AP Calculus FRQ Worksheet with Solutions (AB & BC) 2026
AP Calculus free response questions (FRQs) require you to show full work and justify every step. This worksheet covers the most frequently tested FRQ types on both AP Calculus AB and BC with complete step-by-step solutions.
Use our AP Calculus AB Score Calculator or AP Calculus BC Score Calculator to see how your FRQ scores translate to an AP grade.
How AP Calculus FRQs Are Scored
Each AP Calculus exam has 6 FRQs worth a total of 54 raw points (9 points each). The FRQ section counts for 50% of your composite score.
| Section | Questions | Points Each | Total |
|---|---|---|---|
| Part A (calculator allowed) | 2 | 9 | 18 |
| Part B (no calculator) | 4 | 9 | 36 |
| Total | 6 | 54 |
Points are awarded for method (setting up correctly), answer (correct result), and justification (explaining why). A wrong final answer can still earn most points if your setup is correct.
FRQ 1: Related Rates (No Calculator)
Problem: A spherical balloon is being inflated. The radius is increasing at a rate of 2 cm/s. At the moment when the radius is 5 cm, at what rate is the volume increasing?
Solution:
Volume of a sphere: $V = \frac{4}{3}\pi r^3$
Differentiate both sides with respect to time $t$:
$$\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$$
Substitute $r = 5$ and $\frac{dr}{dt} = 2$:
$$\frac{dV}{dt} = 4\pi (5)^2 (2) = 4\pi (25)(2) = 200\pi \approx 628.3 \text{ cm}^3/\text{s}$$
Scoring notes: 1 pt for correct formula, 1 pt for implicit differentiation, 1 pt for substitution, 1 pt for correct answer with units.
FRQ 2: Area Between Curves (Calculator Allowed)
Problem: Let $f(x) = x^2$ and $g(x) = 2x$. Find the area of the region enclosed by the two curves.
Solution:
Step 1: Find intersection points.
$$x^2 = 2x \implies x^2 - 2x = 0 \implies x(x-2) = 0$$
Intersections at $x = 0$ and $x = 2$.
Step 2: Determine which function is on top on $[0, 2]$.
At $x = 1$: $g(1) = 2 > f(1) = 1$, so $g(x) \geq f(x)$ on $[0, 2]$.
Step 3: Set up and evaluate the integral.
$$A = \int_0^2 [g(x) - f(x)], dx = \int_0^2 (2x - x^2), dx$$
$$= \left[x^2 - \frac{x^3}{3}\right]_0^2 = \left(4 - \frac{8}{3}\right) - 0 = \frac{12 - 8}{3} = \frac{4}{3}$$
Scoring notes: 1 pt for correct limits, 1 pt for correct integrand (top minus bottom), 1 pt for antiderivative, 1 pt for correct answer.
FRQ 3: Particle Motion (No Calculator)
Problem: A particle moves along the x-axis. Its velocity at time $t \geq 0$ is given by $v(t) = t^2 - 4t + 3$.
(a) Find all times when the particle is at rest. (b) Find the total distance traveled from $t = 0$ to $t = 3$. (c) Is the particle speeding up or slowing down at $t = 2$? Justify.
Solution:
(a) Particle at rest when $v(t) = 0$:
$$t^2 - 4t + 3 = 0 \implies (t-1)(t-3) = 0$$
Particle is at rest at $t = 1$ and $t = 3$.
(b) Total distance (not displacement) requires integrating $|v(t)|$.
Sign analysis of $v(t)$: positive on $(0,1)$, negative on $(1,3)$.
$$\text{Distance} = \int_0^1 v(t), dt + \left|\int_1^3 v(t), dt\right|$$
$$\int_0^1 (t^2 - 4t + 3), dt = \left[\frac{t^3}{3} - 2t^2 + 3t\right]_0^1 = \frac{1}{3} - 2 + 3 = \frac{4}{3}$$
$$\int_1^3 (t^2 - 4t + 3), dt = \left[\frac{t^3}{3} - 2t^2 + 3t\right]_1^3 = (9 - 18 + 9) - \left(\frac{1}{3} - 2 + 3\right) = 0 - \frac{4}{3} = -\frac{4}{3}$$
Total distance $= \frac{4}{3} + \frac{4}{3} = \frac{8}{3}$
(c) At $t = 2$: $v(2) = 4 - 8 + 3 = -1$ (negative, moving left)
Acceleration $a(t) = v'(t) = 2t - 4$, so $a(2) = 0$.
At $t = 2$, $v = -1$ and $a = 0$. Since acceleration is zero, the particle is neither speeding up nor slowing down at exactly $t = 2$ — it is at an instantaneous constant speed.
Note: AP readers accept "slowing down" if justified as $v$ and $a$ having the same sign in a neighborhood — check your scoring guideline.
FRQ 4: Accumulation / Definite Integral Application (Calculator Allowed)
Problem: Water flows into a tank at the rate $R(t) = 20\sin\left(\frac{t}{4}\right) + 30$ gallons per hour, where $t$ is hours after midnight. At $t = 0$, the tank contains 100 gallons.
(a) How much water enters the tank from $t = 0$ to $t = 6$? (b) What is the amount of water in the tank at $t = 6$?
Solution:
(a) Water entered $= \int_0^6 R(t), dt = \int_0^6 \left(20\sin\frac{t}{4} + 30\right) dt$
$$= \left[-80\cos\frac{t}{4} + 30t\right]_0^6$$
$$= \left(-80\cos\frac{6}{4} + 180\right) - (-80\cos 0 + 0)$$
$$= -80\cos(1.5) + 180 + 80 \approx -80(0.0707) + 260 \approx -5.66 + 260 \approx 254.3 \text{ gallons}$$
(b) Total at $t = 6$: $100 + 254.3 \approx 354.3$ gallons.
FRQ 5: Differential Equations (No Calculator)
Problem: Consider the differential equation $\frac{dy}{dx} = xy$.
(a) Find the general solution. (b) Find the particular solution satisfying $y(0) = 3$.
Solution:
(a) Separate variables:
$$\frac{dy}{y} = x, dx$$
Integrate both sides:
$$\ln|y| = \frac{x^2}{2} + C$$
$$y = Ae^{x^2/2} \quad \text{(where } A = \pm e^C\text{)}$$
(b) Apply initial condition $y(0) = 3$:
$$3 = Ae^0 = A$$
Particular solution: $y = 3e^{x^2/2}$
Scoring notes: 1 pt for separating variables, 1 pt for integrating both sides correctly, 1 pt for general solution form, 1 pt for applying initial condition, 1 pt for particular solution.
FRQ 6: AP Calculus BC — Polar Area (BC Only)
Problem: Find the area enclosed by the polar curve $r = 2\cos\theta$.
Solution:
The curve $r = 2\cos\theta$ is a circle. It traces once for $\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.
Polar area formula: $A = \frac{1}{2}\int_\alpha^\beta r^2, d\theta$
$$A = \frac{1}{2}\int_{-\pi/2}^{\pi/2} (2\cos\theta)^2, d\theta = \frac{1}{2}\int_{-\pi/2}^{\pi/2} 4\cos^2\theta, d\theta$$
Use identity $\cos^2\theta = \frac{1 + \cos 2\theta}{2}$:
$$= 2\int_{-\pi/2}^{\pi/2} \frac{1 + \cos 2\theta}{2}, d\theta = \int_{-\pi/2}^{\pi/2} (1 + \cos 2\theta), d\theta$$
$$= \left[\theta + \frac{\sin 2\theta}{2}\right]_{-\pi/2}^{\pi/2} = \left(\frac{\pi}{2} + 0\right) - \left(-\frac{\pi}{2} + 0\right) = \pi$$
Area $= \pi$ square units (which matches the area of a circle with radius 1). ✓
Common AP Calculus FRQ Mistakes
| Mistake | Fix |
|---|---|
| Forgetting $+C$ in indefinite integrals | Write $+C$ every time |
| Not showing units in applied problems | Always include units (cm³/s, gallons, etc.) |
| Using displacement instead of total distance | Integrate $|v(t)|$, not $v(t)$ |
| Not justifying with sign analysis | State the sign of $v$ and $a$ explicitly |
| Rounding too early with calculator | Keep full decimals until final answer |