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, exact answers, and every integrand is continuous where it is used. Notation: ∫ (a to x) f(t) dt means the definite integral of f(t) from t = a to t = x. The context in Question 7 is invented. This set is for both Calculus AB and Calculus BC students.
Question 1 (multiple choice · foundation)
Let g(x) = ∫ (4 to x) √(t + 5) dt. What is g′(4)?
- (A) 0
- (B) 1/6
- (C) 3
- (D) −3
Answer and explanation
Answer: (C). √(t + 5) is continuous for t ≥ −5, so by the theorem g′(x) = √(x + 5). Then g′(4) = √9 = 3.
- (A) is g(4), not g′(4). The interval from 4 to 4 has no area, but the slope there is not 0.
- (B) differentiates the integrand: d/dt √(t + 5) = 1/(2√(t + 5)), which is 1/6 at t = 4. The theorem gives back the integrand itself, not its derivative.
- (D) adds a minus sign that belongs only when x is the lower limit.
Question 2 (multiple choice · core)
What is d/dx ∫ (x to 2) cos(t²) dt?
- (A) cos(x²)
- (B) −cos(x²)
- (C) −2x sin(x²)
- (D) cos 4 − cos(x²)
Answer and explanation
Answer: (B). Swap the limits: ∫ (x to 2) cos(t²) dt = −∫ (2 to x) cos(t²) dt. By the theorem its derivative is −cos(x²).
- (A) misses the minus sign that comes from x being the lower limit.
- (C) is the derivative of cos(x²). It differentiates the integrand instead of applying the theorem.
- (D) substitutes the limits into the integrand and subtracts. The result still contains a constant term, cos 4, which a derivative of this kind never has.
Question 3 (multiple choice · core)
What is d/dx ∫ (0 to 2x) e^(t²) dt?
- (A) e^(4x²)
- (B) 2e^(4x²)
- (C) e^(4x²) − 1
- (D) 8x e^(4x²)
Answer and explanation
Answer: (B). The upper limit is u(x) = 2x, so u′(x) = 2. Substitute u into the integrand and multiply by u′: e^((2x)²) · 2 = 2e^(4x²).
- (A) forgets the chain rule factor u′(x) = 2.
- (C) substitutes both limits into the integrand and subtracts (e^(4x²) − e⁰). That is not how the theorem works.
- (D) is the derivative of e^(4x²): it differentiates the integrand after substituting.
Question 4 (multiple choice · core)
Let f(t) = 3 − t and g(x) = ∫ (1 to x) f(t) dt. What is g(4)?
- (A) −1
- (B) 3/2
- (C) 5/2
- (D) 4
Answer and explanation
Answer: (B). The line y = 3 − t crosses the axis at t = 3. From t = 1 to 3 it is above the axis: a triangle with base 2 and height 2, area 2. From 3 to 4 it is below: a triangle with base 1 and height 1, area 1/2, counted as negative. So g(4) = 2 − 1/2 = 3/2.
- (A) is f(4) = g′(4), a slope, not the accumulated value.
- (C) adds the two areas without signs: 2 + 1/2.
- (D) starts the area at t = 0 instead of t = 1: ∫ (0 to 4) (3 − t) dt = 9/2 − 1/2 = 4.
Question 5 (graph · core)
The graph of f on −2 ≤ t ≤ 5 is shown above. Let g(x) = ∫ (0 to x) f(t) dt.
(a) Find g(2), g(4) and g(5). (b) Find g(−2), and explain the sign of your answer. (c) Find g′(1) and g′(4.5). (d) Write an equation of the tangent line to the graph of g at x = 4.
Worked solution and suggested Marlbridge rubric
(a) From 0 to 2: a rectangle, 2 × 2 = 4, so g(2) = 4. From 2 to 4 the graph falls from 2 to 0: a triangle with area (1/2)(2)(2) = 2, so g(4) = 6. From 4 to 5 it is below the axis: a triangle with area (1/2)(1)(1) = 1/2, so g(5) = 6 − 1/2 = 11/2.
(b) g(−2) = ∫ (0 to −2) f(t) dt = −∫ (−2 to 0) f(t) dt. On [−2, 0] there is a triangle below the axis from −2 to −1 (signed area −1) and a triangle above from −1 to 0 (signed area +1). So ∫ (−2 to 0) f(t) dt = 0 and g(−2) = 0. The two triangles cancel, so reversing the direction makes no difference here.
(c) f is continuous, so g′ = f. g′(1) = f(1) = 2. On [2, 5], f(t) = 4 − t, so g′(4.5) = −0.5.
(d) g(4) = 6 and g′(4) = f(4) = 0. Tangent line: y = 6 (horizontal).
| Point | What earns it |
|---|---|
| 1 | g(2) = 4, g(4) = 6 and g(5) = 11/2, using signed areas |
| 1 | g(−2) = 0 with the reversal of limits shown or explained |
| 1 | g′(1) = 2 and g′(4.5) = −1/2, using g′ = f |
| 1 | Tangent line y = 6, from the point (4, 6) and slope 0 |
Question 6 (constructed response · core)
Let F(x) = ∫ (0 to x) (t − 1)/(t² + 1) dt.
(a) Explain why F is differentiable for every real x, and find F′(x). (b) Write an equation of the tangent line to the graph of F at x = 0. (c) Let G(x) = ∫ (0 to x²) (t − 1)/(t² + 1) dt. Find G′(x) and G′(2). (d) A student says G′(1) = 0 “because the upper limit is 1 and the integrand is 0 at t = 1”. The value is right. Is the reason complete? Explain.
Worked solution and suggested Marlbridge rubric
(a) The integrand is a quotient of polynomials whose denominator t² + 1 is never 0, so it is continuous for every real t. By the theorem, F is differentiable everywhere and F′(x) = (x − 1)/(x² + 1).
(b) F(0) = ∫ (0 to 0) … = 0 and F′(0) = (0 − 1)/(0 + 1) = −1. Tangent line: y = −x.
(c) G(x) = F(x²), so by the chain rule G′(x) = F′(x²) · 2x:
G′(x) = 2x(x² − 1)/(x⁴ + 1), and G′(2) = 4 · 3/17 = 12/17.
(d) Not complete. G′(1) = F′(1) · 2(1) = 0 · 2 = 0. The chain-rule factor 2x must be included; here the result is 0 only because the integrand is 0 at u(1) = 1. In general the factor u′(x) changes the answer, as G′(2) shows.
| Point | What earns it |
|---|---|
| 1 | Continuity of the integrand stated, and F′(x) = (x − 1)/(x² + 1) |
| 1 | y = −x, using F(0) = 0 and F′(0) = −1 |
| 1 | G′(x) with the factor 2x, and G′(2) = 12/17 |
| 1 | Explains that the chain-rule factor belongs in the derivative, even when the product is 0 |
Question 7 (constructed response · stretch)
Rainwater flows into a tank at a rate of r(t) = 20 + 6 sin(t/2) litres per hour, for 0 ≤ t ≤ 12, where t is in hours. At t = 0 the tank holds 500 litres. For 0 ≤ x ≤ 12, let
W(x) = 500 + ∫ (0 to x) r(t) dt
(a) Explain what W(x) represents, with units. (b) Find W′(x) and W′(π). Interpret W′(π) in context. (c) Using a short time interval from x to x + h, explain in words why W′(x) = r(x). (d) Let P(x) = ∫ (x to 12) r(t) dt. Say what P(x) means, find P′(π), and explain why its sign makes sense.
Worked solution and suggested Marlbridge rubric
(a) ∫ (0 to x) r(t) dt is the water (in litres) that flows in from time 0 to time x hours. Adding the starting 500 litres, W(x) is the volume of water in the tank, in litres, at time x hours.
(b) r is continuous, and the derivative of the constant 500 is 0, so W′(x) = r(x) = 20 + 6 sin(x/2). Then W′(π) = 20 + 6 sin(π/2) = 26 litres per hour. At t = π hours the volume of water in the tank is increasing at 26 litres per hour.
(c) From x to x + h, the extra water is W(x + h) − W(x). Over a short interval the inflow rate stays close to r(x), so the extra water is about r(x) · h litres. Dividing by h gives about r(x) litres per hour, and the approximation becomes exact as h → 0. So W′(x) = r(x).
(d) P(x) is the water, in litres, that will flow in from time x until t = 12 hours. Since x is the lower limit, P′(x) = −r(x), so P′(π) = −26 litres per hour. The sign makes sense: as time passes, less of the period is left, so the amount still to come decreases, at the rate water is arriving.
| Point | What earns it |
|---|---|
| 1 | W(x) as the amount of water in the tank at time x, in litres |
| 1 | W′(x) = r(x) and W′(π) = 26 litres per hour, with interpretation |
| 1 | Short-interval argument: extra water ≈ r(x) · h, divide by h, let h → 0 |
| 1 | Meaning of P(x), P′(π) = −26, and a reason for the negative sign |
How did you do?
- Q1 or Q6(a) wrong: reread “The Fundamental Theorem of Calculus (accumulation form)” in the study guide.
- Q2, Q3 or Q6(c) wrong: redo Worked example 2 and the “Variations you must recognise” table.
- Q4 or Q5 wrong: redo Worked example 1 with Figure 2, paying attention to signs.
- Q6(b) or Q7 wrong: see Worked example 3 and “Why it is true” with Figure 1.
Then tick off the topic checklist.