2022 AP Calculus BC — FRQ Complete Guide
6 questions · 54 points total · Click any question tab to begin
Q1
Integral / Rate
Q2
Parametric
Q3
Graph of f′
Q4
Table / Cone
Q5
Area / Volume
Q6
Power Series
Calculator Policy
CALC Q1, Q2 — graphing calculator REQUIRED    NO CALC Q3 – Q6 — no calculator allowed
Perfect-Score Strategy
1
Read each sub-part carefully — the question tells you exactly what to do
2
Show ALL setup work, even if a calculator gives the answer
3
Write units whenever asked ("vehicles per hour", "cm/day", etc.)
4
Justify by citing the theorem (FTC, MVT, IVT, Alternating Series Test…)
5
Round final numerical answers to 3 decimal places unless told otherwise
Question 1 — Toll Plaza Arrival Rate CALCULATOR
A(t) = 450√sin(0.62t), vehicles/hour, t = hours after 5 AM, interval [0, 5]
Core Concepts
Definite Integral as Accumulation Average Value of a Function Sign of A′(t) FTC — Net Change Optimizing N(t)
(a)
Write the integral for total vehicles, t = 1 to t = 5
2 pts
▼
Key Concept: Rate × Time = Accumulation

Integrating a rate gives total quantity. Do NOT evaluate — just write the expression.

Total vehicles = ∫₁⁵ A(t) dt = ∫₁⁵ 450√(sin(0.62t)) dt
⚠ Common Trap
Don't evaluate the integral — the question says "write, but do not evaluate." Evaluating wastes time and won't earn extra points.
Model Answer
The total number of vehicles that arrive from t = 1 to t = 5 is given by
∫₁⁵ 450√(sin(0.62t)) dt
(b)
Average value of A(t) from t = 1 to t = 5
2 pts
▼
Average Value Formula
f_avg = (1/(b−a)) · ∫ₐᵇ f(t) dt = (1/(5−1)) · ∫₁⁵ 450√(sin(0.62t)) dt = (1/4) · ∫₁⁵ 450√(sin(0.62t)) dt
TI-Nspire CX II Steps
Home → Calculator (scratchpad or new doc) Press [menu] → 4: Calculus → 2: Numerical Integral (∫) Enter: (1/4)·∫(450·√(sin(0.62·x)),x,1,5) Or type directly: (1/4)*∫(450*√(sin(0.62*x)),x,1,5) then [enter]
Result ≈ 362.240 vehicles per hour
⚠ Common Trap
Must divide by (b − a) = 4, not just compute the integral. And include UNITS: vehicles per hour.
Model Answer
Average value = (1/4)·∫₁⁵ 450√(sin(0.62t)) dt ≈ 362.240 vehicles per hour
(c)
Is A(t) increasing or decreasing at t = 1? Justify.
2 pts
▼
Check the sign of A′(1)
A′(t) = d/dt [450·(sin(0.62t))^(1/2)] = 450 · (1/2)·(sin(0.62t))^(−1/2)·cos(0.62t)·0.62 A′(1) = 450·0.62·cos(0.62) / (2√(sin(0.62)))
TI-Nspire CX II Steps
Define a(x)=450·√(sin(0.62·x)) Then: derivative(a(x),x)|x=1 OR d/dx(450·√(sin(0.62·x)))|x=1 [menu] → 4: Calculus → 1: Derivative → enter → substitute x=1
Result ≈ 109.0 > 0 → INCREASING
Model Answer
A′(1) ≈ 109.0 > 0. Since A′(1) > 0, the rate at which vehicles arrive is increasing at t = 1 (6 AM).
(d)
Greatest number of vehicles in line on [a, 4]. Justify.
4 pts
▼
Strategy: Find a, then maximize N(t) = ∫ₐᵗ (A(x)−400) dx
1
Find a: solve A(t) = 400 → 450√(sin(0.62t)) = 400. On [0,5], find first time A(t) reaches 400.
2
N(t) is increasing when A(t) > 400 (integrand positive), decreasing when A(t) < 400.
3
On [a, 4]: A(t) ≥ 400 throughout (verify), so N is increasing on [a,4] → maximum at t = 4.
4
Evaluate N(4) = ∫ₐ⁴ (A(x) − 400) dx using calculator.
TI-Nspire CX II Steps
Step 1 — find a: use [menu] → 6:Analyze Graph → 1:Zero on A(x)−400 Or: solve(450·√(sin(0.62·x))=400,x) → a ≈ 0.4557 Step 2 — evaluate N(4): ∫(450·√(sin(0.62·x))−400,x,0.4557,4)
N(4) ≈ 163 vehicles (round to nearest whole number)
⚠ Common Trap
You MUST justify why t = 4 gives the maximum — state that N′(t) = A(t) − 400 ≥ 0 on [a, 4], so N is non-decreasing and maximum occurs at t = 4. Don't just compute — justify!
Model Answer
A(t) = 400 first at a ≈ 0.4557. N′(t) = A(t) − 400 ≥ 0 for all t ∈ [a, 4], so N is non-decreasing on [a, 4]. Therefore the greatest number of vehicles in line occurs at t = 4:
N(4) = ∫₀.₄₅₅₇⁴ (450√(sin(0.62x)) − 400) dx ≈ 163 vehicles
Question 2 — Parametric Particle Motion CALCULATOR
dx/dt = √(1+t²), dy/dt = ln(2+t²), position (1,5) at t=4
Core Concepts
Parametric Slope = (dy/dt)/(dx/dt) Speed = √((dx/dt)²+(dy/dt)²) Acceleration vector FTC — position from velocity Arc Length integral
(a)
Slope of tangent at t = 4
2 pts
▼
dy/dx = (dy/dt) ÷ (dx/dt)
dy/dx|_{t=4} = ln(2+16) / √(1+16) = ln(18) / √17
TI-Nspire CX II
ln(2+4²)/√(1+4²) then [enter] → ≈ 0.696
Model Answer
dy/dx = ln(18)/√17 ≈ 0.696
(b)
Speed at t = 4, and acceleration vector at t = 4
3 pts
▼
Speed and Acceleration Formulas
Speed = √((dx/dt)² + (dy/dt)²) Acceleration = ⟨d²x/dt², d²y/dt²⟩ d²x/dt² = d/dt[√(1+t²)] = t/√(1+t²) d²y/dt² = d/dt[ln(2+t²)] = 2t/(2+t²)
TI-Nspire CX II
Speed: √((√(1+4²))²+(ln(2+4²))²) → ≈ 4.313 ax: 4/√(1+16) → ≈ 0.970 ay: 2·4/(2+16) → = 8/18 = 4/9 ≈ 0.444
Model Answer
Speed at t=4: √(17 + (ln18)²) ≈ 4.313
Acceleration vector: ⟨4/√17, 4/9⟩ ≈ ⟨0.970, 0.444⟩
(c)
y-coordinate at t = 6
3 pts
▼
Use FTC: y(6) = y(4) + ∫₄⁶ (dy/dt) dt
y(6) = 5 + ∫₄⁶ ln(2+t²) dt
TI-Nspire CX II
5 + ∫(ln(2+x²),x,4,6) → ≈ 5 + 7.919 = 12.919 [menu] → 4:Calculus → 2:Numerical Integral
⚠ Common Trap
Start from the KNOWN position y(4) = 5, not from zero. Must add the initial condition.
Model Answer
y(6) = 5 + ∫₄⁶ ln(2+t²) dt ≈ 12.919
(d)
Total distance from t = 4 to t = 6
2 pts
▼
Arc Length (= total distance for parametric curve)
Distance = ∫₄⁶ √((dx/dt)² + (dy/dt)²) dt = ∫₄⁶ √(1+t² + (ln(2+t²))²) dt
TI-Nspire CX II
∫(√(1+x²+(ln(2+x²))²),x,4,6) → ≈ 9.417
⚠ Common Trap
Distance ≠ displacement. Arc length is always ∫|velocity| dt. No matter what, integrate the magnitude of the velocity vector.
Model Answer
Total distance = ∫₄⁶ √(1+t² + [ln(2+t²)]²) dt ≈ 9.417
Question 3 — Graph of f′ (Semicircle + Line Segments) NO CALCULATOR
f(4) = 3; f′ graph = semicircle on [0,3] + line segments on [3,4] and [4,7]
Reading the Graph of f′

From the graph: semicircle centered at (2,0) radius 2 on [0,4] (below x-axis, so f′ < 0). Line segment from (4,0) to (6,2) on [4,6]. Line segment from (6,2) to (7,1) on [6,7].

Area of semicircle (radius 2) = (1/2)π(2²) = 2π f(4) - f(0) = ∫₀⁴ f′(x)dx = −2π (semicircle is BELOW x-axis → negative) f(0) = f(4) + 2π = 3 + 2π f(5) - f(4) = ∫₄⁵ f′(x)dx = area of triangle base=1 height=1 = 1/2 f(5) = 3 + 1/2 = 7/2
(a)
Find f(0) and f(5)
2 pts
▼
FTC Part 2: f(b) = f(a) + ∫ₐᵇ f′(x) dx
1
f(0): work backwards from f(4). ∫₀⁴ f′(x)dx = −(area of semicircle) = −2π. So f(0) = f(4) − (−2π) = 3 + 2π.
2
f(5): ∫₄⁵ f′(x)dx = area of triangle with vertices (4,0),(5,0),(5,1) = 1/2. So f(5) = 3 + 1/2 = 7/2.
⚠ Common Trap
The semicircle is BELOW the x-axis (f′ < 0 on [0,4]), so the integral is −2π, not +2π. Always check the sign from the graph.
Model Answer
f(0) = 3 + 2π   |   f(5) = 7/2
(b)
x-coordinates of all inflection points for 0 < x < 7
3 pts
▼
Inflection of f ↔ f′ changes monotonicity ↔ f″ changes sign

Inflection points of f occur where f″ changes sign, i.e., where f′ changes from increasing to decreasing or vice versa (local extrema of f′).

From the graph of f′: — f′ has a local minimum at x = 2 (bottom of semicircle) → f″ changes sign → inflection — f′ changes slope at x = 6 (line segment changes from slope 1 to slope −1) → f″ changes sign → inflection
⚠ Common Trap
x = 4 is where f′ = 0 (zero of f′), which is an extremum of f, NOT an inflection point of f. Students confuse zeros of f′ with inflection points.
Model Answer
f has inflection points at x = 2 and x = 6. At x = 2, f′ changes from decreasing to increasing (f″ changes − to +). At x = 6, f′ changes from increasing to decreasing (f″ changes + to −).
(c)
Where is g(x) = f(x) − x decreasing on [0,7]?
3 pts
▼
g′(x) = f′(x) − 1 < 0 when f′(x) < 1

From the graph, f′(x) < 1 on:

— [0, 4]: semicircle, f′ ≤ 0 < 1 throughout → g′ < 0 → g decreasing — [4, 5]: f′ goes 0 to 1, so f′ < 1 on (4, 5) → g decreasing on (4,5) — At x = 5: f′(5) = 1 → g′(5) = 0 — [5, 6]: f′ goes 1 to 2, so f′ > 1 → g increasing — [6, 7]: f′ goes 2 to 1, so f′ > 1 → g increasing (barely) Wait — at x=7: f′(7)=1 → g′(7)=0 So g′(x) < 0 on (0, 5) and g is decreasing on [0, 5]
Model Answer
g′(x) = f′(x) − 1. Since f′(x) < 1 for 0 ≤ x < 5 (the semicircle gives f′ ≤ 0 on [0,4], and the line segment gives f′ < 1 on (4,5)), g′(x) < 0 on (0, 5). Therefore g is decreasing on [0, 5].
(d)
Absolute minimum of g on [0,7]. Justify.
3 pts
▼
Compare g at critical points and endpoints
Critical points where g′ = 0: x = 5 and x = 7 Endpoints: x = 0 and x = 7 g(0) = f(0) − 0 = 3 + 2π ≈ 9.28 g(5) = f(5) − 5 = 7/2 − 5 = −3/2 g(7) = f(7) − 7 f(7) = f(4) + ∫₄⁷ f′(x)dx = 3 + (triangle [4,6] area) + (triangle [6,7] area) = 3 + (1/2·2·2) + (1/2·1·(2+1)) Wait — segment [4,6]: from (4,0) to (6,2), triangle area = 1/2·2·2 = 2 Segment [6,7]: from (6,2) to (7,1), area = 1/2·1·(2+1) = 3/2 f(7) = 3 + 2 + 3/2 = 3 + 7/2 = 13/2 g(7) = 13/2 − 7 = −1/2
⚠ Common Trap
Must check ALL critical points AND endpoints. g′ = 0 at x=5 and x=7, and also check x=0 (endpoint). The minimum at x=5 is a local min — but it IS the global min here.
Model Answer
g(0) = 3+2π, g(5) = −3/2, g(7) = −1/2.
The absolute minimum value of g on [0,7] is −3/2, occurring at x = 5.
Question 4 — Melting Ice Cone (Table) NO CALCULATOR
r′(t) given in table at t = 0,3,7,10,12. V = (1/3)πr²h
Core Concepts
Difference Quotient ≈ 2nd Derivative Intermediate Value Theorem Right Riemann Sum Related Rates (implicit differentiation)
(a)
Approximate r″(8.5) using average rate of change of r′ on [7,10]
2 pts
▼
r″(8.5) ≈ [r′(10) − r′(7)] / (10 − 7)
r″(8.5) ≈ (r′(10) − r′(7)) / (10 − 7) = (−3.8 − (−4.4)) / 3 = 0.6 / 3 = 0.2 cm/day²
⚠ Common Trap
Include UNITS: cm/day² (centimeters per day per day). Missing units loses a point.
Model Answer
r″(8.5) ≈ (−3.8 − (−4.4))/3 = 0.6/3 = 0.2 cm/day²
(b)
Is there t ∈ [0,3] where r′(t) = −6? Justify.
2 pts
▼
Use the Intermediate Value Theorem on r′
r is twice-differentiable → r′ is continuous on [0,3]. r′(0) = −6.1 and r′(3) = −5.0 −6.1 < −6 < −5.0 By IVT, there exists c ∈ (0,3) such that r′(c) = −6.
⚠ Common Trap
Must explicitly name the theorem: "By the Intermediate Value Theorem." Also must verify r′ is continuous (it is, since r is twice-differentiable).
Model Answer
Since r is twice-differentiable, r′ is continuous on [0,3]. r′(0) = −6.1 < −6 < −5.0 = r′(3). By the Intermediate Value Theorem, there exists c ∈ (0,3) with r′(c) = −6. Yes.
(c)
Right Riemann sum to approximate ∫₀¹² r′(t) dt
2 pts
▼
Right Riemann Sum: use RIGHT endpoint of each subinterval
Subintervals from table: [0,3], [3,7], [7,10], [10,12] Right endpoints: t = 3, 7, 10, 12 Widths (Δt): 3, 4, 3, 2 RRS = r′(3)·3 + r′(7)·4 + r′(10)·3 + r′(12)·2 = (−5.0)(3) + (−4.4)(4) + (−3.8)(3) + (−3.5)(2) = −15.0 + (−17.6) + (−11.4) + (−7.0) = −51.0
⚠ Common Trap
Unequal subinterval widths! Use the actual Δt values (3, 4, 3, 2), NOT assuming uniform width. This is a classic trap.
Model Answer
∫₀¹² r′(t) dt ≈ (−5.0)(3) + (−4.4)(4) + (−3.8)(3) + (−3.5)(2) = −51.0 cm
(d)
dV/dt at t = 3. (r=100, h=50, dr/dt=−5, dh/dt=−2)
4 pts
▼
Implicit differentiation of V = (1/3)πr²h
V = (1/3)π r²h dV/dt = (1/3)π · [2r·(dr/dt)·h + r²·(dh/dt)] At t=3: r=100, h=50, dr/dt = r′(3) = −5.0, dh/dt = −2 dV/dt = (1/3)π · [2(100)(−5)(50) + (100²)(−2)] = (1/3)π · [−50000 + (−20000)] = (1/3)π · (−70000) = −70000π/3 cm³/day
⚠ Common Trap
Use dr/dt = r′(3) = −5.0 from the TABLE (not −6.1 or any other value). Product rule on r²h: d/dt[r²h] = 2r(dr/dt)h + r²(dh/dt).
Model Answer
dV/dt = (π/3)[2(100)(−5)(50) + (10000)(−2)] = (π/3)(−70000) = −70000π/3 cm³/day
Question 5 — Area and Volume (Improper Integrals) NO CALCULATOR
R: region under 1/x, x∈[1,5]. W: region under 1/x², x≥3
Core Concepts
∫(1/x)dx = ln|x| + C Integration by Parts for xe^(x/5) Disk Method: V = π∫[f(x)]² dx Improper Integrals: lim as b→∞
(a)
Area of region R
2 pts
▼
Area = ∫₁⁵ (1/x) dx = [ln x]₁⁵ = ln5 − ln1 = ln5
Model Answer
Area = ∫₁⁵ (1/x) dx = ln5
(b)
Volume of solid with cross sections of area xe^(x/5)
5 pts
▼
V = ∫₁⁵ (cross-section area) dx — then integrate by parts
V = ∫₁⁵ x·e^(x/5) dx Integration by Parts: u = x, dv = e^(x/5)dx du = dx, v = 5e^(x/5) = [5x·e^(x/5)]₁⁵ − ∫₁⁵ 5e^(x/5) dx = [5x·e^(x/5)]₁⁵ − [25e^(x/5)]₁⁵ = (25e¹ − 5e^(1/5)) − (25e¹ − 25e^(1/5)) = 25e − 5e^(1/5) − 25e + 25e^(1/5) = 20e^(1/5)
⚠ Common Trap
When integrating e^(x/5), the result is 5e^(x/5) (multiply by 5 — chain rule in reverse). This is a very common arithmetic error.
Model Answer
V = ∫₁⁵ xe^(x/5) dx = [5xe^(x/5) − 25e^(x/5)]₁⁵ = 25e − 5e^(1/5) − 25e + 25e^(1/5) = 20e^(1/5)
(c)
Volume when W is revolved about x-axis
4 pts
▼
Disk Method + Improper Integral
V = π ∫₃^∞ (1/x²)² dx = π ∫₃^∞ x⁻⁴ dx = π · lim_{b→∞} ∫₃ᵇ x⁻⁴ dx = π · lim_{b→∞} [−(1/3)x⁻³]₃ᵇ = π · lim_{b→∞} [−1/(3b³) + 1/(3·27)] = π · (0 + 1/81) = π/81
⚠ Common Trap
Must use LIMIT notation for improper integrals. Write lim_{b→∞} explicitly — just writing ∞ as an endpoint without the limit will lose a point. Also, (1/x²)² = 1/x⁴, not 1/x².
Model Answer
V = π lim_{b→∞} [−x⁻³/3]₃ᵇ = π(0 + 1/81) = π/81
Question 6 — Power Series NO CALCULATOR
f(x) = x − x³/3 + x⁵/5 − x⁷/7 + ··· + (−1)ⁿx^(2n+1)/(2n+1) + ···
Recognize This Series!

This is the arctan(x) Maclaurin series. f(x) = arctan(x). This makes part (d) elegant.

Ratio Test Alternating Series Estimation Theorem Term-by-term differentiation Geometric Series Sum
(a)
Interval of convergence using Ratio Test
4 pts
▼
Ratio Test: lim|a_{n+1}/a_n| < 1
aₙ = (−1)ⁿ x^(2n+1) / (2n+1) aₙ₊₁ = (−1)^(n+1) x^(2n+3) / (2n+3) |aₙ₊₁/aₙ| = |x^(2n+3)/(2n+3)| · |(2n+1)/x^(2n+1)| = x² · (2n+1)/(2n+3) lim_{n→∞} x² · (2n+1)/(2n+3) = x² Converges when x² < 1 → −1 < x < 1 Check endpoints: x=1: Σ(−1)ⁿ/(2n+1) → converges (Alternating Series Test) x=−1: Σ(−1)^(n+1)/(2n+1) → converges (AST) Interval of convergence: [−1, 1]
⚠ Common Trap
MUST check both endpoints separately. Forgetting endpoint analysis loses points. Show the AST conditions: terms decrease to 0 monotonically.
Model Answer
By the Ratio Test, |L| = x² < 1 gives radius of convergence R=1. At x=±1 the series converges by the Alternating Series Test. Interval: [−1, 1]
(b)
Show |f(1/2) − 1/2| < 1/10
3 pts
▼
Alternating Series Estimation Theorem
f(1/2) = 1/2 − (1/2)³/3 + (1/2)⁵/5 − ··· This is an alternating series with decreasing terms → 0. By the Alternating Series Estimation Theorem: |error| ≤ |first omitted term| If we approximate f(1/2) ≈ 1/2 (i.e., stop at n=0): |f(1/2) − 1/2| ≤ |(1/2)³/3| = (1/8)/3 = 1/24 < 1/10 ✓
⚠ Common Trap
Must CITE the Alternating Series Estimation Theorem by name. Must verify: (i) terms alternate, (ii) |terms| decrease, (iii) |terms| → 0. Don't skip the verification.
Model Answer
Since f(x) is an alternating series at x=1/2 with decreasing terms converging to 0, by the Alternating Series Estimation Theorem, |f(1/2)−1/2| ≤ (1/2)³/3 = 1/24 < 1/10. ✓
(c)
First four terms + general term for f′(x)
3 pts
▼
Differentiate term by term
f(x) = x − x³/3 + x⁵/5 − x⁷/7 + ··· f′(x) = 1 − x² + x⁴ − x⁶ + ··· + (−1)ⁿ x^(2n) + ··· General term: (−1)ⁿ x^(2n)

Note: f′(x) = 1/(1+x²) — this is the geometric series with ratio −x², which converges for |x| < 1.

Model Answer
f′(x) = 1 − x² + x⁴ − x⁶ + ··· + (−1)ⁿx^(2n) + ···
(d)
Find f′(1/6) using the series from (c)
3 pts
▼
Substitute x = 1/6 into the geometric series
f′(x) = 1 − x² + x⁴ − x⁶ + ··· (geometric, ratio r = −x²) Sum = 1/(1−(−x²)) = 1/(1+x²) f′(1/6) = 1/(1+(1/6)²) = 1/(1+1/36) = 1/(37/36) = 36/37
⚠ Common Trap
Recognize that f′(x) = Σ(−x²)ⁿ is a geometric series with first term a=1 and ratio r=−x². Sum = 1/(1+x²) provided |x|<1. For x=1/6 this clearly converges.
Model Answer
f′(x) is a geometric series with a=1, r=−x². Sum = 1/(1+x²).
f′(1/6) = 1/(1+1/36) = 36/37