AP Calculus BC Quick Review

A limit describes the value a function approaches as $x$ approaches a point, independent of whether the function is even defined there.

$_x→ af(x)=L$ means $f(x)$ gets arbitrarily close to $L$ as $x$ gets close to $a$ from both sides. One-sided limits: $_x→ a^-f(x)$ (from the left), $_x→ a^+f(x)$ (from the right). $_x→ af(x)$ exists if and only if $_x→ a^-f(x)=_x→ a^+f(x)$.

The limit $_x→ af(x)$ vs. the function value $f(a)$ --- they can differ (removable discontinuity) or one can exist without the other.

Concluding a limit does not exist just because $f(a)$ is undefined --- the limit only depends on behavior near $a$, not at $a$.

Limit exists $$ left limit $=$ right limit; limit $≠$ function value in general.

Most limits are evaluated by direct substitution; when that gives $0/0$, algebraic manipulation or known trig limits resolve the indeterminate form.

Factor and cancel common factors; rationalize by multiplying by the conjugate; $_x→0/ xx=1$, $_x→0/1- xx=0$.

An indeterminate form $0/0$ (requires more work, may still have a finite limit) vs. a genuinely undefined limit (e.g., a nonzero number over $0$, which signals a vertical asymptote or a two-sided mismatch).

Plugging in the value too early and declaring ``undefined'' without attempting to factor or simplify first --- $0/0$ is a signal to keep working, not a final answer.

$0/0$ $arrow$ factor, rationalize, or use $/ xx→1$; never stop at ``undefined.''

A function is continuous at a point only when the limit exists, the function is defined there, and the two values match; the Intermediate Value Theorem relies on this continuity.

$f$ is continuous at $a$ iff (1) $f(a)$ is defined, (2) $_x→ af(x)$ exists, (3) $_x→ af(x)=f(a)$. Discontinuity types: removable (a hole; limit exists but $ f(a)$ or $f(a)$ undefined), jump (one-sided limits differ), infinite (a vertical asymptote). IVT: if $f$ is continuous on $[a,b]$ and $k$ is between $f(a)$ and $f(b)$, there exists $c$ with $f(c)=k$.

Removable discontinuity (limit exists, just a ``hole'') vs. jump discontinuity (left/right limits both exist but are unequal) vs. infinite discontinuity (a one- or two-sided limit is $±∞$).

Applying IVT (or any continuity-based theorem) to a function that is NOT continuous on the given closed interval --- always verify continuity first, especially for piecewise functions.

Continuous = limit exists AND equals $f(a)$; IVT needs continuity on a closed interval.

Limits at infinity describe end behavior and horizontal asymptotes, while the Squeeze Theorem pins down limits that are hard to evaluate directly by trapping them between two known functions.

For rational functions, compare degrees: if $(num)<(denom)$, the limit is 0; if equal, the limit is the ratio of leading coefficients; if $(num)>(denom)$, the limit is $±∞$. Squeeze Theorem: if $g(x)≤ f(x)≤ h(x)$ near $a$ and $_x→ ag(x)=_x→ ah(x)=L$, then $_x→ af(x)=L$.

A horizontal asymptote (describes end behavior as $x→±∞$; the function CAN cross it) vs. a vertical asymptote (occurs at a finite $x$ where the function is unbounded, from a zero denominator).

Comparing the wrong degrees, or forgetting that $sqrtx^2=|x|$ flips sign as $x→-∞$, when finding a limit at infinity of an expression with a square root.

Compare degrees for rational-function limits at $∞$; $sqrtx^2=|x|$ flips sign as $x→-∞$.

The derivative is defined as the limit of a difference quotient, representing instantaneous rate of change; differentiability is a stronger condition than continuity.

$$f'(a)=_h→0/f(a+h)-f(a)h=_x→ a/f(x)-f(a)x-a$$ If $f$ is differentiable at $a$, then $f$ is continuous at $a$ (but not conversely).

Differentiable (a well-defined tangent line slope) vs. merely continuous (no breaks, but can have corners, cusps, or vertical tangents where the derivative fails to exist).

Assuming continuity implies differentiability --- a function can be continuous at a point (e.g., $|x|$ at $x=0$) yet fail to be differentiable there due to a corner, cusp, or vertical tangent.

Differentiable $⇒$ continuous, but continuous $⇒$ differentiable (corners, cusps, vertical tangents).

A small set of rules --- the power rule plus the six trig derivatives --- let you differentiate any polynomial or trig expression instantly, with no limit computation needed.

$/ddxx^n=nx^n-1$; $/ddx x= x$, $/ddx x=- x$, $/ddx x=^2x$, $/ddx x=-^2x$, $/ddx x= x x$, $/ddx x=- x x$.

$/ddx x= x$ vs. $/ddx x=- x$ --- the negative sign belongs only with cosine's derivative, a frequent sign slip.

Forgetting the negative sign on the derivatives of $ x$, $ x$, and $ x$ --- exactly the three ``co-'' functions carry a minus sign.

Power rule: bring down exponent, subtract 1. ``Co-'' functions (cos, cot, csc) $arrow$ negative derivative.

Derivatives of products and quotients of functions do NOT simply multiply or divide the individual derivatives --- they require the product and quotient rules.

Product rule: $(fg)'=f'g+fg'$. Quotient rule: $≤ft(/fg)'=/f'g-fg'g^2$.

$(fg)' f'g'$ (a common false shortcut) --- the product rule requires BOTH cross terms, $f'g+fg'$.

Reversing the order of terms in the numerator of the quotient rule (writing $fg'-f'g$ instead of $f'g-fg'$) --- a sign error, since subtraction is not commutative.

Product: ``first times derivative of second, plus second times derivative of first.'' Quotient: ``low d-high minus high d-low, over low squared.''

Exponential functions with base $e$ are their own derivative, and this special property extends to general exponential and logarithmic bases through the constant $ a$.

$/ddxe^x=e^x$; $/ddxa^x=a^x a$; $/ddx x=/1x$; $/ddx_a x=/1x a$.

$/ddxe^x=e^x$ (base $e$ is unique, unchanged) vs. $/ddxa^x=a^x a$ (any other base picks up a factor of $ a$).

Forgetting the extra factor of $ a$ when differentiating a general exponential like $2^x$, or forgetting to chain-rule the exponent when it is not just $x$ (e.g., $/ddxe^3x=3e^3x$, not $e^3x$).

$e^x→ e^x$ (no change); $a^x→ a^x a$; $ x→ /1x$; always chain-rule the exponent/argument.

Repeated differentiation produces higher-order derivatives that describe how a rate of change is itself changing, and the first derivative gives the slope used to write a tangent (or normal) line.

Notation: $f'(x), f''(x), f'''(x), f^(n)(x)$ or $/d^2ydx^2$. Tangent line at $x=a$: $y-f(a)=f'(a)(x-a)$. Normal line (perpendicular to the tangent): slope $=-/1f'(a)$.

The tangent line's slope ($f'(a)$, used directly) vs. the normal line's slope (the NEGATIVE RECIPROCAL of $f'(a)$).

Using $f'(a)$ as the normal line's slope instead of $-1/f'(a)$, or plugging the derivative's value in for $y$ instead of the original function's value when writing the point $(a,f(a))$.

Tangent: $y-f(a)=f'(a)(x-a)$; normal slope $=-1/f'(a)$.

The chain rule differentiates composite functions by multiplying the derivative of the ``outside'' function (evaluated at the inside) by the derivative of the ``inside'' function.

$$/ddxf(g(x))=f'(g(x))· g'(x)$$ Applies repeatedly for nested compositions (chain the chain).

Differentiating a composite function $f(g(x))$ (needs the chain rule) vs. differentiating a product $f(x)g(x)$ (needs the product rule) --- these structures are often confused.

Forgetting to multiply by the inner derivative $g'(x)$ (``dropping the chain'') --- e.g., $/ddx(3x)=3(3x)$, not just $(3x)$.

Outside derivative (leave inside alone) $×$ inside derivative --- never ``drop the chain.''

When $y$ cannot easily be solved for in terms of $x$, differentiate both sides of the equation with respect to $x$, treating $y$ as an implicit function of $x$.

Differentiate each term with respect to $x$; every time you differentiate a $y$-term, multiply by $/dydx$ (chain rule); then algebraically solve for $/dydx$.

$/ddx(y^2)=2y/dydx$ (implicit; $y$ is a function of $x$) vs. $/ddx(x^2)=2x$ (explicit; no extra factor needed).

Forgetting to attach $/dydx$ to every differentiated $y$-term, or forgetting the product rule when a term mixes $x$ and $y$ (e.g., $/ddx(xy)=x/dydx+y$).

Differentiate both sides; every $y$ gets a $/dydx$ tag; then solve algebraically for $/dydx$.

The derivative of an inverse function at a point is the reciprocal of the original function's derivative, evaluated at the corresponding swapped point.

If $g$ is the inverse of $f$, $$(f^-1)'(a) = /1f'(f^-1(a))$$

$(f^-1)'(a)$ (derivative of the inverse function) vs. $/1f'(a)$ (reciprocal of the derivative at the SAME input $a$) --- you must evaluate $f'$ at $f^-1(a)$, not at $a$ itself.

Plugging $a$ directly into $f'$ instead of first finding $f^-1(a)$ and plugging THAT value in --- this is the most common error on inverse-derivative FRQs.

$(f^-1)'(a) = /1f'(f^-1(a))$ --- find $f^-1(a)$ first, then plug into $f'$.

Each inverse trig function has a specific derivative formula built from the Pythagorean identity, obtained by implicitly differentiating the corresponding trig equation.

$/ddx x=/1sqrt1-x^2$; $/ddx x=/-1sqrt1-x^2$; $/ddx x=/11+x^2$; $/ddxarccot x=/-11+x^2$.

$/ddx x = /1sqrt1-x^2$ (positive) vs. $/ddx x = /-1sqrt1-x^2$ (negative) --- same denominator, opposite signs.

Confusing the $$ derivative's denominator ($1+x^2$, no square root) with the $$ derivative's denominator ($sqrt1-x^2$, has a square root).

arcsin $→ /1sqrt1-x^2$ (+); arccos $→$ same but $-$; arctan $→ /11+x^2$, no root.

Velocity is the derivative of position and acceleration is the derivative of velocity, so the signs of $v(t)$ and $a(t)$ reveal whether an object is speeding up or slowing down.

$v(t)=s'(t)$, $a(t)=v'(t)=s''(t)$. Speed $=|v(t)|$. An object speeds up when $v(t)$ and $a(t)$ have the SAME sign; it slows down when they have OPPOSITE signs. Total distance traveled on $[a,b]$: $_a^b|v(t)| dt$.

Displacement ($_a^b v(t) dt$, net change in position, can be negative) vs. total distance traveled ($_a^b|v(t)| dt$, always non-negative).

Computing total distance by integrating $v(t)$ directly instead of $|v(t)|$ --- if the object reverses direction, this cancels distance traveled backward instead of adding it.

$v=s'$, $a=v'$; same sign $→$ speeding up; distance $=|v(t)|dt $ displacement $= v(t)dt$.

Related-rates problems connect the rates of change of two or more quantities linked by an equation, found by differentiating that equation implicitly with respect to time.

Method: (1) identify variables and write an equation relating them; (2) differentiate both sides with respect to $t$ (chain rule; every variable gets a $/d(·)dt$); (3) substitute known values ONLY AFTER differentiating; (4) solve for the desired rate.

Substituting a specific numeric value too early (BEFORE differentiating) vs. correctly plugging numbers in only after the implicit differentiation step.

Plugging in given numbers before differentiating --- this incorrectly ``freezes'' the variable, making its rate term vanish from the differentiated equation.

Write equation $arrow$ differentiate w.r.t. $t$ (chain rule) $arrow$ THEN substitute numbers $arrow$ solve.

Near a point, a differentiable function can be approximated by its tangent line, giving a fast estimate of nearby function values.

Linearization: $$L(x)=f(a)+f'(a)(x-a)$$ For a function concave up near $a$, the tangent line lies below the curve, so linearization UNDERESTIMATES; for concave down, it OVERESTIMATES.

Local linearization (uses the tangent line, a first-derivative-only approximation) vs. a full Taylor polynomial (uses more derivative terms for a better approximation) --- linearization is just the degree-1 Taylor polynomial.

Assuming a tangent-line approximation is always an overestimate --- the direction of error is governed by concavity, and accuracy degrades as $x$ moves away from $a$.

$L(x)=f(a)+f'(a)(x-a)$; concave up $→$ underestimate; concave down $→$ overestimate.

When a limit of a quotient produces an indeterminate form, L'H\^opital's Rule allows differentiating the numerator and denominator separately to resolve it.

If $_x→ a/f(x)g(x)$ is $/00$ or $/∞∞$, then $$_x→ a/f(x)g(x)=_x→ a/f'(x)g'(x)$$ provided the right-hand limit exists (can be reapplied if still indeterminate).

L'H\^opital's Rule (differentiate numerator and denominator SEPARATELY) vs. the quotient rule (used to differentiate a quotient AS A SINGLE function) --- different tools for different tasks.

Applying L'H\^opital's Rule to a limit that is NOT in indeterminate form --- verify the form at every step, especially after reapplying the rule once.

$0/0$ or $∞/∞$ only $→$ differentiate top and bottom SEPARATELY, recheck the form each time.

The Extreme Value Theorem guarantees that a continuous function on a closed interval attains an absolute max and min, and these can only occur at critical points or endpoints.

EVT: if $f$ is continuous on $[a,b]$, $f$ attains an absolute maximum and minimum on $[a,b]$. Critical point: where $f'(x)=0$ or $f'(x)$ is undefined. Candidates test: evaluate $f$ at all critical points AND both endpoints; the largest/smallest value is the absolute max/min.

A critical point (where $f'=0$ or DNE) vs. an extremum (an actual max/min value) --- not every critical point is an extremum.

Forgetting to check the ENDPOINTS of a closed interval when hunting for absolute extrema --- the absolute max/min can occur at an endpoint even if it is not a critical point.

Absolute extrema on $[a,b]$: check ALL critical points AND both endpoints, compare $f$-values.

The Mean Value Theorem guarantees a point where the instantaneous rate of change equals the average rate of change over an interval.

MVT: if $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, there exists $c(a,b)$ with $f'(c)=/f(b)-f(a)b-a$. Rolle's Theorem: the special case where $f(a)=f(b)$, guaranteeing some $c$ with $f'(c)=0$.

Rolle's Theorem (requires $f(a)=f(b)$, concludes $f'(c)=0$) vs. the general MVT (no such requirement, concludes $f'(c)=$ average rate of change).

Applying MVT/Rolle's without verifying BOTH hypotheses (continuity on the closed interval AND differentiability on the open interval).

MVT: $f'(c)=/f(b)-f(a)b-a$; Rolle's = MVT when $f(a)=f(b)$, so $f'(c)=0$.

The sign of the first derivative directly indicates whether the original function is increasing or decreasing, and sign changes at critical points classify local extrema.

$f'(x)>0$ on an interval $⇒ f$ increasing there; $f'(x)<0 ⇒$ decreasing. First Derivative Test: if $f'$ changes from $+$ to $-$ at $c$, $f(c)$ is a local max; from $-$ to $+$, a local min; no sign change, not an extremum.

The sign of $f$ (whether the function's values are positive/negative) vs. the sign of $f'$ (whether the function is increasing/decreasing).

Reading a graph of $f'(x)$ and describing where $f$ is increasing/decreasing as if the picture directly showed $f$ --- when given $f'$'s graph, $f$ increases wherever the $f'$-graph is ABOVE the $x$-axis.

$f'>0→ f$ increasing; $f'$ changes $+→-$ at $c$ $→$ local max at $c$ (and vice versa for a min).

The sign of the second derivative reveals concavity, and a change in concavity marks an inflection point.

$f''(x)>0 ⇒$ concave up; $f''(x)<0 ⇒$ concave down. Inflection point: where concavity CHANGES (i.e., $f''$ changes sign), not merely where $f''=0$. Second Derivative Test: if $f'(c)=0$ and $f''(c)>0$, local min at $c$; if $f''(c)<0$, local max at $c$; if $f''(c)=0$, inconclusive.

A point where $f''(x)=0$ (a candidate) vs. a true inflection point (requires $f''$ to actually CHANGE SIGN there --- e.g., $f(x)=x^4$ at $x=0$ has $f''(0)=0$ but no inflection).

Declaring an inflection point wherever $f''(x)=0$ without checking that concavity actually switches sign on either side.

$f''>0$ concave up (min-friendly); inflection = concavity CHANGES sign, not just $f''=0$.

The graphs of $f$, $f'$, and $f''$ are interlocked, and the same sign analysis powers optimization problems that ask for a maximum or minimum value.

Curve-sketching chart connects the sign of $f'$ (increasing/decreasing) and the sign of $f''$ (concavity) across intervals. Optimization procedure: write a function for the quantity to optimize in terms of ONE variable (using a constraint equation), find critical points, then justify max/min with the First or Second Derivative Test (or EVT on a closed domain).

Optimizing an UNBOUNDED domain (must justify the extremum with a derivative test) vs. a CLOSED interval (must also check both endpoints per the Candidates Test).

Finding a critical point of the optimization function and stopping --- full credit requires justifying WHY that point is a max or min, not just solving $f'(x)=0$.

Optimization: one-variable function $→ f'(x)=0$ $→$ JUSTIFY with a derivative test, don't just solve and stop.

A definite integral is the limit of Riemann sums --- the accumulated signed area between a curve and the $x$-axis --- and can be approximated with rectangles or trapezoids from tabular data.

Left/right/midpoint Riemann sums use left endpoints, right endpoints, or midpoints of each subinterval. Trapezoidal sum: $$T=/ x2≤ft[f(x_0)+2f(x_1)+·s+2f(x_n-1)+f(x_n)]$$ For an increasing $f$, the left sum underestimates and the right sum overestimates the integral (reverse for decreasing); the trapezoidal sum overestimates for concave up, underestimates for concave down.

Left/right Riemann sum error (depends on whether $f$ is INCREASING or DECREASING) vs. trapezoidal sum error (depends on CONCAVITY).

Using unequal-width subintervals from a data table but still applying the equal-width formula --- with unequal widths, each rectangle/trapezoid must be computed individually using its own $ x$.

Increasing $f$: left sum under, right sum over. Concave up: trapezoid over, midpoint under.

The Fundamental Theorem of Calculus links differentiation and integration: accumulating a rate of change gives net change, and differentiating an accumulation function recovers the original rate.

FTC Part 1: $/ddx_a^x f(t) dt = f(x)$; with a variable upper limit $g(x)$, $/ddx_a^g(x) f(t) dt = f(g(x))· g'(x)$. FTC Part 2: $$_a^b f(x) dx = F(b)-F(a), F'=f$$

FTC Part 1 (differentiating an integral --- gives back the integrand) vs. FTC Part 2 (evaluating a definite integral --- gives net accumulated change).

Forgetting to multiply by $g'(x)$ when the upper limit of an accumulation function is not plain $x$ --- e.g., $/ddx_0^x^2f(t)dt = f(x^2)· 2x$, not just $f(x^2)$.

$/ddx_a^x f(t)dt=f(x)$; if the bound is $g(x)$, multiply by $g'(x)$ too.

Antidifferentiation reverses differentiation, and every indefinite integral carries an arbitrary constant because a whole family of functions shares the same derivative.

$ x^n dx=/x^n+1n+1+C\ (n-1)$; $ /1xdx=|x|+C$; $ e^x dx = e^x+C$; $ x dx=- x+C$; $ x dx= x+C$; $ ^2x dx= x+C$.

An indefinite integral $ f(x) dx$ (a FAMILY of functions, $+C$ required) vs. a definite integral $_a^b f(x) dx$ (a single NUMBER, no $+C$).

Dropping the ``$+C$'' on an indefinite integral, or carrying a ``$+C$'' into a definite integral's final numeric answer.

Indefinite integral $→$ always $+C$; definite integral $→$ a number, no $C$.

$u$-substitution reverses the chain rule, turning a composite integrand into a simpler one by replacing an inner expression and its differential with a single new variable $u$.

Method: let $u=g(x)$, compute $du=g'(x) dx$, rewrite the integral entirely in terms of $u$, integrate, then substitute back. For a DEFINITE integral, either convert the limits to $u$-values or substitute back to $x$ before evaluating.

Changing the limits of integration to match $u$ (evaluate directly in $u$) vs. forgetting to change the limits (then MUST substitute back to $x$ before using the original $x$-bounds).

Plugging the original $x$-limits into an antiderivative that is still written in terms of $u$.

$u=$ inside, $du=$ its derivative $· dx$; for definite integrals, convert the limits to $u$ OR substitute back before evaluating.

Integration by parts reverses the product rule, breaking an integral of a product into a simpler piece by strategically choosing which factor to differentiate and which to integrate.

$$ u dv = uv - v du$$ Choose $u$ using LIPET priority (Logarithmic, Inverse trig, Polynomial, Exponential, Trig) as a guide for what to differentiate first; $dv$ is the rest, which you integrate.

Integration by parts (for a PRODUCT of two different types of functions, e.g., $x x$) vs. $u$-substitution (for a composite function with its derivative present, e.g., $(x^2)· 2x$).

Choosing $u$ and $dv$ poorly (e.g., picking $u=e^x$ instead of $u=x$ in $ xe^x dx$), which produces a MORE complicated integral instead of a simpler one.

$ u dv=uv- v du$; pick $u$ via LIPET (Log, Inverse trig, Polynomial, Exp, Trig).

A rational function whose denominator factors into distinct linear pieces can be rewritten as a sum of simpler fractions, each easy to integrate using logarithms.

For $/P(x)(x-a)(x-b)$ (degree of $P$ less than the denominator's), decompose as $/Ax-a+/Bx-b$, solve for $A$ and $B$, then integrate: $ /Ax-adx = A|x-a|+C$.

Partial fraction decomposition (requires the numerator's degree to be LESS than the denominator's --- otherwise divide first) vs. direct $u$-substitution (works when the numerator is already the derivative of the denominator).

Attempting partial fractions on an improper rational function without first performing polynomial long division --- decomposition only applies to the proper-fraction remainder.

Improper degree? Divide first. Then split into $/Ax-a+/Bx-b$, integrate to logs.

An improper integral --- one with an infinite bound or an unbounded integrand --- is evaluated as a limit, and it converges only if that limit exists as a finite number.

$$_a^∞ f(x) dx = _b→∞_a^b f(x) dx$$ For a discontinuity at $c$ within $[a,b]$: split the integral at $c$ and take a limit approaching $c$ from each side. The integral converges if the limit is finite; it diverges if the limit is infinite or does not exist.

A convergent improper integral (the limit exists and is finite, even over an infinite region) vs. a divergent one (the limit is $±∞$ or does not exist).

Treating an improper integral like an ordinary definite integral and plugging in $∞$ directly, or ignoring a discontinuity inside the interval, instead of setting up the required limit.

Improper integral $=_b→∞_a^b$ (or a limit approaching the discontinuity); finite limit $→$ converges.

A slope field visualizes a first-order differential equation by drawing a short line segment of the correct slope at many points, letting you sketch or match solution curves without solving the equation.

At each point $(x,y)$, the segment's slope equals $/dydx$ evaluated from the given differential equation at that point. Solution curves run tangent to the local segments and never cross (for well-behaved DEs).

A slope field for $/dydx=f(x)$ (slope depends only on $x$, so it is identical along any vertical line) vs. $/dydx=f(y)$ (slope depends only on $y$, identical along any horizontal line).

Matching a differential equation to the wrong slope field by checking only one or two points instead of checking behavior along whole rows/columns.

Slope field segment slope $=/dydx$ from the DE at that point; solution curves run tangent to segments, never cross.

A separable differential equation can be solved exactly by algebraically separating all $y$-terms to one side and all $x$-terms to the other, then integrating both sides.

Rewrite $/dydx=f(x)g(y)$ as $/dyg(y)=f(x) dx$, integrate both sides, then solve for $y$ explicitly if possible (don't forget $+C$; use an initial condition to solve for $C$ if given).

The general solution (contains an arbitrary constant $C$, a family of curves) vs. the particular solution (uses a given initial condition to solve for the exact value of $C$).

Losing the constant of integration during separation, or forgetting to apply the initial condition at the very end to solve for $C$.

Separate $y$'s and $x$'s $→$ integrate both sides $→ +C$ $→$ use the initial condition to pin down $C$.

Any quantity whose rate of change is proportional to its current amount follows an exponential model, obtained by solving the differential equation $/dydt=ky$.

$$/dydt=ky y=y_0e^kt$$ where $y_0$ is the initial amount and $k$ is the constant of proportionality ($k>0$ growth, $k<0$ decay).

The differential equation $/dydt=ky$ (a statement about the RATE of change) vs. its solution $y=y_0e^kt$ (the actual function of $t$).

Assuming any exponential-looking context automatically means $y=y_0e^kt$ without confirming the rate of change is proportional to the CURRENT amount $y$ (as opposed to, e.g., logistic growth).

$/dydt=ky y=y_0e^kt$; $k>0$ growth, $k<0$ decay.

Logistic growth models a quantity that grows exponentially at first but levels off as it approaches a carrying capacity, because its growth rate slows as the population fills its limit.

$$/dPdt=kP≤ft(1-/PL)$$ where $L$ is the carrying capacity. The population grows fastest (inflection point of $P(t)$) when $P=/L2$. As $t→∞$, $P→ L$. Solution: $P(t)=/L1+Ae^-kt$ for some constant $A$.

Exponential growth ($/dydt=ky$, unbounded) vs. logistic growth ($/dPdt=kP(1-P/L)$, bounded by carrying capacity $L$).

Assuming the logistic growth rate is maximized as $P→ L$ --- the growth RATE is actually maximized at $P=L/2$ and DECREASES to 0 as $P$ approaches $L$.

Logistic: bounded by $L$; fastest growth at $P=L/2$; $P→ L$ as $t→∞$.

The average value of a function over an interval is the height of the rectangle with the same base and the same area as the region under the curve.

$$f_avg=/1b-a_a^b f(x) dx$$

Average VALUE of a function (computed with the integral formula above) vs. average RATE of change (computed as $/f(b)-f(a)b-a$, using function values, NOT an integral).

Forgetting to divide by $(b-a)$ --- computing only $_a^b f(x)dx$ and reporting that as the ``average value'' instead of dividing by the interval length.

Average value $=/1b-a_a^b f(x)dx$; NOT the same as average rate of change $/f(b)-f(a)b-a$.

The area between two curves is the definite integral of the (top minus bottom) difference of the functions, and requires knowing which function is greater on the interval.

$$A=_a^b [f(x)-g(x)] dx, f(x)≥ g(x) on [a,b]$$ If curves cross within the interval, split into subintervals and take the difference in the correct order on each piece. For regions bounded in $y$: $A=_c^d[right(y)-left(y)] dy$.

Integrating with respect to $x$ (top minus bottom, vertical strips) vs. integrating with respect to $y$ (right minus left, horizontal strips).

Using one fixed ``top minus bottom'' order across the whole interval when the curves actually cross --- find the intersection point(s) first and split the integral, flipping the subtraction order on each piece.

Area $=(top-bottom)dx$ (or right$-$left in $y$); curves cross $→$ split and flip order.

The volume of a solid of revolution or a solid with known cross sections is found by integrating the cross-sectional area function along the axis of the solid.

Disk method (no gap): $$V=_a^b [R(x)]^2 dx$$ Washer method (gap/hole): $$V=_a^b ≤ft([R(x)]^2-[r(x)]^2)dx$$ Known cross sections (perpendicular to an axis, built on a base region): $$V=_a^b A(x) dx$$ where $A(x)$ is the cross-sectional area formula in terms of the base's width.

The disk method (solid touches the axis of revolution, a single squared radius) vs. the washer method (region does NOT touch the axis, leaving a hole --- requires SUBTRACTING the inner radius squared from the outer).

Squaring the difference of the radii, $[R(x)-r(x)]^2$, instead of taking the difference of the squares, $[R(x)]^2-[r(x)]^2$.

Disk: $ R^2 dx$. Washer: $ (R^2-r^2) dx$ (difference of squares, not square of difference). Cross sections: $ A(x) dx$.

The arc length of a curve is found by integrating the length of infinitesimal straight-line segments approximating the curve, built from the Pythagorean theorem applied to $dx$ and $dy$.

$$L=_a^b sqrt1+[f'(x)]^2 dx$$

The arc length formula (uses the DERIVATIVE inside the square root) vs. simply integrating $f(x)$ itself --- arc length is never computed by integrating the function directly.

Forgetting to differentiate $f(x)$ before squaring it and adding 1 --- squaring $f(x)$ itself instead of $f'(x)$ inside the radical.

Arc length $=_a^bsqrt1+[f'(x)]^2 dx$ --- always the DERIVATIVE under the root.

For a curve defined parametrically by $x(t)$ and $y(t)$, the slope $/dydx$ is found by dividing the rate of change of $y$ by the rate of change of $x$, both with respect to the parameter $t$.

$$/dydx=/dy/dtdx/dt /d^2ydx^2=//ddt≤ft(/dydx)dx/dt$$

The second derivative $/d^2ydx^2$ (take $/ddt$ of $/dydx$, THEN divide by $dx/dt$ again) vs. simply differentiating $/dydx$ directly with respect to $x$.

Computing $/d^2ydx^2$ by differentiating $/dydx$ with respect to $t$ and stopping, forgetting to divide that result by $dx/dt$ a second time.

$/dydx=/dy/dtdx/dt$; for the 2nd derivative, take $/ddt$ of that, then divide by $dx/dt$ AGAIN.

Arc length for a parametrically defined curve integrates the magnitude of the velocity vector (speed) over the parameter interval, generalizing the single-variable arc length formula.

$$L=_t_1^t_2sqrt≤ft(/dxdt)^2+≤ft(/dydt)^2 dt$$ This equals $_t_1^t_2speed(t) dt$ for a vector-valued position function.

Parametric/vector arc length (uses BOTH $dx/dt$ and $dy/dt$ under the root, integrates with respect to $t$) vs. function arc length (uses only $f'(x)$, integrates with respect to $x$).

Using the single-variable arc length formula on a parametric curve, or forgetting to square and add BOTH $dx/dt$ and $dy/dt$ under the radical.

Parametric arc length $=(dx/dt)^2+(dy/dt)^2 dt$ $=$ speed $ dt$.

A vector-valued position function packages an object's $x$- and $y$-coordinates as functions of time, and differentiating it component-wise gives velocity and acceleration vectors.

Position: $r(t)= x(t),y(t)$. Velocity: $v(t)= x'(t),y'(t)$. Acceleration: $a(t)= x''(t),y''(t)$. Speed $=|v(t)|=sqrt[x'(t)]^2+[y'(t)]^2$. Total distance traveled on $[t_1,t_2]$: $_t_1^t_2|v(t)| dt$.

Velocity (a VECTOR, components $ x'(t),y'(t)$) vs. speed (a SCALAR, the magnitude of velocity, $|v(t)|$).

Reporting the velocity vector's components as ``the speed,'' or computing distance traveled by integrating the components separately instead of integrating the magnitude of the velocity vector.

$v(t)= x'(t),y'(t)$ (vector); speed $=|v(t)|$ (scalar); distance $=|v(t)|dt$.

A polar curve $r=f()$ can be treated as a parametric curve in $$ by converting to Cartesian coordinates first, then applying the parametric derivative formula.

Conversion: $x=r=f()$, $y=r=f()$. Slope: $$/dydx=/dy/ddx/d$$ found by differentiating the converted $x$ and $y$ (product rule required, since both factors depend on $$).

$/drd$ (how the RADIUS changes with angle, NOT the slope of the tangent line) vs. $/dydx$ (the actual Cartesian slope of the tangent line to the polar curve).

Using $/drd$ as if it were the tangent line's slope --- always convert to $x()$ and $y()$ first and use $/dy/ddx/d$.

Polar slope $=/dy/ddx/d$, from $x=r$, $y=r$ --- never just $/drd$.

The area enclosed by a polar curve is computed by integrating one-half the square of the radius over the angle swept, since a thin polar ``slice'' approximates a triangular sector.

$$A=/12_^ [r()]^2 d$$ For the area between two polar curves $r=f()$ (outer) and $r=g()$ (inner): $$A=/12_^ ≤ft([f()]^2-[g()]^2)d$$

The polar area formula (uses $/12 r^2 d$, integrates over an ANGLE) vs. the Cartesian area-between-curves formula (uses $ (f-g) dx$, integrates over a LENGTH).

Forgetting the factor of $/12$ in the polar area formula, or using the wrong bounds of $$ --- must find where the curves intersect or where the loop begins and ends.

Polar area $=/12_^ r^2 d$ (never forget the $/12$).

A sequence converges if its terms approach a single finite limit as the index $n$ grows without bound; sequence convergence is the foundation for the rest of the series unit.

$_n→∞a_n=L$ means the sequence converges to $L$. Many sequence limits are evaluated with the same techniques as function limits at infinity (L'H\^opital's Rule can be applied by treating $n$ as continuous).

A sequence ($a_n$, an ordered LIST of numbers) vs. a series ($ a_n$, the SUM of the terms of a sequence) --- a sequence can converge while the series formed from it diverges.

Confusing ``the sequence of terms converges to 0'' with ``the series converges'' --- necessary but NOT sufficient, which is exactly why the $n$th-term test can only prove divergence, never convergence.

Sequence = list, converges if $a_n→ L$. Series = sum; $a_n→0$ is necessary but NOT sufficient for the series to converge.

The $n$th-term test gives a quick divergence check, and the geometric series is the one series family whose sum can always be computed exactly in closed form.

$n$th-Term (Divergence) Test: if $_n→∞a_n 0$ (or DNE), then $ a_n$ DIVERGES (if the limit is 0, the test is inconclusive). Geometric series: $$_n=0^∞ ar^n = /a1-r, converges iff |r|<1$$

The $n$th-term test proving DIVERGENCE (valid when $a_n→0$) vs. trying to use it to prove CONVERGENCE (invalid --- $a_n→0$ never proves a series converges).

Concluding a series converges because its terms go to zero --- the harmonic series $/1n$ has terms $→0$ but DIVERGES.

$a_n→0 ⇒$ diverges; $a_n→0 ⇒$ test is inconclusive. Geometric: converges iff $|r|<1$, sum $=/a1-r$.

The Integral Test connects a series' convergence to the convergence of a related improper integral, and this directly establishes the $p$-series family as a key benchmark.

Integral Test: if $f$ is positive, continuous, and decreasing for $x≥ N$ with $f(n)=a_n$, then $ a_n$ and $_N^∞ f(x) dx$ either BOTH converge or BOTH diverge. $p$-series: $$ /1n^p converges iff p>1$$

A $p$-series with $p>1$ (converges, e.g., $/1n^2$) vs. the harmonic series, $p=1$ (DIVERGES, $/1n$).

Applying the Integral Test to a series whose terms are NOT positive, continuous, and decreasing --- all three conditions must be verified before the test can be used.

$p$-series converges iff $p>1$ (harmonic series, $p=1$, diverges). Integral Test needs positive, continuous, decreasing.

Comparison tests determine convergence by relating an unfamiliar series to a known benchmark series (usually geometric or $p$-series) whose behavior is already established.

Direct Comparison: if $0≤ a_n≤ b_n$ for all $n$ and $ b_n$ converges, then $ a_n$ converges; if $a_n≥ b_n≥0$ and $ b_n$ diverges, then $ a_n$ diverges. Limit Comparison: if $_n→∞/a_nb_n=c$ for some finite $c>0$, then $ a_n$ and $ b_n$ either both converge or both diverge.

Direct Comparison (requires a strict term-by-term inequality in the RIGHT direction) vs. Limit Comparison (just needs the ratio of terms to approach a positive finite constant).

Using Direct Comparison with an inequality that goes the WRONG way for the conclusion desired --- Limit Comparison avoids this pitfall entirely.

Direct Comparison needs the right-direction inequality; Limit Comparison just needs $/a_nb_n→$ finite positive $c$.

An alternating series converges if its terms shrink steadily to zero, and when it does, the error from truncating the series is bounded by the size of the very next omitted term.

Alternating Series Test: $(-1)^n b_n$ converges if $b_n>0$, $b_n$ is (eventually) decreasing, and $_n→∞b_n=0$. Alternating Series Error Bound: $$|R_n|≤ b_n+1$$ (the absolute value of the first omitted term); the true sum also lies between any two consecutive partial sums.

The alternating series error bound (only valid for TRUE alternating series meeting all three conditions) vs. the Lagrange error bound (valid for any Taylor series remainder, involves a maximum of a higher derivative).

Applying the alternating series error bound without first confirming the terms are DECREASING in absolute value.

Alternating, decreasing, $→0$ $⇒$ converges; error $|R_n|≤$ first omitted term $b_n+1$.

The Ratio Test is the go-to tool for series involving factorials or exponentials, and every series is classified as absolutely convergent, conditionally convergent, or divergent.

Ratio Test: let $L=_n→∞≤ft|/a_n+1a_n|$. If $L<1$, $ a_n$ converges absolutely; if $L>1$ (or $L=∞$), diverges; if $L=1$, inconclusive. Absolute convergence: $|a_n|$ converges (implies $ a_n$ converges). Conditional convergence: $ a_n$ converges but $|a_n|$ diverges.

Absolute convergence ($|a_n|$ converges) vs. conditional convergence ($ a_n$ converges only from alternating-sign cancellation, while $|a_n|$ diverges) --- e.g., the alternating harmonic series converges conditionally, not absolutely.

Concluding a series diverges just because the Ratio Test gives $L=1$ --- this case is INCONCLUSIVE and requires a different test.

Ratio test: $L<1$ converges, $L>1$ diverges, $L=1$ inconclusive. Absolute conv. $⇒$ conv.; conditional = converges but not absolutely.

A power series converges absolutely within a symmetric interval centered at its center of expansion, found using the Ratio Test, but the two endpoints must be checked individually.

For $ c_n(x-a)^n$, apply the Ratio Test to $≤ft|/c_n+1(x-a)^n+1c_n(x-a)^n|$ and solve the resulting inequality $<1$ for $|x-a|$ to find the radius of convergence $R$. The interval of convergence is $(a-R,a+R)$, and EACH endpoint must be tested SEPARATELY (plug back into the original series), since the Ratio Test is inconclusive exactly there.

The radius of convergence $R$ (a single number) vs. the interval of convergence (may include NEITHER, ONE, or BOTH endpoints depending on separate testing).

Assuming the interval of convergence is automatically the open interval $(a-R,a+R)$ without separately testing each endpoint with a different test (often alternating series or $p$-series).

Ratio Test $→$ radius $R$; interval $=(a-R,a+R)$, but test BOTH endpoints separately.

A Taylor series represents a function as an infinite sum built from its derivatives at a single point, and the Lagrange error bound quantifies how good a finite truncation (Taylor polynomial) is as an approximation.

Taylor series centered at $a$: $$f(x)=_n=0^∞/f^(n)(a)n!(x-a)^n$$ Key Maclaurin series: $$e^x=_n=0^∞ /x^nn!, x=_n=0^∞ /(-1)^n x^2n+1(2n+1)!, x=_n=0^∞/(-1)^n x^2n(2n)!, /11-x=_n=0^∞ x^n\ (|x|<1)$$ Lagrange Error Bound: if $|f^(n+1)(z)|≤ M$ for all $z$ between $a$ and $x$, then $$|R_n(x)|≤ /M(n+1)!|x-a|^n+1$$

The Lagrange error bound (general, valid for ANY Taylor series, requires bounding the $(n+1)$th derivative) vs. the alternating series error bound (only for alternating series, just the first omitted term).

Using the degree of the LAST term included instead of the NEXT term's degree ($n+1$) in the Lagrange error bound's factorial and exponent, or forgetting to justify the bound $M$ on the derivative over the correct interval.

Taylor: $f(x)=/f^(n)(a)n!(x-a)^n$; Lagrange error $≤ /M(n+1)!|x-a|^n+1$, with $n+1$ = next term's order.

Know the boundary values cold: $p=1$, $|r|=1$, $L=1$ --- these are the exact lines between converge and diverge.

When two terms sound alike, ask: is this about the function's VALUE or its RATE? Its SUM or its SEQUENCE? Absolute or just barely convergent?

Always try the $n$th-Term Test first (fastest divergence check), then match the series' shape to the test above.

No formula sheet is handed out on exam day --- every formula in this box must be memorized cold.

Name the theorem $→$ verify its hypotheses $→$ state the conclusion in context --- skipping any step loses points even with the right final answer.

42 MCQ (29 no-calc + 13 calc, 50%, 100 min) + 6 FRQ (2 calc + 4 no-calc, 50%, 90 min) --- 3h10m total, plus a separate AB subscore.

$n$th-term $→$ geometric/$p$-series $→$ alternating $→$ ratio $→$ comparison $→$ integral --- work down this list in order.

Sketch $→$ find bounds from intersections $→$ pick the right formula for the method $→$ integrate $→$ label units.