Difficulty: Intermediate to Advanced | Prerequisites: Integral techniques, FTC, derivative applications
The AP exam uses integrals for far more than computing areas. You will need to find areas between curves, volumes of solids with known cross-sections, average values of functions, accumulation from rate functions, and net change from velocity graphs. Both the calculator and no-calculator sections test these ideas, and at least one free-response question each year centres on integral applications. The questions reward setup more than computation: knowing which integral to write is worth more than grinding through the arithmetic.
Area between curves is ∫(top – bottom) dx. Volume with known cross-sections is ∫(cross-sectional area) dx. Average value is (1/(b – a)) ∫ₐᵇ f(x) dx. Net change equals the integral of the rate function. If you can set up the correct integral, the calculator handles the rest.
Area between curves
The integral ∫ₐᵇ [f(x) – g(x)] dx, where f(x) ≥ g(x) on [a, b]. In simple terms, integrate the vertical gap between the top curve and the bottom curve.
Volume by cross-sections
If a solid has a known cross-sectional shape perpendicular to the x-axis, its volume is ∫ₐᵇ A(x) dx, where A(x) is the area of the cross-section at position x. Think of it as stacking thin slices and adding up their volumes.
Average value of a function
(1/(b – a)) ∫ₐᵇ f(x) dx. In simple terms, the height of a rectangle with the same base and the same total area as the region under f.
Net change / accumulation
∫ₐᵇ f'(t) dt = f(b) – f(a). The integral of a rate of change gives the total change over the interval. This applies to position from velocity, temperature from a cooling rate, or volume from a flow rate.
Trapezoidal approximation
An estimate of a definite integral using trapezoids. For subintervals of width Δx: (Δx/2)[f(x₀) + 2f(x₁) + 2f(x₂) + ... + 2f(xₙ₋₁) + f(xₙ)]. When subintervals have different widths, compute each trapezoid individually.
Riemann sum
An approximation of a definite integral using rectangles. Left, right, or midpoint versions use different sample points within each subinterval.
Worked example (from Q77): The base is bounded by x = 1, x = 3, y = 2, and y = ln x. Cross-sections perpendicular to the x-axis are squares.
The side length of each square is the vertical distance between y = 2 and y = ln x, i.e. s(x) = 2 – ln x.
Volume = ∫₁³ (2 – ln x)² dx ≈ 3.85
The answer is A) 3.85.
Worked example (from Q79): Which graph of f has (1/2) ∫₁³ f(x) dx = 2?
This means ∫₁³ f(x) dx = 4. The graph must have a total area of 4 between x = 1 and x = 3. The graph in option A satisfies this condition.
Worked example (from Q88): The rate of change of a helicopter's altitude is r(t) = –t³ + 5t² – 8. Which expression gives the change in altitude during the time the altitude is increasing?
The altitude is increasing where r(t) > 0. Solving –t³ + 5t² – 8 = 0 numerically gives roots at approximately t = 1.515 and t = 4.626 (within the 6-second window). The change in altitude during the increasing phase is ∫₁.₅₁₅^4.626 r(t) dt.
The answer is C) ∫₁.₅₁₅^4.626 r(t) dt.
Key distinction: the question asks for the change in altitude (integrate r(t), the rate), not the change in rate (which would involve r'(t)).
Worked example (from Q89): A potato at 400°F cools with rate r(t) = –82e⁻⁰·²⁵ᵗ. Temperature after 4 minutes?
Temperature = initial temperature + ∫₀⁴ r(t) dt = 400 + ∫₀⁴ (–82e⁻⁰·²⁵ᵗ) dt
The integral evaluates to (–82)(–4)(e⁻¹ – 1) = 328(1 – e⁻¹) ≈ 328(0.6321) ≈ 207.3
So temperature ≈ 400 – 207 ≈ 193°F
The answer is B) 193.
Worked example (from Q92): Regions A, B, C have areas 1, 2, 3 respectively. Region B is below the x-axis. What is ∫₋₁⁴ (f(x) + 2) dx?
Split the integral: ∫₋₁⁴ f(x) dx + ∫₋₁⁴ 2 dx
∫₋₁⁴ f(x) dx = Area A – Area B + Area C = 1 – 2 + 3 = 2 (B is negative because it is below the axis)
∫₋₁⁴ 2 dx = 2(4 – (–1)) = 2(5) = 10
Total = 2 + 10 = 12
The answer is D) 12.
Worked example (from Q91): v(t) = 2eˣ + x sin(x). Average velocity from t = 0 to t = 4.
Average velocity = (1/4) ∫₀⁴ v(t) dt
Evaluating on a calculator gives approximately 27.26.
The answer is C) 27.26.
The table gives: t = 0, 2, 3, 5 with h = 0, 280, 330, 240.
Using three subintervals [0,2], [2,3], [3,5] with trapezoids:
(1/2)(0 + 280)(2) = 280
(1/2)(280 + 330)(1) = 305
(1/2)(330 + 240)(2) = 570
Total ≈ 1155
Average height = 1155/5 = 231 metres
For ∫₀⁵ g(x) dx under-approximated by a trapezoidal sum and over-approximated by a left Riemann sum, g must be concave down and decreasing. A decreasing concave-down function has a trapezoidal sum that underestimates (the curve sags below the trapezoid edges, but for concave down and decreasing, the trapezoids lie below the curve) and a left Riemann sum that overestimates (left rectangles sit above a decreasing function).
The answer is E).
Area between curves: ∫ₐᵇ [top(x) – bottom(x)] dx
Volume by square cross-sections: ∫ₐᵇ [side length(x)]² dx
Volume by semicircular cross-sections: ∫ₐᵇ (π/8)[diameter(x)]² dx
Average value: (1/(b – a)) ∫ₐᵇ f(x) dx
Net change: ∫ₐᵇ rate dt = total change
Trapezoidal sum (unequal subintervals): sum of (1/2)(yₖ + yₖ₊₁)(Δxₖ)
Computing the area between two curves models the difference in cumulative outcomes (e.g. revenue vs. cost over time). Volumes by cross-sections arise in engineering whenever a solid has a known profile, from aircraft wing design (as in FRQ 2 of this exam) to architectural columns. Average value is used in physics (average temperature over a day) and economics (average price over a period).
Students set up volume integrals using the side length alone, forgetting to square it (for square cross-sections) or apply the relevant area formula. The integrand is the cross-sectional area, not the side length.
When computing net change from a rate that is sometimes negative, students sometimes take the absolute value. The integral of the rate gives net change, which accounts for both increases and decreases. Total distance (always positive) requires integrating |v(t)|.
In signed-area problems, regions below the x-axis contribute negatively to the integral. Students sometimes add all areas as positive.
The average value formula divides by the length of the interval (b – a), not by the number of data points.
⚠️ At least one free-response question each year involves area, volume, or accumulation. Setting up the integral correctly is worth the bulk of the marks; computation is secondary.
⚠️ Trapezoidal approximations with unequal subintervals appear regularly. Compute each trapezoid separately.
⚠️ The distinction between net change (∫ rate dt) and total accumulated quantity (initial + ∫ rate dt) matters. Read the question carefully.
⚠️ Over-approximation and under-approximation depend on the function's monotonicity and concavity. Know which sums overestimate and which underestimate for each combination.
True or false: The area between two curves can be negative.
Fill in the blank: If cross-sections perpendicular to the x-axis are squares with side length s(x), then the volume is ∫ ______ dx.
True or false: The average value of f on [1, 5] is (1/4) ∫₁⁵ f(x) dx.
Fill in the blank: ∫₀⁴ r(t) dt gives the ______ over the interval [0, 4].
True or false: A left Riemann sum overestimates the integral of a decreasing function.
Answers: 1. False; area is always non-negative (take top minus bottom). The integral can be negative, but area is the absolute value interpretation. 2. [s(x)]². 3. True. 4. Net change. 5. True.
Q: Regions A, B, C under f have areas 1, 2, 3 (B below the axis). What is ∫₋₁⁴ (f(x) + 2) dx?
A: 12. Split into ∫ f(x) dx (= 1 – 2 + 3 = 2) plus ∫ 2 dx (= 10).
Q: A potato cools from 400°F with rate r(t) = –82e⁻⁰·²⁵ᵗ. What is the temperature after 4 minutes?
A: Approximately 193°F. Compute 400 + ∫₀⁴ r(t) dt on a calculator.
Q: What expression gives the helicopter's altitude change while it is increasing, given r(t) = –t³ + 5t² – 8?
A: ∫₁.₅₁₅^4.626 r(t) dt. These are the bounds where r(t) > 0.
Q: v(t) = 2eᵗ + t sin(t). What is the average velocity from t = 0 to t = 4?
A: Approximately 27.26. Compute (1/4) ∫₀⁴ v(t) dt.
Q: The base of a solid is bounded by x = 1, x = 3, y = 2, and y = ln x. Cross-sections perpendicular to the x-axis are squares. What is the volume?
A: Approximately 3.85. The side length is 2 – ln x, so volume = ∫₁³ (2 – ln x)² dx.
Applications of integrals build on everything covered in derivative applications (you need to determine where rate functions are positive or negative) and integral techniques (you need to evaluate or set up the integral correctly). In the free-response section, these problems often combine with differential equations (e.g. finding total accumulated change from a given rate function) and data-table problems (trapezoidal approximation of real-world measurements).
area between curves, volume by cross-sections, average value, net change, accumulation, trapezoidal approximation, Riemann sum, left sum, right sum, overestimate, underestimate, signed area, AP Calculus AB integrals applications