GELE Mathematics — Integral CalculusExam Answer Templates

Marks

1

Topic

Basic Integration — Power Rule

Difficulty

easy

Template Id

T1

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

Question Type

very_short_answer

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

Marks

1

Topic

Basic Integration — Logarithmic Rule

Difficulty

easy

Template Id

T2

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

Question Type

very_short_answer

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

Marks

2

Topic

Basic Integration — Polynomial Functions

Difficulty

easy

Template Id

T3

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

Question Type

short_answer

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

Marks

2

Topic

Definite Integrals

Difficulty

easy

Template Id

T4

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

Question Type

numerical

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

Marks

2

Topic

Definite Integrals

Difficulty

easy

Template Id

T5

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

Question Type

numerical

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

Marks

3

Topic

Area Between Curves

Difficulty

medium

Template Id

T6

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

Question Type

numerical

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

Marks

3

Topic

Definite Integrals and Area

Difficulty

medium

Template Id

T7

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

Question Type

numerical

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

Marks

3

Topic

Integration Techniques — Integration by Parts

Difficulty

medium

Template Id

T8

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

Question Type

numerical

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

Marks

3

Topic

Volumes of Revolution — Disk Method

Difficulty

medium

Template Id

T9

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

Question Type

numerical

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

Marks

3

Topic

Volumes of Revolution — Disk Method

Difficulty

medium

Template Id

T10

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

Question Type

numerical

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

Marks

5

Topic

Volumes of Revolution — Shell Method

Difficulty

hard

Template Id

T11

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 ✓

Question Type

long_answer

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

Marks

5

Topic

Centroids by Integration

Difficulty

hard

Template Id

T12

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)

Question Type

long_answer

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

Marks

3

Topic

Integration Techniques — Integration by Parts

Difficulty

medium

Template Id

T13

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

Question Type

numerical

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

Marks

3

Topic

Integration Techniques — Substitution

Difficulty

medium

Template Id

T14

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

Question Type

numerical

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

Marks

5

Topic

Arc Length by Integration

Difficulty

hard

Template Id

T15

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

Question Type

long_answer

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