These are original Marlbridge practice questions, not past exam questions. The rubrics are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. Assumptions for every question: no calculator, angles in radians and exact answers unless stated. Notation: ∫ (a to b) f(x) dx is the definite integral from a to b, and [F(x)] (a to b) means F(b) − F(a). The context in Question 7 is invented. This set is for both Calculus AB and Calculus BC students.
Question 1 (multiple choice · foundation)
What is ∫ (1 to 2) (6x² − 4x + 1) dx?
- (A) −9
- (B) 9
- (C) 10
- (D) 14
Answer and explanation
Answer: (B). An antiderivative is F(x) = 2x³ − 2x² + x (check: F′(x) = 6x² − 4x + 1). Then F(2) − F(1) = (16 − 8 + 2) − (2 − 2 + 1) = 10 − 1 = 9.
- (A) subtracts in the wrong order: F(1) − F(2) = −9.
- (C) is F(2) alone. It forgets to subtract F(1), which is not 0 here.
- (D) substitutes into the integrand instead of an antiderivative: f(2) − f(1) = 17 − 3 = 14.
Question 2 (multiple choice · core)
What is ∫ (0 to π/3) 4 sin x dx?
- (A) −6
- (B) −2
- (C) 2
- (D) 2√3
Answer and explanation
Answer: (C). An antiderivative of 4 sin x is −4 cos x. So the integral is [−4 cos x] (0 to π/3) = (−4 · 1/2) − (−4 · 1) = −2 + 4 = 2.
- (A) drops the brackets around the second value: −4 cos(π/3) − 4 cos 0 = −2 − 4 = −6.
- (B) uses 4 cos x as the antiderivative (a sign error): 4 · 1/2 − 4 · 1 = −2. Differentiating 4 cos x gives −4 sin x, so the check fails.
- (D) substitutes into the integrand: 4 sin(π/3) − 4 sin 0 = 2√3.
Question 3 (multiple choice · core)
f is continuous for all real numbers. F is an antiderivative of f with F(5) = 3 and F(9) = 11. Let G(x) = ∫ (5 to x) f(t) dt. What is G(9)?
- (A) −8
- (B) 8
- (C) 11
- (D) 14
Answer and explanation
Answer: (B). G(9) = ∫ (5 to 9) f(t) dt. f is continuous and F is an antiderivative, so this is F(9) − F(5) = 11 − 3 = 8. (Equivalently, G is the antiderivative of f with G(5) = 0, so G(x) = F(x) − 3.)
- (A) subtracts in the wrong order.
- (C) assumes G and F are the same antiderivative. They differ by a constant: G(5) = 0 but F(5) = 3.
- (D) adds the two values instead of subtracting.
Question 4 (multiple choice · core)
What is lim (n → ∞) Σ (i = 1 to n) (2/n) e^(2i/n)?
- (A) e² − 1
- (B) e²
- (C) 2e² − 2
- (D) 1 − e²
Answer and explanation
Answer: (A). Δx = 2/n and xᵢ = 2i/n, so x runs from 0 to 2 and f(x) = eˣ. The limit is ∫ (0 to 2) eˣ dx = [eˣ] (0 to 2) = e² − e⁰ = e² − 1 (about 6.389). With n = 1000 the sum is about 6.395, which agrees.
- (B) treats e⁰ as 0, or forgets the lower limit.
- (C) multiplies the integral by 2 as well. The factor 2/n is Δx, which becomes the dx in the integral; it is not an extra constant.
- (D) subtracts in the wrong order.
Question 5 (constructed response · core)
A student writes: ∫ (π/4 to 3π/4) sec²x dx = [tan x] (π/4 to 3π/4) = tan(3π/4) − tan(π/4) = −1 − 1 = −2.
(a) Explain why the answer −2 cannot be correct, using the sign of the integrand. (b) State which condition of the Fundamental Theorem of Calculus fails, and where. (c) Evaluate ∫ (−π/4 to π/3) sec²x dx exactly, and explain why the theorem does apply this time.
Worked solution and suggested Marlbridge rubric
(a) sec²x = 1/cos²x is positive wherever it is defined. A region that lies entirely above the axis has a positive signed area, so the integral cannot be negative. (In fact the region is unbounded, so no finite value is possible.)
(b) The theorem needs sec²x to be continuous on the whole of [π/4, 3π/4]. It is not: cos(π/2) = 0, so sec²x is undefined at x = π/2, which lies inside the interval. The subtraction tan(3π/4) − tan(π/4) has no meaning as the value of this integral.
(c) On [−π/4, π/3], cos x ≥ 1/2 > 0, so sec²x is continuous there. tan x is an antiderivative (d/dx tan x = sec²x). So
∫ (−π/4 to π/3) sec²x dx = [tan x] (−π/4 to π/3) = √3 − (−1) = √3 + 1.
| Point | What earns it |
|---|---|
| 1 | Integrand is positive (where defined), so the integral cannot be negative |
| 1 | Identifies that sec²x is not continuous on [π/4, 3π/4], undefined at x = π/2 |
| 1 | Justifies continuity on [−π/4, π/3] (cos x is not 0 there) |
| 1 | Correct antiderivative and value √3 + 1 |
Question 6 (constructed response · core)
Let f(x) = 3x² − 6x.
(a) Evaluate ∫ (0 to 3) f(x) dx. (b) The answer to (a) is 0, but f(x) is not 0 on (0, 3). Explain, using ∫ (0 to 2) f(x) dx and ∫ (2 to 3) f(x) dx. (c) Find the total area of the regions between the graph of f and the x-axis for 0 ≤ x ≤ 3.
Worked solution and suggested Marlbridge rubric
(a) F(x) = x³ − 3x² is an antiderivative (F′ = 3x² − 6x). f is a polynomial, so it is continuous.
∫ (0 to 3) f(x) dx = [x³ − 3x²] (0 to 3) = (27 − 27) − 0 = 0.
(b) f(x) = 3x(x − 2) is negative on (0, 2) and positive on (2, 3).
∫ (0 to 2) f(x) dx = F(2) − F(0) = (8 − 12) − 0 = −4. ∫ (2 to 3) f(x) dx = F(3) − F(2) = 0 − (−4) = 4.
The region below the axis and the region above have equal areas, so their signed values cancel.
(c) Total area counts each region as positive: 4 + 4 = 8.
| Point | What earns it |
|---|---|
| 1 | Correct antiderivative x³ − 3x² and value 0 |
| 1 | Sign of f on each subinterval, from the factorisation or a sign check |
| 1 | Values −4 and 4, with the cancellation explained |
| 1 | Total area 8 |
Question 7 (constructed response · stretch)
The depth of water in a garden pond is D(t) centimetres, where t is in days, 0 ≤ t ≤ 9. The depth changes at the rate
D′(t) = 0.6√t − 0.3t centimetres per day.
At t = 0 the depth is 120 cm.
(a) Find ∫ (0 to 9) D′(t) dt. Give its meaning in context, with units. (b) Find D(9). (c) Find D(4). (d) Find the greatest depth of the pond for 0 ≤ t ≤ 9, and justify your answer.
Worked solution and suggested Marlbridge rubric
(a) Write D′(t) = 0.6t^(1/2) − 0.3t. An antiderivative is 0.4t^(3/2) − 0.15t² (check: 0.4 × 1.5 = 0.6 and 0.15 × 2 = 0.3). D′ is continuous on [0, 9].
∫ (0 to 9) D′(t) dt = [0.4t^(3/2) − 0.15t²] (0 to 9) = 0.4(27) − 0.15(81) − 0 = 10.8 − 12.15 = −1.35
Between day 0 and day 9, the depth of the pond decreases by a net 1.35 cm.
(b) D(9) = D(0) + ∫ (0 to 9) D′(t) dt = 120 − 1.35 = 118.65 cm.
(c) ∫ (0 to 4) D′(t) dt = 0.4(8) − 0.15(16) = 3.2 − 2.4 = 0.8, so D(4) = 120.8 cm.
(d) D′(t) = 0 when 0.6√t = 0.3t, so √t = 2 (or t = 0), giving t = 4. D′(1) = 0.3 > 0 and D′(9) = −0.9 < 0, so D increases on (0, 4) and decreases on (4, 9). Compare the candidates: D(0) = 120, D(4) = 120.8, D(9) = 118.65. The greatest depth is 120.8 cm, at t = 4 days.
| Point | What earns it |
|---|---|
| 1 | Correct antiderivative and −1.35 |
| 1 | Interpretation: net decrease of 1.35 cm in depth over the 9 days |
| 1 | D(9) = 118.65 and D(4) = 120.8, each as 120 plus an integral |
| 1 | Critical point t = 4 with a sign change or candidates comparison, and maximum 120.8 cm |
Acceptable alternative for (d): a candidates test (Topic 5.5) comparing D at 0, 4 and 9, without the sign analysis.
How did you do?
- Q1 or Q2 wrong: redo Worked example 1 and Worked example 2 in the study guide, using brackets for F(a).
- Q3 wrong: reread “Where the theorem comes from”.
- Q4 wrong: see “Reading a limit of Riemann sums as an integral”.
- Q5 wrong: reread “When the theorem does not apply”.
- Q6 or Q7 wrong: redo Worked example 3 and the net change section, paying attention to signs and units.
Then tick off the topic checklist.