What this is for. Use this diagnostic to find which topics of Unit 8, Applications of Integration, to revisit. There is one question per topic. These are original Marlbridge practice questions, not past exam questions; contexts are invented. 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 1 to 12 are for both courses, and Question 13 is BC only (Topic 8.13, arc length).
Question 1 (multiple choice · 8.1)
What is the average value of f(x) = cos x on the interval [0, π/3]?
- (A) √3/2
- (B) 3√3/(2π)
- (C) −3/(2π)
- (D) 3/4
Answer and explanation
Answer: (B). f_avg = (3/π) ∫ (0 to π/3) cos x dx = (3/π)(√3/2) = 3√3/(2π), about 0.827.
- (A) forgets to divide by the length π/3.
- (C) is the average rate of change, (cos(π/3) − cos 0)/(π/3).
- (D) averages the two end values, which works only for linear functions.
If you missed this: Topic 8.1 study guide.
Question 2 (multiple choice · 8.2)
A particle moves along a line with velocity v(t) = t² − 4t + 3 for 0 ≤ t ≤ 3. At t = 0 it is at x = 2. What is the total distance it travels over 0 ≤ t ≤ 3?
- (A) 0
- (B) 4/3
- (C) 2
- (D) 8/3
Answer and explanation
Answer: (D). v(t) = (t − 1)(t − 3) changes sign at t = 1, so split there: ∫ (0 to 1) v dt = 4/3 and ∫ (1 to 3) v dt = −4/3. Total distance = 4/3 + 4/3 = 8/3.
- (A) is the displacement; forward and backward parts cancel.
- (B) counts only the first part.
- (C) is the final position x(3) = 2 + 0.
If you missed this: Topic 8.2 study guide.
Question 3 (short answer · 8.3)
The volume of water in a reservoir is V(t) thousand cubic metres, t days after a storm, for 0 ≤ t ≤ 8. V(0) = 200, and V changes at the rate r(t) = 30 − 6t thousand m³ per day.
(a) Find V(8). (b) Find the greatest volume on 0 ≤ t ≤ 8. Justify. (c) Find ∫ (5 to 8) r(t) dt and explain its meaning in context.
Worked answer
(a) V(8) = 200 + ∫ (0 to 8) (30 − 6t) dt = 200 + (240 − 192) = 248 thousand m³.
(b) r changes from positive to negative at t = 5, so V rises then falls. Candidates: V(0) = 200, V(5) = 275 and V(8) = 248. The greatest volume is 275 thousand m³, at t = 5.
(c) ∫ (5 to 8) (30 − 6t) dt = −27: from day 5 to day 8, the volume of water fell by 27 thousand cubic metres. It is a net change, not an amount.
If you missed this: Topic 8.3 study guide.
Question 4 (multiple choice · 8.4)
What is the area of the region enclosed by y = √x and y = x³?
- (A) 5/12
- (B) −5/12
- (C) 11/12
- (D) 2/3
Answer and explanation
Answer: (A). The curves meet at x = 0 and x = 1, and √x is on top between them (test x = 1/4). Area = ∫ (0 to 1) (√x − x³) dx = 2/3 − 1/4 = 5/12.
- (B) subtracts top from bottom.
- (C) adds the curves instead of subtracting.
- (D) is the area under √x only; it forgets the bottom curve.
If you missed this: Topic 8.4 study guide.
Question 5 (multiple choice · 8.5)
R is the region bounded by y = √x, the line y = x − 2 and the x-axis. Using horizontal strips, what is the area of R?
- (A) 10/3
- (B) 9/2
- (C) 16/3
- (D) −10/3
Answer and explanation
Answer: (A). As x in terms of y, the curve is x = y² (left) and the line is x = y + 2 (right). They meet where y² = y + 2, at y = 2 (y = −1 is below the x-axis). Area = ∫ (0 to 2) [(y + 2) − y²] dy = 10/3, one integral where vertical strips need two.
- (B) uses −1 to 2 as limits, ignoring the x-axis.
- (C) is the area under √x from 0 to 4, triangle included.
- (D) uses left minus right.
If you missed this: Topic 8.5 study guide.
Question 6 (multiple choice · 8.6)
What is the total area of the regions enclosed by y = x³ − x and y = 3x?
- (A) 0
- (B) 4
- (C) 8
- (D) −4
Answer and explanation
Answer: (C). x³ − x = 3x gives x(x² − 4) = 0: crossings at x = −2, 0 and 2. ∫ (−2 to 0) (x³ − 4x) dx = 4 and ∫ (0 to 2) (x³ − 4x) dx = −4. Area = 4 + 4 = 8.
- (A) is the net integral; the pieces cancel.
- (B) finds only one region.
- (D) is one region with bottom minus top.
If you missed this: Topic 8.6 study guide.
Question 7 (multiple choice · 8.7)
The base of a solid is the region bounded by y = 2√x, the line y = 4 and the y-axis. Cross sections perpendicular to the y-axis are squares. What is the volume?
- (A) 16/3
- (B) 32
- (C) 64/5
- (D) 64π/5
Answer and explanation
Answer: (C). Slices perpendicular to the y-axis are horizontal, so work in y. The curve is x = y²/4, so the side is s(y) = y²/4. V = ∫ (0 to 4) (y²/4)² dy = (1/16)(4⁵/5) = 64/5.
- (A) is the base area; the side is not squared.
- (B) slices across x with side 2√x: wrong direction and wrong region.
- (D) adds π, which squares do not have.
If you missed this: Topic 8.7 study guide.
Question 8 (short answer · 8.8)
The base of a solid is the region R enclosed by y = 3x − x² and y = x.
(a) Find the limits of integration and the length s(x) of a vertical segment across R. (b) Cross sections perpendicular to the x-axis are semicircles with diameters in R. Find the volume. (c) Cross sections perpendicular to the x-axis are instead isosceles right triangles, each with its hypotenuse in R. Find the volume.
Worked answer
(a) 3x − x² = x gives x = 0 and x = 2. The parabola is on top, so s(x) = 2x − x².
(b) A semicircle with diameter s has area (1/2)π(s/2)² = (π/8)s². ∫ (0 to 2) (2x − x²)² dx = 16/15, so V = (π/8)(16/15) = 2π/15.
(c) Each leg is s/√2, so the area is s²/4 and V = (1/4)(16/15) = 4/15.
Using s as the radius gives 8π/15, four times too big.
If you missed this: Topic 8.8 study guide.
Question 9 (multiple choice · 8.9)
The region bounded by y = ln x, the y-axis, the x-axis and the line y = 1 is revolved around the y-axis. What is the volume?
- (A) π(e − 1)
- (B) π(e² − 1)/2
- (C) π(e − 2)
- (D) (e² − 1)/2
Answer and explanation
Answer: (B). The axis is vertical, so work in y. The radius is x = eʸ. V = π ∫ (0 to 1) (eʸ)² dy = π[e^(2y)/2] (0 to 1) = π(e² − 1)/2.
- (A) forgets to square the radius.
- (C) is π ∫ (1 to e) (ln x)² dx, discs about the x-axis.
- (D) forgets π.
If you missed this: Topic 8.9 study guide.
Question 10 (multiple choice · 8.10)
The region bounded by y = 1/x, the line y = 1/2 and the line x = 1 is revolved around the line y = 1/2. What is the volume?
- (A) π/2
- (B) π/4
- (C) π(3/4 − ln 2)
- (D) π(3/4 + ln 2)
Answer and explanation
Answer: (C). The curve meets y = 1/2 at x = 2, and the radius is 1/x − 1/2. V = π ∫ (1 to 2) (1/x − 1/2)² dx = π ∫ (1 to 2) (1/x² − 1/x + 1/4) dx = π(1/2 − ln 2 + 1/4) = π(3/4 − ln 2), about 0.179.
- (A) uses radius 1/x, as if spinning about the x-axis.
- (B) squares each term separately.
- (D) loses the minus sign on the middle term.
If you missed this: Topic 8.10 study guide.
Question 11 (multiple choice · 8.11)
The region enclosed by y = x and y = √x is revolved around the y-axis. What is the volume?
- (A) 2π/15
- (B) π/30
- (C) π/6
- (D) 2/15
Answer and explanation
Answer: (A). Work in y. For 0 < y < 1, x = y is farther from the axis than x = y², so R(y) = y and r(y) = y². V = π ∫ (0 to 1) (y² − y⁴) dy = π(1/3 − 1/5) = 2π/15.
- (B) uses π(R − r)² instead of π(R² − r²).
- (C) revolves around the x-axis.
- (D) forgets π.
If you missed this: Topic 8.11 study guide.
Question 12 (short answer · 8.12)
The region enclosed by y = 2x and y = x² is revolved around the line x = 3.
(a) Write each boundary as x in terms of y, and say which is on the left. (b) Find R(y) and r(y), with a reason. (c) Write and evaluate an integral for the volume.
Worked answer
(a) x = y/2 and x = √y, for 0 ≤ y ≤ 4. At y = 1 these give 1/2 and 1, so x = y/2 is on the left.
(b) The axis x = 3 is to the right, so the left curve is farther away: R(y) = 3 − y/2 and r(y) = 3 − √y.
(c) V = π ∫ (0 to 4) [(3 − y/2)² − (3 − √y)²] dy = π ∫ (0 to 4) (6√y − 4y + y²/4) dy = π(32 − 32 + 16/3) = 16π/3.
If you got −16π/3, you swapped R and r.
If you missed this: Topic 8.12 study guide.
Question 13 (multiple choice · 8.13) (BC only)
Which integral gives the length of the curve y = e^(2x) from x = 0 to x = ln 2?
- (A) ∫ (0 to ln 2) √(1 + 4e^(4x)) dx
- (B) ∫ (0 to ln 2) √(1 + e^(4x)) dx
- (C) ∫ (0 to ln 2) √(1 + 2e^(2x)) dx
- (D) ∫ (1 to 4) √(1 + 4e^(4x)) dx
Answer and explanation
Answer: (A). The length is ∫ √(1 + (f′(x))²) dx, with f′(x) = 2e^(2x), so (f′(x))² = 4e^(4x). (Its value is about 3.091.)
- (B) drops the chain rule factor 2.
- (C) forgets to square f′(x).
- (D) uses the y-values 1 and 4 as limits of a dx integral.
If you missed this: Topic 8.13 study guide.
Your next step
| Topic | Question | If you missed it, read |
|---|---|---|
| 8.1 Average value | 1 | Guide 8.1 |
| 8.2 Motion with integrals | 2 | Guide 8.2 |
| 8.3 Accumulation in context | 3 | Guide 8.3 |
| 8.4 Area, functions of x | 4 | Guide 8.4 |
| 8.5 Area, functions of y | 5 | Guide 8.5 |
| 8.6 Curves meeting more than twice | 6 | Guide 8.6 |
| 8.7 Squares and rectangles | 7 | Guide 8.7 |
| 8.8 Triangles and semicircles | 8 | Guide 8.8 |
| 8.9 Discs about the axes | 9 | Guide 8.9 |
| 8.10 Discs about other lines | 10 | Guide 8.10 |
| 8.11 Washers about the axes | 11 | Guide 8.11 |
| 8.12 Washers about other lines | 12 | Guide 8.12 |
| 8.13 Arc length (BC only) | 13 | Guide 8.13 |
How to use your result
- Mark each topic secure, shaky or gap.
- Fix gaps in Topics 8.3 and 8.4 first. Net change (8.3) and “top minus bottom” (8.4) sit under every later topic.
- Look for a pattern in Questions 7 to 12: the wrong variable, a missing square or R and r swapped.
- Check your reasons in Questions 3, 8 and 12, not just the values.
- For a gap, read the guide, then do its practice set.
- Then try the Unit 8 mixed review, which combines topics.