line.

Model answer

Using Integration by Parts: ∫ u dv = uv − ∫ v du

Let: u = x → du = dx
dv = sin(x) dx → v = −cos(x)

∫ x·sin(x) dx = x·(−cos x) − ∫ (−cos x) dx
= −x·cos x + ∫ cos x dx
= −x·cos x + sin x + C

Answer structure

• Line 1: State IBP formula and choose u = x, dv = sin(x)dx [1 mark]
• Line 2: Derive du = dx, v = −cos(x) and substitute into formula [1 mark]
• Line 3: Integrate ∫cos(x)dx = sin(x) and write final answer with C [1 mark]

Scoring breakdown

Common mark deductions

• Taking v = cos(x) instead of v = −cos(x) — wrong sign on antiderivative of sin(x)
• Sign error in the double negative: −∫(−cos x)dx written as −∫cos x dx

Key phrases to include

• u = x, dv = sin(x)dx
• v = −cos(x)
• LIATE rule
• −x·cos(x) + sin(x) + C

Examiner tip

After applying u-substitution, verify that ALL x-terms including dx are replaced by u and du before integrating. A mixed-variable integral cannot be evaluated and signals an incomplete substitution.

Model answer

Let u = x² + 1 → du = 2x dx

Substitute: ∫ 2x·(x² + 1)⁴ dx = ∫ u⁴ du
= u⁵/5 + C
= (x² + 1)⁵/5 + C

Answer structure

• Line 1: Choose substitution u = x² + 1, derive du = 2x dx [1 mark]
• Line 2: Rewrite integral entirely in terms of u: ∫u⁴ du [1 mark]
• Line 3: Integrate to get u⁵/5 + C and back-substitute [1 mark]

Scoring breakdown

Common mark deductions

• Substituting u but not replacing dx — leaving the integral with mixed variables
• Forgetting to back-substitute and leaving the final answer as u⁵/5 + C
• Not writing + C

Key phrases to include

• u-substitution
• u = x² + 1
• du = 2x dx
• back-substitute
• (x² + 1)⁵/5 + C

Examiner tip

Remember 4^(3/2) = (4^(1/2))³ = 2³ = 8. Board exam questions on arc length almost always include a base that simplifies cleanly after substitution — this is by design. If your substitution does not simplify nicely, revisit your derivative.

Model answer

Given: y = (2/3)x^(3/2), x from 0 to 3

Arc length formula: L = ∫ₐᵇ √[1 + (dy/dx)²] dx

Step 1 — Differentiate:
dy/dx = (2/3)·(3/2)x^(1/2) = x^(1/2) = √x

Step 2 — Square the derivative:
(dy/dx)² = (√x)² = x

Step 3 — Set up the integral:
L = ∫₀³ √(1 + x) dx

Step 4 — Evaluate using substitution:
Let u = 1 + x → du = dx; when x = 0, u = 1; when x = 3, u = 4

L = ∫₁⁴ √u du = ∫₁⁴ u^(1/2) du
= [u^(3/2) / (3/2)]₁⁴
= (2/3)[u^(3/2)]₁⁴
= (2/3)[4^(3/2) − 1^(3/2)]
= (2/3)[8 − 1]
= (2/3)(7)
= 14/3

∴ L = 14/3 units ≈ 4.667 units

Answer structure

• Step 1: State arc length formula L = ∫√[1+(dy/dx)²]dx [1 mark]
• Step 2: Differentiate y to find dy/dx = √x and square it [1 mark]
• Step 3: Set up simplified integral ∫₀³ √(1+x) dx [1 mark]
• Step 4: Apply u-substitution with changed limits [1 mark]
• Step 5: Evaluate and state final answer 14/3 units [1 mark]

Scoring breakdown

Common mark deductions

• Using the formula without squaring dy/dx — writing √(1 + dy/dx) instead of √(1 + (dy/dx)²)
• Not changing the limits of integration when using u-substitution
• Computing 4^(3/2) incorrectly: 4^(3/2) = (√4)³ = 2³ = 8, not 4·4^(1/2)

Key phrases to include

• L = ∫√[1+(dy/dx)²]dx
• dy/dx = √x
• (dy/dx)² = x
• u = 1 + x, du = dx
• 14/3 units

Do

• Write + C for all indefinite integral answers
• State the numerical value clearly for definite integral VSAs
• Use correct notation: ln|x|, not ln(x), for logarithmic integrals
• Double-check the exponent in power rule answers

Don't

• Do not spend more than 1 minute on a 1-mark item
• Do not leave blank — attempt always; partial answer may still earn the mark
• Do not use decimal approximations when an exact form (e.g., π/5) is simpler and more correct

Do

• Show the antiderivative explicitly in bracket notation before substituting limits
• Write F(b) − F(a) as a separate line
• Include units (sq. units, cu. units) in area and volume answers

Don't

• Do not skip the intermediate step and jump to the final number
• Do not write F(a) − F(b); always upper limit minus lower limit
• Do not use approximate values mid-calculation; keep exact fractions

Do

• State which integration technique you are using before starting
• For area problems: identify upper and lower curves explicitly
• For IBP: write out u, dv, du, v as a 2×2 table or list before substituting
• Show the final numerical answer in a box or underline it

Don't

• Do not mix up the disk formula (π∫R²dx) with the volume formula for a cylinder
• Do not use y instead of y/2 for centroid of horizontal strip
• Do not skip the limit-substitution step — it earns its own mark

Do

• Write 'Given:', 'Required:' and 'Solution:' headings
• Number each step: Step 1, Step 2, Step 3...
• State your formula before every calculation
• Include a verification step for volumes (use alternate method) or check dimensions
• Box the final answer clearly

Don't

• Do not write a wall of unseparated algebra — examiners cannot trace your reasoning
• Do not assume limits without finding intersection points first
• Do not rush — a 5-mark question is worth the full 10–12 minutes
• Do not omit units — cubic metres or cubic units for volume, square units for area