Applications of integration practice problems AP Calculus, Unit 8
Average value, displacement and distance, areas between curves, and volumes with cross-sections, discs and washers. Try each question before you look: use the hint if you are stuck, then reveal the worked solution one step at a time.
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Key equations for applications of integration
| Area = ∫ab (top − bottom) dx | area between curves |
|---|---|
| V = π∫ [R(x)]2 dx | disc method |
| V = π∫ ([R]2 − [r]2) dx | washer method |
| Distance = ∫ |v(t)| dt | total distance travelled |
Questions marked No calculator should be done by hand, as in Part A of the multiple-choice section. You can type fractions, such as 3/4. The full sheet is in our AP Calculus formula sheet.
Warm-up questions
Quick checks on the core ideas. If these feel easy, go straight to the exam-style questions.
Hint
(1/(b − a)) × the integral.
Worked solution
- ∫04 2x dx = 16.
- 16/4 = 4.
Hint
Which curve is on top between 0 and 1?
Worked solution
- On [0, 1], x ≥ x2.
- ∫01 (x − x2) dx = 1/2 − 1/3.
- = 1/6.
Hint
Distance counts movement in both directions as positive.
Worked solution
- ∫ v(t) dt is the displacement, where backward motion cancels forward motion.
- Distance needs |v(t)|. The answer is B.
Exam-style questions
Questions at the level of the AP Calculus exam. The ones marked Challenge combine more than one idea.
Hint
Disc method: V = π∫ y2 dx.
Worked solution
- V = π∫04 x dx.
- = π × 8 = 8π.
- ≈ 25.13.
Hint
v changes sign at t = 2.
Worked solution
- From 0 to 2 it moves back: area 2.
- From 2 to 4 it moves forward: area 2.
- Total distance = 4 (the displacement is 0).
Hint
The curve crosses the x-axis at x = ±2.
Worked solution
- ∫−22 (4 − x2) dx.
- = [4x − x3/3] = (8 − 8/3) − (−8 + 8/3) = 32/3.
- ≈ 10.667.
Hint
Each square has side √x, so area x.
Worked solution
- Area of a cross-section: (√x)2 = x.
- V = ∫04 x dx = 8.
Hint
Washer method: outer radius x, inner radius x2.
Worked solution
- V = π∫01 (x2 − x4) dx.
- = π(1/3 − 1/5) = 2π/15.
- ≈ 0.4189.
Free-response practice
One original question in the style of the exam's free-response section. Answer it in full, with working, before you open the scoring guide.
Calculator-active free-response style · 6 points · about 15 minutes
Let R be the region bounded by y = 2x and y = x2.
- (a) Find the area of R.
- (b) Find the volume when R is rotated about the x-axis.
- (c) R is the base of a solid with square cross-sections perpendicular to the x-axis. Find its volume.
Write your full answer on paper first, then open the scoring guide and tick each point you earned.
Scoring guide and model answer
Part (a)
The curves meet at x = 0 and 2. Area = ∫02 (2x − x2) dx = 4 − 8/3 = 4/3.
Part (b)
π∫02 (4x2 − x4) dx = π(32/3 − 32/5) = 64π/15 ≈ 13.404.
Part (c)
∫02 (2x − x2)2 dx = 16/15 ≈ 1.067.
Your mark: 0 of 6
Common mistakes in applications of integration
- Subtracting radii before squaring in the washer method.
- Using displacement when the question asks for distance.
- Not finding the intersection points to get the limits.
Keep going
Revise the ideas behind these questions:
Then try real exam questions: AP Calculus past papers, with official scoring guidelines. · All AP Calculus units