AP Calculus AB/BC | 20 Essential Problems

Unit 1 · Limits & Continuity

Core Concept

Limits & L'Hôpital's Rule

A limit describes the value a function approaches as the input approaches a point. For indeterminate forms \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), apply L'Hôpital's Rule.

\[\lim_{x\to a}\frac{f(x)}{g(x)} = \lim_{x\to a}\frac{f'(x)}{g'(x)} \quad \text{(if indeterminate)}\]

📌 Must Memorize

\(\lim_{x\to 0}\dfrac{\sin x}{x} = 1\)
\(\lim_{x\to 0}\dfrac{1-\cos x}{x} = 0\)
\(\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x = e\)

Quick Example

Find \(\lim_{x\to 0}\dfrac{\sin 3x}{x}\)

= 3 · lim(sin 3x / 3x) = 3 · 1 = 3

Question 01

AB/BC Medium

What is \(\displaystyle\lim_{x \to 0} \frac{\tan x - \sin x}{x^3}\)?

Using Taylor series: \(\tan x = x + \frac{x^3}{3} + \cdots\) and \(\sin x = x - \frac{x^3}{6} + \cdots\)
So \(\tan x - \sin x = \frac{x^3}{3} + \frac{x^3}{6} + \cdots = \frac{x^3}{2} + \cdots\)
Therefore \(\dfrac{\tan x - \sin x}{x^3} \to \dfrac{1}{2}\) as \(x\to 0\).

Question 02

AB/BC Easy

Which of the following conditions is NOT sufficient to guarantee that \(f\) is continuous at \(x = a\)?

Continuity requires three conditions: (1) \(f(a)\) is defined, (2) \(\lim_{x\to a}f(x)\) exists, and (3) the limit equals \(f(a)\).
Option C only guarantees the two-sided limit exists, but does NOT require \(f(a)\) to equal that limit (or even be defined). So C alone is insufficient for continuity.

Unit 2–3 · Derivatives

Core Concept

Differentiation Rules

Master the chain rule, product rule, and implicit differentiation — the backbone of the FRQ section.

\[\frac{d}{dx}[f(g(x))] = f'(g(x))\cdot g'(x)\]

📌 Must Memorize

\(\dfrac{d}{dx}[\ln u] = \dfrac{u'}{u}\)
\(\dfrac{d}{dx}[a^u] = a^u \ln a \cdot u'\)
\(\dfrac{d}{dx}[\arctan u] = \dfrac{u'}{1+u^2}\)
\(\dfrac{d}{dx}[\arcsin u] = \dfrac{u'}{\sqrt{1-u^2}}\)

Quick Example

Find \(\dfrac{d}{dx}[x^2 e^{3x}]\)

= 2x·e^(3x) + x²·3e^(3x) = e^(3x)(2x + 3x²)

Question 03

AB/BC Medium

If \(y = \ln(\cos^2 x)\), then \(\dfrac{dy}{dx} =\)

Use the chain rule: \(y = \ln(\cos^2 x) = 2\ln(\cos x)\)
\(\dfrac{dy}{dx} = 2\cdot\dfrac{1}{\cos x}\cdot(-\sin x) = \dfrac{-2\sin x}{\cos x} = -2\tan x\)

Question 04

AB/BC Medium

If \(x^2 + xy + y^2 = 7\), then \(\dfrac{dy}{dx}\) at the point \((1, 2)\) is

Differentiate implicitly: \(2x + y + x\dfrac{dy}{dx} + 2y\dfrac{dy}{dx} = 0\)
\(\dfrac{dy}{dx}(x + 2y) = -(2x + y)\)
At \((1,2)\): \(\dfrac{dy}{dx} = \dfrac{-(2+2)}{1+4} = \dfrac{-4}{5}\)

Unit 4 · Derivative Applications

Core Concept

Related Rates & MVT

The Mean Value Theorem (MVT): if \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\), then there exists \(c\) such that \(f'(c) = \dfrac{f(b)-f(a)}{b-a}\).

📌 Must Memorize

Critical points: \(f'(c) = 0\) or \(f'(c)\) DNE
First Derivative Test: sign change of \(f'\)
Second Derivative Test: \(f''(c) > 0\) → local min; \(f''(c) < 0\) → local max

Quick Example

A ladder 10 ft long slides down a wall. When the base is 6 ft from the wall, it moves at 2 ft/s. How fast is the top sliding?

x²+y²=100 → 2x(dx/dt)+2y(dy/dt)=0 → dy/dt = -(6/8)·2 = −1.5 ft/s

Question 05

AB/BC Medium

The radius of a sphere is increasing at a rate of 3 cm/s. How fast is the volume increasing (in cm³/s) when the radius is 5 cm?
(Volume of sphere: \(V = \frac{4}{3}\pi r^3\))

\(\dfrac{dV}{dt} = 4\pi r^2\dfrac{dr}{dt}\)
At \(r = 5\), \(\dfrac{dr}{dt} = 3\):
\(\dfrac{dV}{dt} = 4\pi(25)(3) = 300\pi\) cm³/s

Question 06

AB/BC Hard

Let \(f(x) = x^3 - 3x^2 - 9x + 5\). On which interval is \(f\) both decreasing and concave up?

\(f'(x) = 3x^2 - 6x - 9 = 3(x-3)(x+1)\)
Decreasing when \(f' < 0\): on \((-1, 3)\)
\(f''(x) = 6x - 6\); concave up when \(f'' > 0\): \(x > 1\)
Intersection: \((1, 3)\)

Unit 5 · Integrals & Antiderivatives

Core Concept

Fundamental Theorem of Calculus

FTC links differentiation and integration. Part 1 gives the derivative of an accumulation function; Part 2 evaluates definite integrals.

\[\frac{d}{dx}\int_a^{g(x)} f(t)\,dt = f(g(x))\cdot g'(x)\]

📌 Must Memorize

\(\int x^n\,dx = \dfrac{x^{n+1}}{n+1} + C\quad (n\neq -1)\)
\(\int \dfrac{1}{x}\,dx = \ln|x| + C\)
\(\int e^x\,dx = e^x + C\)
\(\int \sin x\,dx = -\cos x + C\)
\(\int \cos x\,dx = \sin x + C\)

Quick Example

Find \(\dfrac{d}{dx}\displaystyle\int_0^{x^2}\sin t\,dt\)

= sin(x²) · 2x = 2x sin(x²)

Question 07

AB/BC Medium

If \(F(x) = \displaystyle\int_1^{x^3} \ln(t^2+1)\,dt\), then \(F'(x) =\)

By FTC Part 1 with chain rule: \(F'(x) = \ln((x^3)^2+1)\cdot 3x^2 = 3x^2\ln(x^6+1)\)
Note: the upper limit \(x^3\) is substituted into \(t\), giving \(t^2 = (x^3)^2 = x^6\).

Question 08

AB/BC Medium

\(\displaystyle\int_0^{\pi/2} \sin^2 x\,\cos x\,dx =\)

Let \(u = \sin x\), \(du = \cos x\,dx\). Limits: \(x=0\to u=0\), \(x=\pi/2\to u=1\)
\(\displaystyle\int_0^1 u^2\,du = \left[\frac{u^3}{3}\right]_0^1 = \frac{1}{3}\)

Unit 6 · Integration Techniques

Core Concept

Integration by Parts (BC) & Partial Fractions

Integration by parts: \(\int u\,dv = uv - \int v\,du\). Choose \(u\) using LIATE: Logarithm, Inverse trig, Algebraic, Trig, Exponential.

\[\int u\,dv = uv - \int v\,du\]

Quick Example

Evaluate \(\int x e^x\,dx\)

u = x, dv = eˣdx → xe^x − eˣ + C = eˣ(x−1) + C

Question 09

BC Medium

\(\displaystyle\int x\ln x\,dx =\)

IBP: \(u=\ln x\), \(dv=x\,dx\) → \(du=\frac{1}{x}dx\), \(v=\frac{x^2}{2}\)
\(\int x\ln x\,dx = \frac{x^2\ln x}{2} - \int\frac{x^2}{2}\cdot\frac{1}{x}dx = \frac{x^2\ln x}{2} - \frac{1}{2}\int x\,dx = \frac{x^2\ln x}{2} - \frac{x^2}{4} + C\)
Note: Option D equals \(\frac{x^2\ln x}{2} - \frac{x^2}{2}\) which is incorrect.

Question 10

AB/BC Hard

Which of the following integrals represents the area between \(y=x^2\) and \(y=2x\)?

Intersections: \(x^2 = 2x \Rightarrow x(x-2)=0 \Rightarrow x=0,2\)
On \([0,2]\): \(2x \geq x^2\), so area \(= \displaystyle\int_0^2(2x-x^2)\,dx\)
\(= [x^2 - \frac{x^3}{3}]_0^2 = 4 - \frac{8}{3} = \frac{4}{3}\)

Unit 7 · Differential Equations

Core Concept

Separable Differential Equations & Euler's Method

Separate variables and integrate both sides. Exponential growth/decay: \(\dfrac{dy}{dt} = ky \Rightarrow y = Ce^{kt}\).

📌 Must Memorize

Euler's Method: \(y_{n+1} = y_n + h\cdot f'(x_n, y_n)\)
Logistic: \(\dfrac{dy}{dt} = ky\!\left(1-\dfrac{y}{L}\right)\), inflection at \(y = L/2\)

Question 11

AB/BC Medium

The solution to \(\dfrac{dy}{dx} = \dfrac{y}{x}\) with initial condition \(y(1)=3\) is

Separate: \(\dfrac{dy}{y} = \dfrac{dx}{x}\)
Integrate: \(\ln|y| = \ln|x| + C_1\) → \(y = Cx\)
Apply \(y(1)=3\): \(3 = C(1)\) → \(C=3\), so \(y = 3x\)

Question 12

AB/BC Hard

A population satisfies \(\dfrac{dP}{dt} = 0.4P\left(1-\dfrac{P}{500}\right)\). At which value of \(P\) is the population growing fastest?

For a logistic equation \(\dfrac{dP}{dt} = kP(1-P/L)\), the growth rate is maximized at \(P = \dfrac{L}{2}\).
Here \(L = 500\), so maximum growth occurs at \(P = 250\).
(This is the inflection point of the logistic curve.)

Unit 8 · Applications of Integration

Core Concept

Volumes of Revolution

Disk/Washer Method rotates a region about an axis. Shell Method integrates cylindrical shells — choose whichever avoids solving for the other variable.

\[V_{\text{disk}} = \pi\int_a^b [f(x)]^2\,dx \qquad V_{\text{washer}} = \pi\int_a^b\![R^2\!-\!r^2]\,dx\]

Question 13

AB/BC Hard

The region bounded by \(y = \sqrt{x}\) and \(y = x\) is revolved about the \(x\)-axis. The volume is

Intersections: \(\sqrt{x}=x \Rightarrow x=0,1\). On \([0,1]\), \(\sqrt{x}\geq x\), so outer radius \(R=\sqrt{x}\), inner \(r=x\).
\(V = \pi\displaystyle\int_0^1(x - x^2)\,dx = \pi\!\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1 = \pi\!\left(\frac{1}{2}-\frac{1}{3}\right) = \frac{\pi}{6}\)

Question 14

AB/BC Medium

The average value of \(f(x) = \sin x\) on \([0, \pi]\) is

Average value \(= \dfrac{1}{\pi - 0}\displaystyle\int_0^{\pi}\sin x\,dx = \dfrac{1}{\pi}[-\cos x]_0^{\pi}\)
\(= \dfrac{1}{\pi}(-\cos\pi + \cos 0) = \dfrac{1}{\pi}(1+1) = \dfrac{2}{\pi}\)

Unit 9 · Parametric, Polar & Vectors (BC)

Core Concept

Parametric Calculus

For parametric curves \(x=x(t)\), \(y=y(t)\):

\[\frac{dy}{dx} = \frac{dy/dt}{dx/dt}, \qquad \text{Arc Length} = \int_a^b\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt\]

Polar Area

Area enclosed by \(r = f(\theta)\)

A = ½ ∫[α to β] r² dθ

Question 15

BC Medium

A curve is defined parametrically by \(x = t^2 + 1\) and \(y = t^3 - 3t\). The slope of the tangent at \(t = 2\) is

\(\dfrac{dx}{dt} = 2t\), \(\dfrac{dy}{dt} = 3t^2 - 3\)
At \(t=2\): \(\dfrac{dx}{dt}=4\), \(\dfrac{dy}{dt}= 12-3=9\)
\(\dfrac{dy}{dx} = \dfrac{9}{4}\)

Question 16

BC Hard

The area enclosed by the polar curve \(r = 2\cos\theta\) is

\(r = 2\cos\theta\) is a circle of radius 1 centered at \((1,0)\). Its area = \(\pi r^2 = \pi(1)^2 = \pi\).
Verify: \(A = \frac{1}{2}\displaystyle\int_{-\pi/2}^{\pi/2}(2\cos\theta)^2\,d\theta = 2\int_{-\pi/2}^{\pi/2}\cos^2\theta\,d\theta = 2\cdot\frac{\pi}{2} = \pi\)

Unit 10 · Sequences & Series (BC)

Core Concept

Convergence Tests & Power Series

Key tests: Geometric series, p-series, Ratio Test, Alternating Series Test. Taylor series represent functions as infinite polynomials.

\[e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!}, \quad \sin x = \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{(2n+1)!}, \quad \frac{1}{1-x} = \sum_{n=0}^{\infty}x^n\]

📌 Must Memorize

Geometric: \(\sum ar^n\) converges iff \(|r|<1\), sum \(= \dfrac{a}{1-r}\)
p-series: \(\sum\dfrac{1}{n^p}\) converges iff \(p>1\)
Ratio Test: \(L = \lim\left|\dfrac{a_{n+1}}{a_n}\right|\); converges if \(L<1\)

Question 17

BC Medium

The sum of the series \(\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n}{3^n}\) is

This is a geometric series: \(\displaystyle\sum_{n=0}^{\infty}\left(-\frac{1}{3}\right)^n\) with \(a=1\), \(r=-\frac{1}{3}\)
Sum \(= \dfrac{1}{1-(-1/3)} = \dfrac{1}{4/3} = \dfrac{3}{4}\)

Question 18

BC Hard

The interval of convergence of \(\displaystyle\sum_{n=1}^{\infty}\frac{(x-2)^n}{n\cdot 3^n}\) is

Ratio test: \(L=\lim\left|\dfrac{(x-2)^{n+1}}{(n+1)3^{n+1}}\cdot\dfrac{n\cdot3^n}{(x-2)^n}\right|=\dfrac{|x-2|}{3}\)
Converges when \(|x-2|<3\), i.e. \(-1<x<5\). Check endpoints:
At \(x=-1\): \(\sum\dfrac{(-3)^n}{n\cdot3^n}=\sum\dfrac{(-1)^n}{n}\) → alternating harmonic series → converges
At \(x=5\): \(\sum\dfrac{3^n}{n\cdot3^n}=\sum\dfrac{1}{n}\) → harmonic series → diverges
∴ Interval of convergence: \([-1,\, 5)\)

Unit 9 · Particle Motion & Accumulation

Question 19

AB/BC Medium

A particle moves along the \(x\)-axis with velocity \(v(t) = t^2 - 4t + 3\) for \(t \geq 0\). What is the total distance traveled from \(t=0\) to \(t=4\)?

\(v(t)=t^2-4t+3=(t-1)(t-3)\). Zeros at \(t=1,3\).
Total distance \(=\displaystyle\int_0^1(t^2-4t+3)\,dt + \int_1^3\!(-(t^2-4t+3))\,dt + \int_3^4(t^2-4t+3)\,dt\)
\(=\left[\tfrac{t^3}{3}-2t^2+3t\right]_0^1 + \left[-\tfrac{t^3}{3}+2t^2-3t\right]_1^3 + \left[\tfrac{t^3}{3}-2t^2+3t\right]_3^4\)
\(=(1-2+3)+[(-9+18-9)-(-\frac{1}{3}+2-3)]+(\frac{64}{3}-32+12)-(\frac{27}{3}-18+9)\)
\(= \frac{4}{3} + \frac{4}{3} + \frac{4}{3} + \frac{8}{3} = \frac{20}{3}\)

Unit 10 · Taylor Series (BC)

Question 20

BC Hard

The coefficient of \(x^4\) in the Taylor series for \(f(x) = e^{x^2}\) centered at \(x = 0\) is

Since \(e^u = 1 + u + \dfrac{u^2}{2!} + \dfrac{u^3}{3!} + \cdots\), substitute \(u = x^2\):
\(e^{x^2} = 1 + x^2 + \dfrac{x^4}{2!} + \dfrac{x^6}{3!} + \cdots\)
The coefficient of \(x^4\) is \(\dfrac{1}{2!} = \dfrac{1}{2}\).

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AP Calculus AB & BC

Solution Key & Explanations