What this is for. Use this diagnostic to find which topics of Unit 6, Integration and Accumulation of Change, to revisit: one question per topic. These are original Marlbridge practice questions, not past exam questions, with invented data. They are not calibrated, and your result is not a predicted score.
Rules. No calculator; about 30 minutes. Answer everything before opening the answers. The unit is shared by Calculus AB and Calculus BC. Questions 11, 12 and 13 are BC only (Topics 6.11 to 6.13); Calculus AB students skip them.
Question 1 (multiple choice · 6.1)
The depth of snow on a roof changes at a rate of s(t) cm per hour, t hours after midnight. The graph of s is made of straight segments joining (0, 0), (3, 6), (5, 6), (6, 0) and (8, −4). At t = 0 the snow is 10 cm deep. How deep is it at t = 8?
- (A) 20 cm
- (B) 30 cm
- (C) 38 cm
- (D) 34 cm
Answer and explanation
Answer: (B). Above the axis: 9 + 12 + 3 = 24. From t = 6 to 8 the region is below the axis, with area 4, so it counts as −4. Net change = 20 cm, so the depth is 10 + 20 = 30 cm.
- (A) leaves out the starting depth.
- (C) adds the area below the axis.
- (D) stops at t = 6.
If you missed this: Guide 6.1.
Question 2 (multiple choice · 6.2)
Water flows into a pool at an increasing rate R(t) litres per minute.
| t (minutes) | 0 | 3 | 4 | 8 |
|---|---|---|---|---|
| R(t) (litres per minute) | 2 | 5 | 6 | 9 |
Which is true of the right Riemann sum for ∫ (0 to 8) R(t) dt with the table’s three subintervals?
- (A) It is 57, and it is an overestimate.
- (B) It is 57, and it is an underestimate.
- (C) It is 35, and it is an overestimate.
- (D) It is 160/3, and it is an overestimate.
Answer and explanation
Answer: (A). The widths are 3, 1 and 4, so the right sum is 5(3) + 6(1) + 9(4) = 57 litres. R is increasing, so each right-hand height is the highest on its strip: 57 is an overestimate.
- (B) reverses the rule.
- (C) is the left sum, an underestimate.
- (D) uses equal widths of 8/3; the table’s gaps are not equal.
If you missed this: Guide 6.2.
Question 3 (multiple choice · 6.3)
What is the value of lim (n → ∞) Σ (i = 1 to n) (2/n) · 1/(1 + 2i/n)²?
- (A) 4/3
- (B) −2/3
- (C) 2/3
- (D) ln 3
Answer and explanation
Answer: (C). Δx = 2/n and xᵢ = 1 + 2i/n runs from 1 to 3, so the limit is ∫ (1 to 3) 1/x² dx = [−1/x] (1 to 3) = 2/3.
- (A) counts the width 2/n twice.
- (B) has a sign slip in the antiderivative.
- (D) integrates 1/x, not 1/x².
If you missed this: Guide 6.3.
Question 4 (multiple choice · 6.4)
Let G(x) = ∫ (1 to x²) √(t² + 9) dt. What is G′(2)?
- (A) 5
- (B) 4√13
- (C) 20 − √10
- (D) 20
Answer and explanation
Answer: (D). By the Fundamental Theorem and the chain rule, G′(x) = √(x⁴ + 9) · 2x, so G′(2) = 5 × 4 = 20.
- (A) leaves out the chain factor 2x.
- (B) puts x, not x², into the integrand.
- (C) subtracts the integrand at the constant lower limit.
If you missed this: Guide 6.4.
Question 5 (short answer · 6.5)
The graph of a continuous function f, for 0 ≤ t ≤ 6, is made of straight segments joining (0, 2), (1, 0), (3, −2), (4, 0) and (6, 2). Let g(x) = 5 + ∫ (1 to x) f(t) dt.
(a) Find g(4) and g(6). (b) Find the x-value of each relative extremum of g on (0, 6), and classify it. Justify. (c) On what interval is the graph of g concave down? Give a reason.
Worked answer
(a) From 1 to 4 the graph is a triangle below the axis with area 3, so g(4) = 5 − 3 = 2. From 4 to 6 a triangle above the axis has area 2, so g(6) = 4.
(b) g′ = f. f changes from positive to negative at x = 1: relative maximum at x = 1. f changes from negative to positive at x = 4: relative minimum at x = 4.
(c) g″ = f′, and f is decreasing on (0, 3), so g is concave down on (0, 3).
If you missed this: Guide 6.5.
Question 6 (multiple choice · 6.6)
f is continuous, ∫ (0 to 5) f(x) dx = 7 and ∫ (2 to 5) f(x) dx = 10. What is ∫ (2 to 0) [f(x) + 4] dx?
- (A) 5
- (B) −5
- (C) −1
- (D) −25
Answer and explanation
Answer: (B). ∫ (0 to 2) f(x) dx = 7 − 10 = −3, so ∫ (0 to 2) [f(x) + 4] dx = −3 + 8 = 5. Reversing the limits gives −5.
- (A) forgets to reverse the sign.
- (C) adds 4, not 4 × 2.
- (D) adds the given integrals.
If you missed this: Guide 6.6.
Question 7 (multiple choice · 6.7)
A pump removes water from a cellar at r(t) = 12 − 3√t litres per minute, t minutes after it starts. How much water does it remove from t = 1 to t = 4?
- (A) 32 litres
- (B) −3 litres
- (C) 22 litres
- (D) 50 litres
Answer and explanation
Answer: (C). An antiderivative is F(t) = 12t − 2t^(3/2). F(4) − F(1) = (48 − 16) − (12 − 2) = 32 − 10 = 22 litres.
- (A) is F(4) only; it leaves out F(1).
- (B) is r(4) − r(1), the change in the rate.
- (D) uses +2t^(3/2) in the antiderivative, so it gets 36 + 14 instead of 36 − 14.
If you missed this: Guide 6.7.
Question 8 (multiple choice · 6.8)
What is ∫ (4/√(1 − x²) − 2ˣ) dx?
- (A) 4 arcsin x − 2ˣ ln 2 + C
- (B) 4 arctan x − 2ˣ/ln 2 + C
- (C) 4 arcsin x − 2^(x + 1)/(x + 1) + C
- (D) 4 arcsin x − 2ˣ/ln 2 + C
Answer and explanation
Answer: (D). d/dx arcsin x = 1/√(1 − x²) and d/dx (2ˣ/ln 2) = 2ˣ.
- (A) uses the derivative rule for 2ˣ.
- (B) mixes up arcsin and arctan.
- (C) uses the power rule on 2ˣ.
If you missed this: Guide 6.8.
Question 9 (short answer · 6.9)
(a) Find ∫ (ln x)³/x dx. (b) Evaluate ∫ (0 to π/3) sin x/cos³x dx exactly. Show the new limits.
Worked answer
(a) Let u = ln x, so du = (1/x) dx. The integral becomes ∫ u³ du = u⁴/4 + C = (ln x)⁴/4 + C.
(b) Let u = cos x, so du = −sin x dx. x = 0 gives u = 1, and x = π/3 gives u = 1/2. The integral becomes −∫ (1 to 1/2) u⁻³ du = ∫ (1/2 to 1) u⁻³ du = [−1/(2u²)] (1/2 to 1) = −1/2 + 2 = 3/2.
If you missed this: Guide 6.9.
Question 10 (multiple choice · 6.10)
What is ∫ (x² + 7)/(x² + 4) dx?
- (A) x + 3 arctan(x/2) + C
- (B) x + (3/2) arctan(x/2) + C
- (C) x + (7/2) arctan(x/2) + C
- (D) x + (3/2) ln(x² + 4) + C
Answer and explanation
Answer: (B). Equal degrees, so divide first: (x² + 7)/(x² + 4) = 1 + 3/(x² + 4), and ∫ 3/(x² + 4) dx = (3/2) arctan(x/2).
- (A) forgets the factor 1/2 from x² + 2².
- (C) uses 7 as the remainder.
- (D) uses ln, but 3 is not a multiple of 2x.
If you missed this: Guide 6.10.
Question 11 (multiple choice · 6.11) (BC only)
What is ∫ (1 to e) x² ln x dx?
- (A) (2e³ − 1)/9
- (B) e³/3
- (C) (2e³ + 1)/9
- (D) 2e³/9
Answer and explanation
Answer: (C). Take u = ln x and dv = x² dx, so du = (1/x) dx and v = x³/3. Then ∫ x² ln x dx = (x³/3) ln x − ∫ x²/3 dx = (x³/3) ln x − x³/9. From 1 to e: (e³/3 − e³/9) − (0 − 1/9) = (2e³ + 1)/9.
- (A) has a sign slip at the lower limit.
- (B) stops after the uv term.
- (D) forgets the value at the lower limit.
If you missed this: Guide 6.11.
Question 12 (multiple choice · 6.12) (BC only)
What is ∫ (x + 7)/(x² − x − 6) dx?
- (A) 2 ln|x + 2| − ln|x − 3| + C
- (B) 2 ln|x − 3| + ln|x + 2| + C
- (C) ln|x² − x − 6| + C
- (D) 2 ln|x − 3| − ln|x + 2| + C
Answer and explanation
Answer: (D). x² − x − 6 = (x − 3)(x + 2). From x + 7 = A(x + 2) + B(x − 3): x = 3 gives A = 2, and x = −2 gives B = −1.
- (A) swaps the constants.
- (B) has a sign slip in B.
- (C) would need the top to be 2x − 1.
If you missed this: Guide 6.12.
Question 13 (multiple choice · 6.13) (BC only)
What is ∫ (e to ∞) 1/(x (ln x)²) dx?
- (A) 1
- (B) −1
- (C) 1/e
- (D) It diverges.
Answer and explanation
Answer: (A). With u = ln x, an antiderivative is −1/ln x. So the integral is lim (b → ∞) [−1/ln b + 1/ln e] = 0 + 1 = 1.
- (B) has the sign of the antiderivative wrong.
- (C) uses e in place of ln e = 1 at the lower limit.
- (D) is true for 1/(x ln x), not for 1/(x (ln x)²).
If you missed this: Guide 6.13.
Question 14 (short answer · 6.14)
For each integral, name the technique you will use and find the integral.
(a) ∫ 6x/(x² + 25) dx (b) ∫ 6/(x² + 25) dx (c) ∫ (2x² + 6x)/(x² + 25) dx
Worked answer
(a) The top is a multiple of the bottom’s derivative, 2x: substitution, u = x² + 25, so 6x dx = 3 du. 3 ln(x² + 25) + C (no absolute value needed, as x² + 25 > 0).
(b) A constant over x² + 5²: the arctan pattern. (6/5) arctan(x/5) + C.
(c) Equal degrees: long division first, giving 2 + (6x − 50)/(x² + 25). Split the fraction: 6x/(x² + 25) by substitution, −50/(x² + 25) by arctan. 2x + 3 ln(x² + 25) − 10 arctan(x/5) + C.
If you missed this: Guide 6.14.
Your next step
| Topic | Question | If you missed it, read |
|---|---|---|
| 6.1 Accumulation of change | 1 | Guide 6.1 |
| 6.2 Riemann sums | 2 | Guide 6.2 |
| 6.3 Sigma and integral notation | 3 | Guide 6.3 |
| 6.4 Accumulation functions | 4 | Guide 6.4 |
| 6.5 Their behaviour | 5 | Guide 6.5 |
| 6.6 Integral properties | 6 | Guide 6.6 |
| 6.7 Evaluating integrals | 7 | Guide 6.7 |
| 6.8 Basic antiderivatives | 8 | Guide 6.8 |
| 6.9 Substitution | 9 | Guide 6.9 |
| 6.10 Division, completing the square | 10 | Guide 6.10 |
| 6.11 Parts (BC only) | 11 | Guide 6.11 |
| 6.12 Partial fractions (BC only) | 12 | Guide 6.12 |
| 6.13 Improper integrals (BC only) | 13 | Guide 6.13 |
| 6.14 Choosing a technique | 14 | Guide 6.14 |
How to use your result
- Mark each topic secure, shaky (unsure or a slip) or gap (wrong).
- Fix Topics 6.1 and 6.4 first. Most of the unit builds on them.
- Then fix techniques in order: 6.8, 6.9, 6.10. Topic 6.14 needs them all.
- Check your reasons in Questions 5, 9 and 14, not just the values.
- For a gap, read the guide and do the topic’s practice set, then try the Unit 6 mixed review, where each question combines topics.