Lesson 2: Accumulation Functions, Areas and VolumesAP Deep Dive

Accumulation questions at the top end combine three ideas that the main course treated separately: the Fundamental Theorem with moving limits at both ends, washers swung around lines that are not axes, and the difference between net accumulation and total accumulation. Each is worked here with the checks that distinguish a 9 from a 6.

FTC with two moving limits

Worked example 1. Let G(x) = ∫ from x to 2x of t² dt. Find G′(x) — twice, by two methods, because agreement between methods is how you police your own work under time pressure.

*Method 1 (antiderivative).* G(x) = [t³/3] from x to 2x = (8x³ − x³)/3 = 7x³/3, so G′(x) = 7x².

*Method 2 (FTC + chain rule).* d/dx ∫ from a(x) to b(x) of f(t) dt = f(b(x))·b′(x) − f(a(x))·a′(x). Here: f(2x)·2 − f(x)·1 = (4x²)(2) − x² = 8x² − x² = 7x² ✓. The minus sign on the lower-limit term is the single most-tested detail in this family — a variable *lower* limit subtracts. A reliable memory device: differentiating an accumulation function evaluates the integrand at each limit, multiplied by that limit's rate of change, with the lower limit entering negatively because increasing it *removes* area.

Apply the same machine to G(x) = ∫ from 2x to x² of (sin t)/t dt and you get G′(x) = [sin(x²)/x²]·2x − [sin(2x)/(2x)]·2 = 2 sin(x²)/x − sin(2x)/x — no antiderivative exists in elementary functions, which is exactly why the FTC method matters.

Washers about a line that is not an axis

Worked example 2. The region under y = √x on [0, 4] is revolved about the line y = −1. Find the volume.

Radii are distances to the axis of rotation, not distances to the x-axis. Outer radius (from y = −1 up to the curve): R = √x + 1. Inner radius (from y = −1 up to the x-axis, the region's lower boundary): r = 1. Washer volume: V = π ∫₀⁴ [(√x + 1)² − 1²] dx = π ∫₀⁴ [x + 2√x] dx = π[x²/2 + (4/3)x^(3/2)]₀⁴ = π(8 + 32/3) = 56π/3 ≈ 58.6. Two checks: the answer exceeds the volume about the x-axis (8π ≈ 25) — it must, since every radius grew by 1; and if the region had touched the axis (r = 0), the same integral would be a disk problem. Students lose this in two standard ways: writing R = √x (forgetting the shift) or squaring the difference, π(R − r)², instead of the difference of squares. The washer formula is π(R² − r²) — an annulus area — never π(R − r)².

Net versus total: displacement and distance

Worked example 3. A particle has velocity v(t) = t² − 4 on [0, 3]. Find its displacement and the total distance travelled.

Displacement is the net accumulation: ∫₀³ (t² − 4) dt = [t³/3 − 4t]₀³ = 9 − 12 = −3 (the particle ends 3 units left of where it started). Distance integrates speed, |v|: v is negative on [0, 2) and positive on (2, 3]. ∫₀² (4 − t²) dt = 8 − 8/3 = 16/3; ∫₂³ (t² − 4) dt = 7/3; total distance = 16/3 + 7/3 = 23/3 ≈ 7.67. Average velocity is displacement/time = −1; average speed is distance/time = 23/9 ≈ 2.56 — different numbers answering different questions, and FRQs ask for both in adjacent parts precisely to test whether you know that. The mechanical rule: to integrate |v|, first find where v changes sign, then split. Writing ∫|v| dt = |∫v dt| is the signature error of this topic.

Cross sections: volume without rotation

Worked example 4. The region between y = x and y = x² (they meet at x = 0 and x = 1) is the base of a solid whose cross sections perpendicular to the x-axis are squares. The side of each square is the vertical gap s(x) = x − x², so A(x) = (x − x²)² and V = ∫₀¹ (x − x²)² dx = ∫₀¹ (x² − 2x³ + x⁴) dx = 1/3 − 1/2 + 1/5 = 1/30. The template generalises by swapping the area formula: semicircles on diameter s give A = (π/8)s², equilateral triangles give A = (√3/4)s². Graders' favourite variant is cross sections perpendicular to the y-axis, forcing you to rewrite the gap as a function of y (right curve minus left curve) — the geometry is identical, the variable is not.

Common mistakes at the top band

• Sign of the lower limit in FTC-with-chain-rule problems (subtract, always).
• Washer errors: π(R − r)², or radii measured from the wrong line when rotating about y = k or x = k.
• Distance = |displacement| — see above; also forgetting to split at *every* sign change.
• Area between curves with [top − bottom] kept fixed across an intersection point where the curves swap; find all intersections first and re-decide which is on top on each subinterval.
• Average value amnesia: the average of f on [a, b] is (1/(b − a))∫f — the 1/(b − a) factor is a scored step, not decoration.

AP Exam Experience

1. (No calculator.) d/dx [ ∫ from 3 to sin x of t² dt ] equals

• (A) sin²x
• (B) sin²x · cos x
• (C) (sin³x)/3
• (D) 2 sin x · cos x

Answer: (B). By FTC Part 1 with the chain rule, the derivative is f(sin x)·(sin x)′ = sin²x · cos x. Choice (A) forgets the chain-rule factor; (C) is the accumulated *value*, not its derivative.

2. The area enclosed by y = x² and y = 2x is

• (A) 2/3
• (B) 4/3
• (C) 8/3
• (D) 16/3

Answer: (B). The curves meet at x = 0 and x = 2; on [0, 2], 2x is on top. ∫₀² (2x − x²) dx = [x² − x³/3]₀² = 4 − 8/3 = 4/3.

3. The average value of f(x) = sin x on [0, π] is

• (A) 1
• (B) 2/π
• (C) 1/π
• (D) 0

Answer: (B). Average = (1/π)∫₀^π sin x dx = (1/π)(2) = 2/π ≈ 0.64. Choice (D) is the trap for students who average over a full period or confuse the mean with the integral's sign pattern on [0, 2π].

Free-response question. Let R be the region enclosed by the graphs of y = x² and y = 2x.

(a) Find the area of R. (b) Find the volume of the solid generated when R is revolved about the x-axis. (c) Find the volume of the solid whose base is R and whose cross sections perpendicular to the x-axis are semicircles with diameters lying in R.

Scoring guideline.

• (a) 2 points. 1 point for the setup ∫₀² (2x − x²) dx with correct limits from the intersections x = 0, 2 and the correct top-minus-bottom order. 1 point for the answer 4/3. An antiderivative shown (x² − x³/3) is expected; a calculator-only decimal (1.333) may earn the answer point only where the rubric allows, but the setup point requires the integral.
• (b) 3 points. 1 point for the washer structure π ∫ [(outer)² − (inner)²] dx with outer radius 2x (the curve farther from the x-axis) and inner radius x². 1 point for the integral π ∫₀² (4x² − x⁴) dx. 1 point for the value π(32/3 − 32/5) = 64π/15 ≈ 13.4. Using disks instead of washers forfeits the structure point: R does not touch the x-axis where x² > 0, so the solid has a genuine hole along the axis that the inner radius accounts for.
• (c) 3 points. 1 point for the cross-section area with diameter d(x) = 2x − x²: A(x) = (1/2)π(d/2)² = (π/8)(2x − x²)² — the π/8 factor (semicircle on diameter d) is the scored detail. 1 point for the setup ∫₀² A(x) dx. 1 point for the value (π/8)(16/15) = 2π/15 ≈ 0.419. Answers using (π/2)(2x − x²)² (treating d as a radius) earn the setup structure only if the error is solely the area formula; graders treat radius/diameter confusion as the targeted misconception of this part.

Reading an accumulation function from a graph of f

Suppose f is piecewise constant: f(x) = 2 on [0, 2], f(x) = −1 on (2, 5], f(x) = 3 on (5, 6], and let G(x) = ∫₀ˣ f(t) dt. Everything about G can be read from accumulated signed area: G(2) = 2 × 2 = 4; G(5) = 4 + (−1)(3) = 1; G(6) = 1 + 3 × 1 = 4. Now the questions FRQs actually ask:

• Where does G increase or decrease? G′(x) = f(x), so G increases on (0, 2) and (5, 6) and decreases on (2, 5) — the sign of f, read directly.
• Where is G largest? At x = 2 (value 4), with the same value reached again at x = 6; G has a local maximum at x = 2 because f changes sign from + to − there.
• Where is G concave up? Where f is increasing — on a piecewise-constant graph, nowhere within the pieces, which is why G is piecewise *linear* here. With a curved f, "G concave up where f increases" is the most frequently tested sentence in this topic.
• Average value of f on [0, 6]: (1/6)G(6) = 4/6 = 2/3 — the accumulation function and the average-value formula are the same object viewed from two sides.

With a smooth graph instead of steps, replace exact areas by counting grid squares or trapezoids and accept an estimate — but the four questions above, and their answers' justifications, are unchanged.