AP Calculus AB/BC – Complete Reference Guide

1. Limits — Definition & Laws

Meaning

$\lim_{x \to a} f(x) = L$: $f(x)$ approaches $L$ as $x$ approaches $a$. The value $f(a)$ is irrelevant — it may differ from $L$ or not exist.

Existence (One-Sided)

$\lim_{x \to a} f(x) = L$ if and only if $\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = L$

Limit Laws

Each law requires that $\lim f$ and $\lim g$ both exist and are finite.

$\lim (f \pm g) = \lim f \pm \lim g$

$\lim (fg) = (\lim f)(\lim g)$

$\lim \frac{f}{g} = \frac{\lim f}{\lim g}$, provided $\lim g \neq 0$

$\lim [f]^n = [\lim f]^n$

If a limit does not exist, the laws do not apply — you cannot split $\lim(f+g)$ when the pieces individually diverge.

2. Evaluating Limits

Direct Substitution

Valid whenever $f$ is continuous at $a$ (all polynomials, and rationals/roots/trig/exp/log at interior points of their domains).

Indeterminate $\frac{0}{0}$

Factor and cancel

Rationalize (multiply by the conjugate)

Combine complex fractions

Limits at Infinity (Rational $f$)

$\deg(\text{num}) < \deg(\text{den}) \Rightarrow$ limit $= 0$

$\deg(\text{num}) = \deg(\text{den}) \Rightarrow$ ratio of leading coefficients

$\deg(\text{num}) > \deg(\text{den}) \Rightarrow \pm\infty$ (no horizontal asymptote)

Must-Know Special Limits

$\lim_{x \to 0} \frac{\sin x}{x} = 1 \qquad \lim_{x \to 0} \frac{1 - \cos x}{x} = 0$

$\frac{0}{0}$ and $\frac{\infty}{\infty}$ are indeterminate (more work needed). A nonzero number over $0$ is not indeterminate — it diverges.

3. L'Hôpital's Rule

All Hypotheses Required

1. $\lim_{x \to a} \frac{f(x)}{g(x)}$ has the indeterminate form $\frac{0}{0}$ or $\frac{\pm\infty}{\pm\infty}$ — check the form first

2. $f$ and $g$ are differentiable near $a$, with $g'(x) \neq 0$ near $a$ (except possibly at $a$)

3. $\lim_{x \to a} \frac{f'(x)}{g'(x)}$ exists or is $\pm\infty$

Then: $\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$

Also valid for $x \to \pm\infty$ and for one-sided limits.

Three ways this goes wrong: (a) applying it to a form that is not indeterminate; (b) differentiating the quotient instead of the numerator and denominator separately; (c) concluding "the limit does not exist" when $\lim f'/g'$ fails to exist — hypothesis 3 failing means the rule is silent, not that the original limit diverges.

Other indeterminate forms — $0 \cdot \infty$, $\infty - \infty$, $1^{\infty}$, $0^0$, $\infty^0$ — must be rewritten as a quotient first (algebra, or take $\ln$) before the rule applies.

4. Continuity & Discontinuities

Continuous at $x = a$: All Three

1. $f(a)$ is defined

2. $\lim_{x \to a} f(x)$ exists

3. $\lim_{x \to a} f(x) = f(a)$

Types of Discontinuity

Removable (hole): the limit exists but $\neq f(a)$, or $f(a)$ undefined

Jump: one-sided limits exist but differ

Infinite: vertical asymptote

Relationship to Differentiability

Differentiable at $a$ $\Rightarrow$ continuous at $a$.

The converse is false: $f(x) = |x|$ is continuous at $0$ but not differentiable there. Continuity never implies differentiability.

5. IVT, EVT & Squeeze Theorem

Intermediate Value Theorem

If $f$ is continuous on the closed interval $[a,b]$ and $N$ is between $f(a)$ and $f(b)$, then there exists $c \in (a,b)$ with $f(c) = N$.

Guarantees existence only — not uniqueness, and no method for finding $c$.

Extreme Value Theorem

If $f$ is continuous on a closed, bounded interval $[a,b]$, then $f$ attains both an absolute maximum and an absolute minimum on $[a,b]$.

Squeeze (Sandwich) Theorem

If $g(x) \leq f(x) \leq h(x)$ for all $x$ near $a$ (except possibly at $a$) and $\lim_{x \to a} g = \lim_{x \to a} h = L$, then $\lim_{x \to a} f = L$.

Hypotheses do real work. IVT: $f(x)=1/x$ on $[-1,1]$ runs from $-1$ to $1$ but never equals $0$ — it is not continuous. EVT: $f(x)=x$ on the open $(0,1)$ attains neither a max nor a min — the interval must be closed.

6. The Derivative — Definition

Limit Definitions

$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

$f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}$

Interpretations

Slope of the tangent line at $x = a$

Instantaneous rate of change of $f$ at $a$

Limit of the average rate of change $\frac{f(b) - f(a)}{b - a}$ as $b \to a$

Tangent & Normal Lines

Tangent: $y - f(a) = f'(a)(x - a)$

Normal: slope $= -\frac{1}{f'(a)}$, provided $f'(a) \neq 0$

Where $f$ Fails to Be Differentiable

Corner or cusp

Vertical tangent

Any discontinuity

7. Differentiation Rules

Core Rules

Constant: $\frac{d}{dx}[c] = 0$

Power: $\frac{d}{dx}[x^n] = nx^{n-1}$

Constant multiple: $\frac{d}{dx}[cf] = cf'$

Sum: $\frac{d}{dx}[f \pm g] = f' \pm g'$

Product Rule

$\frac{d}{dx}[fg] = f'g + fg'$

Quotient Rule

$\frac{d}{dx}\left[\frac{f}{g}\right] = \frac{f'g - fg'}{g^2}$, where $g \neq 0$

Order matters — the numerator is not symmetric.

Chain Rule

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

Leibniz form: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

The chain rule is the most-missed rule on the exam. Any time the inside of a function is not just $x$, the outer derivative must be multiplied by the inner derivative.

8. Derivatives of Elementary Functions

Trigonometric

$\frac{d}{dx}\sin x = \cos x \qquad \frac{d}{dx}\cos x = -\sin x$

$\frac{d}{dx}\tan x = \sec^2 x \qquad \frac{d}{dx}\cot x = -\csc^2 x$

$\frac{d}{dx}\sec x = \sec x \tan x \qquad \frac{d}{dx}\csc x = -\csc x \cot x$

Every co-function derivative carries a minus sign.

Exponential & Logarithmic

$\frac{d}{dx}e^x = e^x \qquad \frac{d}{dx}a^x = a^x \ln a$

$\frac{d}{dx}\ln x = \frac{1}{x} \; (x > 0) \qquad \frac{d}{dx}\log_a x = \frac{1}{x \ln a}$

Inverse Trigonometric

$\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1 - x^2}} \qquad \frac{d}{dx}\arccos x = \frac{-1}{\sqrt{1 - x^2}}$

$\frac{d}{dx}\arctan x = \frac{1}{1 + x^2} \qquad \frac{d}{dx}\text{arccot}\, x = \frac{-1}{1 + x^2}$

Beyond the AP course: $\frac{d}{dx}\text{arcsec}\, x = \frac{1}{|x|\sqrt{x^2 - 1}}$ — the absolute value is required, not cosmetic.

9. Implicit Differentiation & Inverses

Implicit Differentiation

Differentiate both sides with respect to $x$

Every $y$ term picks up a factor of $\frac{dy}{dx}$ (chain rule)

Collect the $\frac{dy}{dx}$ terms and solve

Example: $x^2 + y^2 = 25 \Rightarrow 2x + 2y\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{x}{y}$

Implicit derivatives usually depend on both $x$ and $y$ — substitute a full point, not just an $x$-value.

Second Derivative, Implicitly

Differentiate $\frac{dy}{dx}$ again, then substitute the expression for $\frac{dy}{dx}$ back in.

Derivative of an Inverse Function

If $f$ is one-to-one and differentiable and $g = f^{-1}$, then

$g'(b) = \frac{1}{f'(g(b))}$, provided $f'(g(b)) \neq 0$

To use it you need the point on $f$: if $f(2) = 7$ then $g(7) = 2$, so $g'(7) = 1/f'(2)$.

10. Related Rates

Procedure

1. Sketch; name the variables and the rates

2. Write an equation relating the variables

3. Differentiate both sides with respect to $t$

4. Substitute the known values

5. Solve for the unknown rate; include units

Differentiate first, substitute second. Plugging in a value that is changing before differentiating freezes it as a constant and kills its rate term. Only genuinely constant quantities may be substituted early.

Common Relations

Right triangle: $x^2 + y^2 = z^2$

Similar triangles: proportional sides

Cone: $V = \frac{1}{3}\pi r^2 h$ (use similar triangles to remove a variable)

Sphere: $V = \frac{4}{3}\pi r^3$, $S = 4\pi r^2$

Cylinder: $V = \pi r^2 h$

11. Mean Value Theorem & Rolle's Theorem

Mean Value Theorem

If $f$ is

continuous on the closed interval $[a,b]$, and

differentiable on the open interval $(a,b)$,

then there exists $c \in (a,b)$ with

$f'(c) = \frac{f(b) - f(a)}{b - a}$

Some instantaneous rate equals the average rate over the interval.

Rolle's Theorem

Same two hypotheses plus $f(a) = f(b)$; then there exists $c \in (a,b)$ with $f'(c) = 0$.

Both hypotheses are required. $f(x) = |x|$ on $[-1,1]$ is continuous and has secant slope $0$, but $f'$ is $+1$ or $-1$ wherever it exists — no such $c$ exists, because $f$ is not differentiable at $0$.

Consequence

If $f' = 0$ on an interval, $f$ is constant there. If $f' = g'$ on an interval, then $f - g$ is constant there.

12. Increasing / Decreasing & Extrema

Monotonicity

$f' > 0$ on an interval $\Rightarrow f$ is increasing there

$f' < 0$ on an interval $\Rightarrow f$ is decreasing there

Critical Points

Interior $c$ where $f'(c) = 0$ or $f'(c)$ is undefined.

Every interior extremum occurs at a critical point — but a critical point need not be an extremum ($f(x)=x^3$ at $0$).

First Derivative Test

$f'$ changes $+ \to -$ at $c$: local maximum

$f'$ changes $- \to +$ at $c$: local minimum

No sign change: neither

Absolute Extrema on $[a,b]$

Candidates test (valid because EVT guarantees the extrema exist): evaluate $f$ at every critical point and at both endpoints, then compare the values.

Endpoints are candidates for absolute extrema but are never critical points, and the first derivative test does not apply there.

13. Concavity & Inflection Points

Concavity

$f'' > 0 \Rightarrow$ concave up ($f'$ increasing)

$f'' < 0 \Rightarrow$ concave down ($f'$ decreasing)

Inflection Points

A point where the concavity changes — $f''$ must change sign there, and $f$ must be continuous at the point.

$f''(c) = 0$ alone is not enough: $f(x) = x^4$ has $f''(0) = 0$ but $f'' > 0$ on both sides, so there is no inflection point at $0$. Always check for a sign change.

Second Derivative Test

At a critical point $c$ with $f'(c) = 0$:

$f''(c) < 0 \Rightarrow$ local maximum

$f''(c) > 0 \Rightarrow$ local minimum

$f''(c) = 0 \Rightarrow$ inconclusive — fall back on the first derivative test

Both $x^4$ (a minimum) and $x^3$ (neither) have $f''(0) = 0$, which is exactly why the test cannot decide.

14. Optimization & Linear Approximation

Optimization Procedure

Write the objective function to be optimized

Use the constraint to reduce to one variable

State the domain (physical limits matter)

Find critical points; justify with the first or second derivative test, or the candidates test

An answer is not complete without a justification that the critical point really is the max or min — this is where free-response points are lost.

Linearization

$L(x) = f(a) + f'(a)(x - a)$, valid for $x$ near $a$

This is the tangent line, i.e. the first-degree Taylor polynomial at $a$.

$f$ concave up near $a$: tangent line lies below, so $L$ underestimates

$f$ concave down near $a$: $L$ overestimates

Differentials

$dy = f'(x)\,dx$, so $\Delta y \approx f'(x)\,\Delta x$

15. Particle Motion — Derivative View

The Chain

position $x(t) \;\to\; $ velocity $v(t) = x'(t) \;\to\;$ acceleration $a(t) = v'(t) = x''(t)$

Speed vs Velocity

$\text{speed} = |v(t)|$ — a nonnegative scalar; velocity carries the direction.

Speeding Up or Slowing Down

$v$ and $a$ have the same sign $\Rightarrow$ speeding up

$v$ and $a$ have opposite signs $\Rightarrow$ slowing down

"Speeding up" is not the same as "$a > 0$". A particle with $v < 0$ and $a < 0$ is speeding up while its velocity decreases.

Key Events

At rest: $v(t) = 0$

Changes direction: $v$ changes sign (a zero of $v$ alone is not enough)

Farthest point: check where $v = 0$ and the endpoints

16. Riemann Sums & Trapezoidal Rule

Setup (Uniform Partition)

$\Delta x = \frac{b - a}{n}$, $\;x_i = a + i\Delta x$

The Sums

Left: $\sum_{i=1}^{n} f(x_{i-1})\Delta x$

Right: $\sum_{i=1}^{n} f(x_i)\Delta x$

Midpoint: $\sum_{i=1}^{n} f\!\left(\frac{x_{i-1} + x_i}{2}\right)\Delta x$

Trapezoidal Rule

$T_n = \frac{\Delta x}{2}\left[f(x_0) + 2f(x_1) + 2f(x_2) + \cdots + 2f(x_{n-1}) + f(x_n)\right]$

Interior coefficients are $2$; the two endpoints are not doubled.

Over or Under?

$f$ increasing: left under, right over (reversed if $f$ is decreasing)

$f$ concave up: trapezoid over, midpoint under

Definite Integral as a Limit

$\int_a^b f(x)\,dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*)\Delta x$

17. Definite Integral — Properties

$\int_a^a f(x)\,dx = 0$

$\int_b^a f(x)\,dx = -\int_a^b f(x)\,dx$

$\int_a^b [f(x) \pm g(x)]\,dx = \int_a^b f(x)\,dx \pm \int_a^b g(x)\,dx$

$\int_a^b cf(x)\,dx = c\int_a^b f(x)\,dx$

$\int_a^b f = \int_a^c f + \int_c^b f$ for any $c$

Sign & Symmetry

$f \geq 0$ on $[a,b] \Rightarrow \int_a^b f \geq 0$

The integral counts area below the axis as negative — it is net signed area, not total area

$f$ even: $\int_{-a}^{a} f = 2\int_0^a f$

$f$ odd: $\int_{-a}^{a} f = 0$

There is no product or quotient rule for integrals: $\int fg \neq \left(\int f\right)\left(\int g\right)$.

18. Fundamental Theorem of Calculus

Part 1 — Derivative of an Integral

If $f$ is continuous on $[a,b]$, then $g(x) = \int_a^x f(t)\,dt$ is continuous on $[a,b]$, differentiable on $(a,b)$, and

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

With the Chain Rule

$\frac{d}{dx}\int_a^{u(x)} f(t)\,dt = f(u(x)) \cdot u'(x)$

$\frac{d}{dx}\int_{v(x)}^{u(x)} f(t)\,dt = f(u)u' - f(v)v'$

Part 2 — Evaluation

If $f$ is continuous on $[a,b]$ and $F$ is any antiderivative of $f$ on $[a,b]$, then

$\int_a^b f(x)\,dx = F(b) - F(a)$

Net Change Theorem

$\int_a^b F'(x)\,dx = F(b) - F(a)$

The integral of a rate of change gives the net change in the quantity — the backbone of the FRQ "accumulation" problems.

Continuity on $[a,b]$ is required. Applying Part 2 across an infinite discontinuity (e.g. $\int_{-1}^{1} \frac{dx}{x^2}$) produces a finite-looking but wrong answer; that integral in fact diverges.

19. Basic Antiderivatives

Every indefinite integral needs $+ \, C$.

$\int x^n\,dx = \frac{x^{n+1}}{n+1} + C$, $n \neq -1$

$\int \frac{1}{x}\,dx = \ln|x| + C$

The absolute value is required — without it the formula is undefined for $x < 0$, where $1/x$ is perfectly well behaved. Likewise $\int \frac{u'}{u}\,dx = \ln|u| + C$.

$\int e^x\,dx = e^x + C \qquad \int a^x\,dx = \frac{a^x}{\ln a} + C$

$\int \sin x\,dx = -\cos x + C \qquad \int \cos x\,dx = \sin x + C$

$\int \sec^2 x\,dx = \tan x + C \qquad \int \csc^2 x\,dx = -\cot x + C$

$\int \sec x \tan x\,dx = \sec x + C$

$\int \csc x \cot x\,dx = -\csc x + C$

$\int \tan x\,dx = -\ln|\cos x| + C = \ln|\sec x| + C$

$\int \cot x\,dx = \ln|\sin x| + C$

Inverse-Trig Forms

$\int \frac{dx}{\sqrt{a^2 - x^2}} = \arcsin\frac{x}{a} + C$

$\int \frac{dx}{a^2 + x^2} = \frac{1}{a}\arctan\frac{x}{a} + C$

Note the $\frac{1}{a}$ out front on the arctangent form — dropping it is a standard error.

20. u-Substitution

Method

$\int f(g(x))g'(x)\,dx = \int f(u)\,du$ with $u = g(x)$, $du = g'(x)\,dx$

Example: $\int 2x e^{x^2}\,dx = \int e^u\,du = e^u + C = e^{x^2} + C$

Definite Integrals — Two Valid Routes

Change the limits to $u$-values: $\int_a^b f(g(x))g'(x)\,dx = \int_{g(a)}^{g(b)} f(u)\,du$

Or back-substitute to $x$ before evaluating at the original limits

Evaluating a $u$-integral at the original $x$-limits is one of the most common integration errors. Pick one route and label which variable your limits belong to.

Choosing $u$

The inside of a composition

Something whose derivative also appears (up to a constant)

The denominator, or the exponent

A constant factor can always be adjusted: if $du$ is off by a multiplicative constant, fix it outside the integral.

21. Area Between Curves

Vertical Slices

$A = \int_a^b [\,\text{top} - \text{bottom}\,]\,dx = \int_a^b [f(x) - g(x)]\,dx$ where $f \geq g$ on $[a,b]$

Horizontal Slices

$A = \int_c^d [\,\text{right} - \text{left}\,]\,dy$

Choose the orientation that needs fewer integrals.

Procedure

Find the intersection points — these are the limits

Determine which curve is on top on each subinterval

If the curves cross inside the interval, split the integral at each crossing

$\int_a^b [f - g]\,dx$ is signed. If the curves swap order and you do not split, the pieces cancel and the area comes out too small. Area needs $\int |f - g|\,dx$.

Area is always positive. A negative answer means the subtraction order was reversed somewhere.

22. Volume — Cross Sections, Disks & Washers

Known Cross Sections

$V = \int_a^b A(x)\,dx$, where $A(x)$ is the area of the slice perpendicular to the axis and $s$ is the length of the base slice.

Square, side $s$: $A = s^2$

Semicircle, diameter $s$: $A = \frac{\pi s^2}{8}$

Equilateral triangle, side $s$: $A = \frac{\sqrt{3}}{4}s^2$

Isosceles right triangle, leg $s$: $A = \frac{s^2}{2}$

Isosceles right triangle, hypotenuse $s$: $A = \frac{s^2}{4}$

Read carefully whether $s$ is the diameter or the radius, the leg or the hypotenuse — the factor changes.

Disk Method

$V = \pi\int_a^b [R(x)]^2\,dx$

Washer Method

$V = \pi\int_a^b \left([R(x)]^2 - [r(x)]^2\right)dx$

$R$ and $r$ are distances from the axis of rotation, not from the $x$-axis. Rotating about $y = k$ makes the outer radius $|f(x) - k|$. Also $[R]^2 - [r]^2 \neq (R - r)^2$ — square first, then subtract.

The shell method is not assessed on the AP exam; disks, washers and known cross sections cover every AP volume question.

23. Average Value & MVT for Integrals

Average Value of a Function

$f_{\text{avg}} = \frac{1}{b - a}\int_a^b f(x)\,dx$

Mean Value Theorem for Integrals

If $f$ is continuous on $[a,b]$, then there exists $c \in [a,b]$ with

$f(c) = \frac{1}{b - a}\int_a^b f(x)\,dx$

The function actually attains its average value somewhere on the interval.

Three different "averages" — do not mix them up:

Average value of $f$: $\frac{1}{b-a}\int_a^b f\,dx$

Average rate of change of $f$: $\frac{f(b) - f(a)}{b - a}$

Average velocity: $\frac{1}{b-a}\int_a^b v\,dt = \frac{x(b) - x(a)}{b - a}$

Average value of the rate $f'$ equals the average rate of change of $f$ — that is just the Net Change Theorem.

24. Accumulation & Motion — Integral View

Accumulation Functions

$g(x) = g(a) + \int_a^x g'(t)\,dt$

"Final $=$ initial $+$ accumulated change." Read $g$'s behaviour off the graph of $g'$: $g$ increases where $g' > 0$; $g$ has a local max where $g'$ crosses from $+$ to $-$; $g$ is concave up where $g'$ is increasing.

Position from Velocity

$x(t_2) = x(t_1) + \int_{t_1}^{t_2} v(t)\,dt$

$v(t_2) = v(t_1) + \int_{t_1}^{t_2} a(t)\,dt$

Displacement vs Total Distance

Displacement: $\int_{t_1}^{t_2} v(t)\,dt$

Total distance: $\int_{t_1}^{t_2} |v(t)|\,dt$

These differ whenever $v$ changes sign. To compute total distance by hand, find the zeros of $v$, split the integral there, and add the absolute values of the pieces.

Net change in a quantity given its rate $R(t)$: $\int_{t_1}^{t_2} R(t)\,dt$ — with the units of $R$ times the units of $t$.

25. Differential Equations & Slope Fields

Slope Fields

At each point $(x,y)$ draw a short segment of slope $\frac{dy}{dx} = f(x,y)$. Solution curves follow the segments.

If $\frac{dy}{dx}$ depends only on $x$, segments are constant down each vertical line

If it depends only on $y$, segments are constant along each horizontal line

Separation of Variables

1. Write as $g(y)\,dy = h(x)\,dx$

2. Integrate both sides; one constant $C$ suffices

3. Apply the initial condition to find $C$ before solving for $y$ when possible — the algebra is easier

4. Solve for $y$, choosing the branch that matches the initial condition

Exponential Model

$\frac{dy}{dt} = ky \;\Longrightarrow\; y = y_0 e^{kt}$

$k > 0$ growth, $k < 0$ decay. Half-life and doubling time both come from solving $e^{kt} = \tfrac12$ or $2$.

A solution must satisfy both the equation and the initial condition — always check by substituting back.

26. Euler's Method & Logistic Growth — BC ONLY

BCEuler's Method

Given $\frac{dy}{dx} = f(x,y)$ with $y(x_0) = y_0$ and step size $h$:

$y_{n+1} = y_n + h \cdot f(x_n, y_n), \qquad x_{n+1} = x_n + h$

Each step follows the tangent line, so the approximation underestimates where the solution is concave up and overestimates where it is concave down.

BCLogistic Differential Equation

$\frac{dP}{dt} = kP\left(1 - \frac{P}{L}\right)$, $L$ = carrying capacity

Solution: $P(t) = \frac{L}{1 + Ae^{-kt}}$, where $A = \frac{L - P_0}{P_0}$

BCLogistic Behaviour

$\lim_{t \to \infty} P(t) = L$ for any $P_0 > 0$

$P$ grows fastest when $P = \frac{L}{2}$ — this is the inflection point of the solution curve

Equilibrium solutions: $P = 0$ and $P = L$

Read $L$ straight off the equation: writing $\frac{dP}{dt} = 0.05P(1 - \frac{P}{800})$ makes $L = 800$ and $k = 0.05$.

27. Integration by Parts — BC ONLY

BCFormula

$\int u\,dv = uv - \int v\,du$

Definite form: $\int_a^b u\,dv = \Big[uv\Big]_a^b - \int_a^b v\,du$

BCChoosing $u$ — LIATE

Logarithmic, Inverse trig, Algebraic, Trig, Exponential — pick $u$ as early in this list as possible.

The aim is a $\int v\,du$ that is simpler than what you started with; LIATE is a heuristic, not a theorem.

BCWorked Examples

$\int x e^x\,dx = xe^x - e^x + C$

$\int x\cos x\,dx = x\sin x + \cos x + C$

$\int \ln x\,dx = x\ln x - x + C$ (take $u = \ln x$, $dv = dx$)

$\int \arctan x\,dx = x\arctan x - \tfrac{1}{2}\ln(1 + x^2) + C$

Repeated parts on $\int e^x \sin x\,dx$ returns the original integral — solve for it algebraically rather than continuing.

28. Partial Fractions — BC ONLY

BCAP Scope

On the AP exam, partial fractions is limited to non-repeating linear factors.

$\frac{P(x)}{(x - a)(x - b)} = \frac{A}{x - a} + \frac{B}{x - b}$

BCRequired Condition

The decomposition is only valid when $\deg(P) < \deg(\text{denominator})$. If it is not, do polynomial long division first and decompose the remainder.

BCMethod

Factor the denominator

Multiply through by the denominator

Substitute the roots to isolate each constant

BCExample

$\frac{1}{(x-1)(x+2)} = \frac{1/3}{x-1} - \frac{1/3}{x+2}$

$\int \frac{dx}{(x-1)(x+2)} = \tfrac{1}{3}\ln|x-1| - \tfrac{1}{3}\ln|x+2| + C$

Absolute values on both logarithms.

29. Improper Integrals — BC ONLY

BCType 1 — Infinite Limit

$\int_a^{\infty} f(x)\,dx = \lim_{t \to \infty} \int_a^{t} f(x)\,dx$

$\int_{-\infty}^{\infty} f\,dx = \int_{-\infty}^{c} f\,dx + \int_{c}^{\infty} f\,dx$ — both pieces must converge independently

BCType 2 — Infinite Discontinuity

If $f$ is unbounded at $x = b$: $\int_a^b f\,dx = \lim_{t \to b^-} \int_a^{t} f\,dx$

If the discontinuity is at an interior $c$, split at $c$ and take a one-sided limit on each piece.

BCConvergence

Converges if the limit exists and is finite; otherwise it diverges.

$\int_1^{\infty} \frac{dx}{x^p}$ converges $\iff p > 1$

Compare with the $p$-series, which converges under the same condition.

An interior discontinuity that goes unnoticed is the classic trap: treating $\int_{-1}^{1}\frac{dx}{x^2}$ as an ordinary integral gives $-2$, but the integral actually diverges.

30. Parametric & Vector Functions — BC ONLY

BCFirst Derivative

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$, provided $\frac{dx}{dt} \neq 0$

BCSecond Derivative

$\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\!\left(\frac{dy}{dx}\right)}{dx/dt}$

$\frac{d^2y}{dx^2} \neq \frac{d^2y/dt^2}{d^2x/dt^2}$. Differentiate $\frac{dy}{dx}$ with respect to $t$, then divide by $\frac{dx}{dt}$ again.

BCVector-Valued Motion

Position $\langle x(t), y(t)\rangle$; velocity $\langle x'(t), y'(t)\rangle$; acceleration $\langle x''(t), y''(t)\rangle$

Speed $= \sqrt{[x'(t)]^2 + [y'(t)]^2}$

BCDistance Travelled

$\int_{t_1}^{t_2} \sqrt{[x'(t)]^2 + [y'(t)]^2}\,dt$

This is the arc length of the path — the integral of speed. Displacement is the vector $\left\langle \int x'\,dt, \int y'\,dt \right\rangle$, which is generally shorter.

31. Polar Coordinates — BC ONLY

BCConversion

$x = r\cos\theta$, $y = r\sin\theta$, $r^2 = x^2 + y^2$

The arctangent quadrant trap. $\theta = \arctan\!\frac{y}{x}$ is only correct when $x > 0$, because $\arctan$ returns angles in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. For $x < 0$ you must add $\pi$. Example: the point $(-1, 1)$ lies in quadrant II at $\theta = \frac{3\pi}{4}$, but $\arctan(-1) = -\frac{\pi}{4}$ — off by exactly $\pi$.

BCArea in Polar Form

$A = \frac{1}{2}\int_{\alpha}^{\beta} r^2\,d\theta$

Between two curves with $r_{\text{out}} \geq r_{\text{in}} \geq 0$:

$A = \frac{1}{2}\int_{\alpha}^{\beta} \left(r_{\text{out}}^2 - r_{\text{in}}^2\right)d\theta$

Choose $\alpha, \beta$ so the region is traced exactly once; for roses and cardioids the natural-looking bounds often sweep it twice.

BCSlope of a Polar Curve

$\frac{dy}{dx} = \frac{\frac{dr}{d\theta}\sin\theta + r\cos\theta} {\frac{dr}{d\theta}\cos\theta - r\sin\theta}$

This is just the parametric rule applied to $x = r(\theta)\cos\theta$, $y = r(\theta)\sin\theta$.

32. Arc Length — BC ONLY

BCFunction Form

$L = \int_a^b \sqrt{1 + [f'(x)]^2}\,dx$

In terms of $y$: $L = \int_c^d \sqrt{1 + \left[\frac{dx}{dy}\right]^2}\,dy$

BCParametric Form

$L = \int_{t_1}^{t_2} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\,dt$

Requires $f'$ (or $x'$, $y'$) continuous on the interval.

BCSanity Checks

Arc length is never shorter than the straight-line distance between the endpoints

A straight segment gives exactly the distance formula

These integrands rarely have elementary antiderivatives — most arc length questions appear on the calculator-permitted portion of the exam.

33. Sequences & Series Basics — BC ONLY

BCSequences

$\{a_n\}$ converges if $\lim_{n \to \infty} a_n$ exists and is finite.

Monotone convergence: a sequence that is monotonic and bounded converges. Both conditions are needed.

BCSeries and Partial Sums

$S_N = \sum_{n=1}^{N} a_n$; the series $\sum_{n=1}^{\infty} a_n$ converges by definition when the sequence $\{S_N\}$ converges.

BCGeometric Series

$\sum ar^n$ converges $\iff |r| < 1$, and then

$\text{sum} = \dfrac{\text{first term}}{1 - r}$

Writing $\frac{a}{1-r}$ assumes the series starts at $n = 0$. Starting at $n = 1$ gives $\frac{ar}{1-r}$ — "first term over $1 - r$" is safe either way.

BCTelescoping Series

$\sum_{n=1}^{\infty}\left(\frac{1}{n} - \frac{1}{n+1}\right) = 1$ — write out $S_N$, cancel, then take $\lim_{N \to \infty} S_N$.

34. Convergence Tests I — BC ONLY

BC$n$th-Term Test (Divergence Only)

If $\lim_{n \to \infty} a_n \neq 0$ or does not exist, then $\sum a_n$ diverges.

The converse is false. $\lim a_n = 0$ proves nothing: the harmonic series $\sum \frac{1}{n}$ has $a_n \to 0$ and still diverges. This test can never establish convergence.

BCGeometric Series

$\sum ar^n$ converges $\iff |r| < 1$ (with $a \neq 0$).

BC$p$-Series

$\sum_{n=1}^{\infty} \frac{1}{n^p}$ converges $\iff p > 1$.

$p = 1$ is the harmonic series and diverges — the boundary is excluded.

BCIntegral Test

Requires $f$ to be continuous, positive, and decreasing on $[N, \infty)$ with $a_n = f(n)$.

Then $\sum a_n$ and $\int_N^{\infty} f(x)\,dx$ both converge or both diverge.

The test compares behaviour only — it does not give the sum. $\int_1^{\infty}\frac{dx}{x^2} = 1$ while $\sum \frac{1}{n^2} = \frac{\pi^2}{6} \approx 1.645$.

35. Convergence Tests II — BC ONLY

BCDirect Comparison

Requires $0 \leq a_n \leq b_n$ for all large $n$ (nonnegative terms).

$\sum b_n$ converges $\Rightarrow \sum a_n$ converges

$\sum a_n$ diverges $\Rightarrow \sum b_n$ diverges

BCLimit Comparison

Requires $a_n > 0$ and $b_n > 0$, and $L = \lim_{n \to \infty}\frac{a_n}{b_n}$ with $0 < L < \infty$ (finite and positive).

Then $\sum a_n$ and $\sum b_n$ both converge or both diverge.

BCAlternating Series Test

For $\sum (-1)^n b_n$ with $b_n > 0$, both conditions are required:

1. $b_n$ is decreasing (for all large $n$)

2. $\lim_{n \to \infty} b_n = 0$

Then the series converges.

Alternating signs plus $b_n \to 0$ is not enough on its own — the decreasing condition must be checked and stated.

BCRatio Test

$L = \lim_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right|$

$L < 1$: converges absolutely

$L > 1$ or $L = \infty$: diverges

$L = 1$: inconclusive — try another test

Both $\sum\frac1n$ (diverges) and $\sum\frac1{n^2}$ (converges) give $L = 1$, which is why the case decides nothing.

Root Test — Not in the AP Course Description

$L = \lim_{n \to \infty} \sqrt[n]{|a_n|}$, with the same three conclusions as the ratio test ($L < 1$ absolute convergence, $L > 1$ divergence, $L = 1$ inconclusive).

Listed for completeness only — the AP course description specifies the $n$th-term, integral, comparison, limit comparison, alternating series and ratio tests. You will not need the root test on the exam.

36. Absolute vs Conditional Convergence — BC ONLY

BCDefinitions

Absolutely convergent: $\sum |a_n|$ converges.

Conditionally convergent: $\sum a_n$ converges but $\sum |a_n|$ diverges.

BCKey Theorem

Absolute convergence $\Rightarrow$ convergence.

The converse fails, which is exactly what "conditional" names.

BCCanonical Example

$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n} = \ln 2$ converges by the alternating series test, but $\sum \frac{1}{n}$ diverges — so it is conditionally convergent.

BCStrategy for "Does It Converge?"

Check $\lim a_n \neq 0$ first — it is the fastest disqualifier

Recognise geometric or $p$-series shapes

Factorials or $n$th powers suggest the ratio test

Alternating signs suggest the alternating series test — then test $\sum|a_n|$ separately to classify it

On free response, name the test and verify its hypotheses explicitly; the conditions carry the credit.

37. Power Series — BC ONLY

BCForm

$\sum_{n=0}^{\infty} c_n(x - a)^n$, centred at $x = a$

BCRadius of Convergence

Apply the ratio test to $|c_n(x-a)^n|$ and solve $L < 1$ for $x$.

$R = 0$: converges only at $x = a$

$0 < R < \infty$: converges on $(a - R,\, a + R)$

$R = \infty$: converges for all $x$

BCInterval of Convergence

The ratio test is silent at the endpoints ($L = 1$ there). Substitute $x = a - R$ and $x = a + R$ separately and test each resulting numerical series. The interval may be open, closed, or half-open.

BCTerm-by-Term Operations

Inside the interval of convergence a power series may be differentiated and integrated term by term, and the radius $R$ is unchanged.

The endpoint behaviour can change: integrating can gain an endpoint, differentiating can lose one.

Building new series from known ones — substitute, multiply by a power of $x$, differentiate or integrate — is almost always faster than computing derivatives.

38. Taylor & Maclaurin Series — BC ONLY

BCTaylor Polynomial of Degree $n$

$P_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x - a)^k$

$= f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n$

BCTaylor Series

$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x - a)^n$

Valid where the series converges and the remainder $R_n(x) \to 0$ — a Taylor series can converge without converging to $f$.

BCMaclaurin Series

The Taylor series centred at $a = 0$: $f(x) = \sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n$

BCNotes

$P_1$ is the tangent line (linearization)

$P_n$ matches $f$ and its first $n$ derivatives at $x = a$

The coefficient of $(x-a)^k$ is $\frac{f^{(k)}(a)}{k!}$, so $f^{(k)}(a) = k! \cdot (\text{that coefficient})$

39. Common Maclaurin Series — BC ONLY

Each interval of convergence is part of the formula.

$e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$ (all $x$)

$\sin x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$ (all $x$)

$\cos x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots$ (all $x$)

$\frac{1}{1 - x} = \sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + \cdots$ ($|x| < 1$)

$\ln(1 + x) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}x^n}{n} = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots$ ($-1 < x \leq 1$)

$\arctan x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1} = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots$ ($|x| \leq 1$)

The AP course requires the first four; the last two are standard and frequently useful.

$\ln(1+x)$ is genuinely half-open: at $x = 1$ it converges (to $\ln 2$), at $x = -1$ it diverges. $\sin$ and $\cos$ carry only odd and only even powers respectively.

40. Error Bounds — BC ONLY

BCAlternating Series Error Bound

If the alternating series test conditions hold ($b_n > 0$, decreasing, $b_n \to 0$), then

$|S - S_N| \leq b_{N+1}$

The error is no bigger than the first omitted term. The bound requires the AST hypotheses — it does not apply to an arbitrary alternating series.

BCLagrange Error Bound

If $f$ has $n+1$ derivatives on an interval containing $a$ and $x$, then the remainder $R_n(x) = f(x) - P_n(x)$ satisfies

$|R_n(x)| \leq \dfrac{M\,|x - a|^{\,n+1}}{(n+1)!}$

where $M$ is a bound on $\left|f^{(n+1)}(t)\right|$ for all $t$ between $a$ and $x$.

Two ways this is misstated. (1) The exponent and the factorial are both $n+1$, not $n$ — a degree-$n$ polynomial has an order-$(n+1)$ error term. (2) $M$ must be the maximum of $|f^{(n+1)}|$ over the whole interval, not its value at $a$; using $f^{(n+1)}(a)$ can produce a "bound" smaller than the actual error, which is simply false.

Exact form (Taylor's theorem): $R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$ for some $c$ between $a$ and $x$. Replacing $f^{(n+1)}(c)$ by its maximum $M$ gives the bound above.

41. AP Exam Structure, Calculators & Scoring

Format (Both AB and BC)

Section I — Multiple Choice, divided into a no-calculator part and a graphing-calculator part

Section II — Free Response, six questions; a graphing calculator is required for part of this section

The two sections are weighted equally — each is worth 50% of the exam score

Scores are reported on the 1–5 scale

Check AP Central before you sit the exam. College Board has announced updated question counts and timing effective with the May 2027 administration. The structure above — two equally weighted sections, with calculator and no-calculator portions — is the current shape of the exam, but if you are testing in 2027 or later, confirm the current question counts and section timings on AP Central rather than relying on any printed reference, including this one.

Calculator Capabilities Expected

Plot a function in an arbitrary viewing window

Find the zeros of a function / solve an equation numerically

Evaluate a derivative numerically at a point

Evaluate a definite integral numerically

Free-Response Conventions

Show the setup — an unsupported answer usually earns no credit even when it is correct

Report decimal answers to three decimal places unless told otherwise; do not round intermediate values

Include units when the context supplies them

When you cite a theorem, state that its hypotheses hold — for MVT, say the function is continuous on the closed interval and differentiable on the open one

Scope Reminders

AB covers limits, differentiation and its applications, integration and its applications, and separable differential equations

BC adds everything tagged BC on this sheet: parametric, vector and polar functions; sequences and series; Euler's method; logistic growth; integration by parts; partial fractions; improper integrals; and arc length

BC does not replace AB material — every AB topic is also assessed on BC

AP CALCULUS AB/BC Formula Sheet