AP Calculus AB

What “Approaching a Limit” Actually Means

A limit asks a subtly different question than “what is f(c)?” It asks: as x gets closer and closer to c — from both directions, without ever actually landing on c — what value does f(x) get closer and closer to? This distinction is the entire reason calculus needed to be invented. Algebra could tell you the value of a function AT a point; it had no tool for describing behavior NEAR a point the function might not even be defined at.

Notation: lim(x→c) f(x) = L, read “the limit of f(x) as x approaches c equals L.” A limit exists at x=c only if the function approaches the SAME value L from the left (x slightly less than c) and from the right (x slightly greater than c). If those two one-sided limits disagree, the two-sided limit does not exist (DNE), even if the function is perfectly well-defined everywhere nearby.

Consider f(x) = (x²−1)/(x−1). At x=1 this is 0/0 — undefined. But at x=0.9, f(x)=1.9; at x=0.99, f(x)=1.99; at x=1.01, f(x)=2.01; at x=1.1, f(x)=2.1. From both directions the output closes in on 2. So even though f(1) doesn't exist, lim(x→1) f(x) = 2. Factoring confirms it: (x−1)(x+1)/(x−1) = x+1 for x≠1, and x+1 at x=1 is 2.

The single most common early mistake is treating “the limit doesn't exist” and “the function isn't defined there” as the same thing — they're not. A function can be undefined at a point and still have a well-defined limit there, and it can be defined at a point while its limit there doesn't match that value, or doesn't exist at all. Keep f(c) and lim(x→c) f(x) mentally separate; conflating them is exactly the error continuity (later in this unit) is built to formalize and catch.

Reading Limits from Graphs and Tables

When you don't have an algebraic formula — just a picture or a table — you find a limit the same conceptual way: by watching what the output does as the input approaches the point in question from both sides.

On a graph, trace the curve from the left toward x=c and note the height you're approaching; then trace from the right. If both traces converge on the same height, that's the limit — regardless of any open circle, filled dot, or gap exactly at x=c. On a table, watch the trend in f(x) as the listed x-values get closer to c from both sides; a genuine limit shows the outputs converging tightly to one number, not just getting “close-ish.”

Suppose a table gives f(1.9)=5.61, f(1.99)=5.96, f(1.999)=5.996 approaching from the left, and f(2.001)=6.004, f(2.01)=6.04, f(2.1)=6.41 from the right. Both sides are clearly homing in on 6, so lim(x→2) f(x) = 6, even without knowing f's formula or the value of f(2) itself.

A graph with a visible jump — approaching one height from the left and a different height from the right — means the limit does not exist, even if the function is defined there (often shown with a filled dot at one height, or a third value entirely). Don't average the two sides or assume the filled dot's height is automatically the limit; it usually isn't.

Algebraic Properties of Limits

Once simple building-block functions are established as well-behaved, formal rules let you compute limits of complicated combinations without re-deriving each one from a table.

If lim f(x)=A and lim g(x)=B (both as x→c): lim[f+g]=A+B, lim[f−g]=A−B, lim[f·g]=A·B, lim[k·f]=k·A for any constant k, and lim[f/g]=A/B provided B≠0. Also lim[f(x)]ⁿ = Aⁿ.

Given lim(x→3) f(x)=5 and lim(x→3) g(x)=2, you can immediately find lim(x→3)[3f(x) − g(x)²] = 3(5) − (2)² = 15 − 4 = 11, without knowing anything else about f or g except these two limit values.

The quotient rule requires B≠0 — if the denominator's limit is zero, you cannot just plug in and divide; you're forced into an indeterminate form that needs algebraic simplification, or the limit may not exist as a finite number at all.

Evaluating Limits When Direct Substitution Fails

Direct substitution — plugging x=c into the formula — works for the enormous majority of well-behaved functions, and you should always try it first. It fails specifically when it produces the indeterminate form 0/0, signaling hidden shared structure between numerator and denominator that algebra can expose.

Three standard rescue techniques: factor both numerator and denominator and cancel a shared factor; rationalize by multiplying by a conjugate when a square root is involved; or combine fractions over a common denominator when the expression has stacked fractions. After simplifying, substitute again.

lim(x→4) (√x − 2)/(x−4): substitution gives 0/0. Multiply by the conjugate (√x+2): [(√x−2)(√x+2)]/[(x−4)(√x+2)] = (x−4)/[(x−4)(√x+2)] = 1/(√x+2). Substituting x=4: 1/(2+2) = 1/4.

Students often stop the moment they see 0/0 and conclude “the limit doesn't exist.” A 0/0 result is not an answer — it's a signal more algebra is needed. Reserve “does not exist” for cases where, after full simplification, the one-sided limits genuinely disagree or blow up to infinity in different directions.

Choosing the Right Technique

This lesson is about pattern recognition — the exam expects you to look at an unfamiliar limit and quickly identify which tool applies, often within seconds, since multiple-choice questions don't reward slow trial-and-error.

A quick diagnostic: a square root next to a subtraction suggests rationalizing. Polynomials sharing a root (visible after factoring) suggest factoring. A fraction within a fraction (“complex fraction”) suggests combining over a common denominator first.

For lim(x→0) [1/(x+2) − 1/2] / x, the stacked fractions are the signal to combine first: 1/(x+2) − 1/2 = [2−(x+2)]/[2(x+2)] = −x/[2(x+2)]. Dividing by x gives −1/[2(x+2)], and substituting x=0 gives −1/4.

Don't force a technique that doesn't fit — rationalizing a limit with no radicals wastes exam time. If none of the three techniques seem to apply and direct substitution gives a clean defined number, the limit may simply equal that number; not every limit is a trick question.

The Squeeze (Sandwich) Theorem

Some limits — especially trig functions divided by their own argument — can't be resolved by factoring or rationalizing because there's no algebraic simplification available. The Squeeze Theorem traps the function between two others whose limits you already know.

If g(x) ≤ f(x) ≤ h(x) for all x near c (except possibly at c), and lim(x→c) g(x) = lim(x→c) h(x) = L, then f is squeezed between two functions converging to the same value, so lim(x→c) f(x) = L too — f has no room to go anywhere else.

The classic use proves lim(x→0) sin(x)/x = 1, using cos(x) ≤ sin(x)/x ≤ 1 for x near 0; since both cos(x) and the constant 1 approach 1 as x→0, sin(x)/x is squeezed to 1. You're more likely to be given the bounding inequalities directly than to construct them yourself.

The theorem requires the inequality to hold on both sides for ALL x near c, and both bounding functions must converge to the exact same limit — if g and h approach different values, squeezing tells you nothing about f.

Types of Discontinuity

Not every “broken” point on a graph is broken the same way, and the exam expects you to name the type of discontinuity precisely, because the type determines what (if anything) can be done to fix it.

A removable discontinuity is a hole: the two-sided limit exists, but f(c) is undefined or doesn't equal it — redefining f(c) to equal the limit fixes it. A jump discontinuity is where the left- and right-hand limits both exist as finite numbers but disagree — no redefinition can fix this. An infinite (essential) discontinuity happens at a vertical asymptote, where the function shoots to +∞ or −∞.

f(x) = (x²−4)/(x−2) has a removable discontinuity at x=2 (simplifies to x+2, which equals 4 there, but the original is undefined at x=2 — a hole at (2,4)). A piecewise function x² for x<1 and x+3 for x≥1 has a jump discontinuity at x=1 (left side approaches 1, right side approaches 4). f(x)=1/(x−3) has an infinite discontinuity at x=3.

Don't call every discontinuity a “hole” out of habit — jump and infinite discontinuities look very different on a graph and are graded differently on free-response questions, which often ask you to justify the specific type using limit language, not just point at the picture.

The Formal Definition of Continuity

Continuity formalizes the intuitive idea of a graph you can draw “without lifting your pencil.” It packages together everything the previous lessons built toward: a function's defined value, its limit, and whether those two actually agree.

f is continuous at x=c if and only if all three hold: f(c) is defined; lim(x→c) f(x) exists (left- and right-hand limits agree); and lim(x→c) f(x) = f(c). If even one condition fails, f is discontinuous at c. A function is continuous on an interval if it's continuous at every point in it.

For f(x)=(x²−4)/(x−2) redefined so f(2)=4 (fixing the earlier hole), all three conditions now hold: f(2)=4 is defined, the limit as x→2 is 4, and they match — so this redefined function is continuous at x=2, even though the original wasn't.

Free-response questions frequently ask you to “justify” continuity, meaning write out all three conditions explicitly with their actual values — skipping the formal justification is one of the most common ways students lose points even when they clearly understand the concept.

Limits at Infinity and Asymptotic Behavior

This lesson shifts from “what happens near a specific x-value” to “what happens as x grows without bound” — describing a function's long-run behavior rather than its local behavior at one point.

For rational functions, compare the numerator's degree to the denominator's: denominator higher → limit at infinity is 0 (horizontal asymptote y=0); equal degrees → limit is the ratio of leading coefficients; numerator higher → limit is ±∞ (no horizontal asymptote). Separately, if substitution gives a nonzero number over zero, that signals an infinite limit and a vertical asymptote at that x-value.

For f(x) = (3x²+1)/(x²−4), both degrees equal 2, so lim(x→∞) f(x) = 3/1 = 3 — a horizontal asymptote at y=3. Separately, since the denominator is zero at x=2 and x=−2, f has vertical asymptotes there.

Horizontal asymptotes describe end behavior (x→±∞) while vertical asymptotes describe blow-up at a specific finite x-value — mixing up which kind a given limit describes is a common error, since both eventually involve the word “infinity” in the answer.

The Intermediate Value Theorem (IVT)

The IVT is the first of several “existence theorems” in this course — theorems guaranteeing something exists without telling you exactly where to find it. It's a logical consequence of continuity: an unbroken curve can't skip over a value on its way from one height to another.

If f is continuous on [a,b], and k is any number between f(a) and f(b), then there exists at least one c in [a,b] such that f(c)=k. The theorem says nothing about how many such c exist, or where — just that at least one does.

Suppose f is continuous with f(1)=−3 and f(4)=5. Since 0 is between −3 and 5, the IVT guarantees at least one c in [1,4] where f(c)=0 — f has a root somewhere in that interval, without solving for it explicitly. This is exactly how the IVT is most often used: to justify that an equation has a solution in a given interval.

The IVT requires continuity on the ENTIRE closed interval — if a function has even one discontinuity inside [a,b], the theorem doesn't apply. Free-response answers must explicitly state the continuity condition as part of the justification, not just quote the conclusion.

Average vs. Instantaneous Rate of Change

Before calculus, “rate of change” only ever meant average rate of change — total change in output divided by total change in input, exactly what slope measures for a line. Calculus answers a question algebra structurally can't: what is the rate of change AT a single instant, where there's no interval to measure across?

The average rate of change of f over [a,b] is the slope of the secant line through (a,f(a)) and (b,f(b)): [f(b)−f(a)]/(b−a). To get the instantaneous rate at x=a, shrink the interval toward zero width — letting b approach a — and ask what the secant slope approaches. That limiting value is the instantaneous rate, and geometrically it's the slope of the tangent line at that point.

For f(x)=x², the average rate of change from x=1 to x=3 is [9−1]/(3−1)=4. The instantaneous rate AT x=1 is different and smaller — found from secant slopes over shrinking intervals starting at x=1, which converge to 2 as the interval width shrinks toward zero.

Mixing up “average over an interval” and “instantaneous at a point” is one of the most persistent conceptual errors in the course — they use similar-looking formulas but answer fundamentally different questions, and graders specifically look for language distinguishing the two.

The Formal Definition of the Derivative

The instantaneous rate of change from the last lesson has a name — the derivative — and a precise limit definition that makes “shrinking the interval to zero” mathematically rigorous rather than hand-wavy.

f'(x) = lim(h→0) [f(x+h) − f(x)] / h. Here h is the interval width (playing the role “b−a” played before), and letting h approach — but never equal — zero is the limiting process from the previous lesson. Alternate notations: dy/dx, d/dx[f(x)], or y'.

For f(x)=x²: f'(x) = lim(h→0) [(x+h)²−x²]/h = lim(h→0) [2xh+h²]/h = lim(h→0)(2x+h) = 2x. This confirms the earlier numerical result: f'(1)=2(1)=2.

Plugging h=0 directly into the original difference quotient gives 0/0 — you must algebraically simplify (expand and cancel an h) before taking the limit, exactly like the indeterminate-form techniques from Unit 1. Students who substitute h=0 immediately get stuck on undefined expressions.

Estimating a Derivative Numerically or Graphically

On the exam you won't always be handed a clean algebraic formula — sometimes only a table of data points or a graph — and you still need a reasonable derivative estimate at a specific point.

From a table, use a symmetric difference quotient: average the slope using a point slightly before and slightly after the target x-value, [f(x+h)−f(x−h)]/(2h), generally more accurate than a one-sided estimate. From a graph, f'(a) is the slope of the line tangent to the curve at x=a — draw it as best you can and read two clear points off it.

Given f(1.9)=8.5 and f(2.1)=9.7, a symmetric estimate for f'(2) is (9.7−8.5)/(2.1−1.9) = 1.2/0.2 = 6.

A numerical estimate is only as good as the data spacing — widely spaced points across a sharply curving region give a much less reliable estimate than closely spaced points, and free-response questions sometimes ask you to comment on exactly that limitation.

When a Derivative Fails to Exist

Not every continuous function is differentiable everywhere — some unbroken curves still have points with no well-defined tangent line, and recognizing these by sight is frequently tested.

Four situations break differentiability at a point: a corner (slope approaches different values from each side, like |x| at x=0), a cusp (slope approaches +∞ from one side and −∞ from the other), a vertical tangent (infinite slope from both sides), and any discontinuity (no tangent line makes sense where the function itself breaks).

f(x)=|x| is continuous everywhere, including at x=0, but f'(0) doesn't exist: approaching from the left, the slope is consistently −1; from the right, consistently +1. Since these disagree, there's no single tangent line at the corner.

“Differentiable implies continuous” is always true, but the reverse — “continuous implies differentiable” — is FALSE, and the exam loves testing this exact asymmetry with corner or cusp examples like |x| or a cube-root function at the origin.

The Power Rule and Constant/Sum/Difference Rules

Deriving every function from the formal limit definition would be impossibly slow, so calculus builds a toolkit of shortcut rules — proven once from the definition, then reused freely forever after.

The power rule: d/dx[x^n] = n·x^(n−1), valid for any real exponent n. Constants pull straight out: d/dx[k·f(x)] = k·f'(x). Derivatives distribute over addition/subtraction term by term. The derivative of a plain constant is always 0.

For f(x)=3x⁴−5x²+7x−2, differentiate term by term: f'(x) = 3(4x³) − 5(2x) + 7 − 0 = 12x³ − 10x + 7.

The power rule does NOT apply to exponential functions with a variable exponent (like 2^x) — that's a separate rule. Confusing x^n (power rule) with n^x (exponential rule) is an extremely common mix-up carried over from algebra.

Derivatives of the Basic Transcendental Functions

Beyond polynomial power functions, a handful of “special” functions come up constantly, and their derivatives don't follow the power rule — they simply have to be memorized as known results.

The four essential ones: d/dx[sin x] = cos x, d/dx[cos x] = −sin x, d/dx[e^x] = e^x (the unique function that is its own derivative), and d/dx[ln x] = 1/x. These appear constantly, embedded inside chain rule, product rule, and quotient rule problems throughout the rest of the course.

For f(x) = 3sin(x) + e^x − 2ln(x): f'(x) = 3cos(x) + e^x − 2/x, applying the constant-multiple rule alongside each memorized derivative.

Watch the sign on cosine's derivative specifically — it's negative sine, and forgetting that minus sign is one of the single most common small errors across the free-response section, showing up in optimization, motion, and related-rates problems built on trig functions.

The Product Rule

When two functions are multiplied together, you cannot just differentiate each piece separately and multiply the results — multiplication of functions needs its own rule because the rate of change of a product depends on both functions' values AND both of their rates of change simultaneously.

d/dx[f(x)·g(x)] = f'(x)g(x) + f(x)g'(x) — “derivative of the first times the second, plus the first times the derivative of the second.”

For h(x)=x²·sin(x): let f(x)=x² (f'(x)=2x) and g(x)=sin(x) (g'(x)=cos(x)). Then h'(x) = 2x·sin(x) + x²·cos(x).

The single most common product-rule error is assuming d/dx[f·g] = f'·g' (just multiplying the two derivatives) — this is essentially never correct, and it's exactly the kind of shortcut-that-isn't-actually-a-shortcut graders are trained to catch.

The Quotient Rule

Division of functions needs its own rule, related to but distinct from the product rule, because of how the denominator's changing value affects the ratio.

d/dx[f(x)/g(x)] = [f'(x)g(x) − f(x)g'(x)] / [g(x)]². Note the subtraction in the numerator (unlike the product rule's addition), and the whole thing is divided by the denominator squared.

For h(x)=x/(x²+1): let f(x)=x (f'(x)=1) and g(x)=x²+1 (g'(x)=2x). Then h'(x) = [1·(x²+1) − x·(2x)]/(x²+1)² = (1−x²)/(x²+1)².

Order matters critically in the numerator — f'(x)g(x) must come first, minus f(x)g'(x). Swapping the order flips the sign of the entire answer, which forfeits credit for that step even when every other part of the work is correct.

Derivatives of the Remaining Trig Functions

Tangent, cotangent, secant, and cosecant can each be rewritten in terms of sine and cosine, so their derivatives can be derived using the quotient rule rather than memorized as entirely separate facts — though in practice it's faster to just memorize the four resulting patterns.

d/dx[tan x] = sec²x, d/dx[cot x] = −csc²x, d/dx[sec x] = sec(x)tan(x), d/dx[csc x] = −csc(x)cot(x). Notice the pattern: the “co-” functions always pick up a negative sign, mirroring cosine's negative derivative.

Since tan(x)=sin(x)/cos(x), the quotient rule gives [cos(x)·cos(x) − sin(x)·(−sin(x))]/cos²(x) = [cos²x+sin²x]/cos²x = 1/cos²x = sec²x, confirming the memorized result from first principles.

It's easy to mix up which of the six trig derivatives carry a negative sign — remember that only the three “co-” functions (cosine, cotangent, cosecant) have negative derivatives; sine, tangent, and secant's derivatives are all positive expressions.

The Chain Rule

d/dx[f(g(x))] = f'(g(x))·g'(x). This is the single most heavily tested differentiation rule on the exam because it shows up embedded inside almost every other technique in the course.

Recognizing When to Apply the Chain Rule

Any time you're differentiating a function “of a function” — sin(x²), (3x+1)^5, e^(2x) — you need the chain rule. The giveaway is an expression nested inside another function.

Implicit Differentiation

When y isn't isolated (as in x² + y² = 25), differentiate both sides with respect to x, treating y as a function of x and applying the chain rule to every y-term (which produces a dy/dx factor), then solve algebraically for dy/dx.

Derivatives of Inverse Functions

If g is the inverse of f, then g'(x) = 1 / f'(g(x)). This connects a function's derivative to its inverse's derivative without ever needing an explicit formula for the inverse itself.

Derivatives of Inverse Trigonometric Functions

d/dx[arcsin x] = 1/√(1−x²), d/dx[arctan x] = 1/(1+x²), and similar formulas for arccos, arccot, arcsec, and arccsc — given on the AP formula sheet, but you should recognize them on sight.

Choosing the Right Combination of Rules

Many problems stack multiple rules — a chain rule inside a quotient rule, or implicit differentiation combined with the product rule. Practice identifying the “outermost” structure of an expression first, then work inward.

Higher-Order Derivatives

The second derivative f''(x) is just the derivative of f'(x), and so on for f'''(x) and beyond. Higher-order derivatives are essential for the concavity work in Unit 5.

Interpreting a Derivative in Context

f'(a) is always a rate: “how fast [quantity] is changing with respect to [variable] at [point].” Free-response questions grade heavily on stating the correct units and interpreting the sign correctly, not just computing a number.

Motion: Position, Velocity, and Acceleration

Velocity is the derivative of position; acceleration is the derivative of velocity. An object speeds up when velocity and acceleration share the same sign, and slows down when they have opposite signs.

Rates of Change Outside of Motion

The same logic applies to any changing quantity — how fast a tank fills, a population grows, or a cost changes. The “speeding up/slowing down” vocabulary is really about the sign relationship between a quantity and its own derivative.

Setting Up Related Rates Problems

Identify every quantity that changes with time, write an equation connecting them (often a geometric formula, like the Pythagorean theorem or a volume formula), and differentiate both sides with respect to time before plugging in any numbers.

Solving Related Rates Problems

After implicitly differentiating with respect to t, substitute the known values — including the given rates — and solve for the unknown rate. Plugging numbers in before differentiating is the most common way a related-rates problem goes wrong.

Local Linearization

The tangent line at a point approximates a function's values nearby: f(x) ≈ f(a) + f'(a)(x−a). Because most functions curve away from their tangent lines, this approximation is only reliable very close to the point of tangency.

L'Hôpital's Rule

For limits of indeterminate form (0/0 or ∞/∞), lim f(x)/g(x) = lim f'(x)/g'(x), provided that new limit exists. You must confirm the indeterminate form first — applying it where it doesn't apply is a scoring trap.

The Mean Value Theorem (MVT)

If f is continuous on [a,b] and differentiable on (a,b), there's at least one c in (a,b) where f'(c) equals the average rate of change of f over [a,b]. Like the IVT, this is an existence theorem — it doesn't tell you where c is, only that it exists.

The Extreme Value Theorem and Critical Points

A function continuous on a closed interval is guaranteed an absolute max and min somewhere on it. Critical points — where f'(x) = 0 or f'(x) is undefined — are the only candidates for local extrema.

Increasing/Decreasing Behavior from the First Derivative

f is increasing where f'(x) > 0 and decreasing where f'(x) < 0. A sign change in f' is exactly where a local extremum can occur.

The First Derivative Test

If f' changes from positive to negative at a critical point, that point is a local max; negative to positive is a local min; no sign change means neither.

The Candidates Test for Absolute Extrema

To find the absolute max/min on a closed interval, evaluate f at every critical point inside the interval AND at both endpoints, then compare all the values — the largest and smallest are the absolute extrema.

Concavity and the Second Derivative

f is concave up where f''(x) > 0 and concave down where f''(x) < 0. A point where concavity actually changes is an inflection point, and f''(x) = 0 or is undefined there.

The Second Derivative Test

At a critical point where f'(x) = 0: if f''(x) > 0 it's a local min, if f''(x) < 0 it's a local max, and if f''(x) = 0 the test is inconclusive and you fall back on the first derivative test.

Connecting f, f', and f''

Being able to sketch f' from a graph of f (or reason about f from a graph of f' alone) is one of the most frequently tested skills on both the multiple-choice and free-response sections.

Optimization Problems

Translate the word problem into a function of a single variable (use a constraint equation to eliminate a second variable), find critical points, and justify — via the first/second derivative test or the candidates test — that you've actually found a maximum or minimum, not just a critical point.

Accumulation as the Reverse of a Rate

If you know a rate of change, integrating that rate over an interval gives the total accumulated change over that interval — the conceptual bridge between derivatives and integrals.

Riemann Sums

Approximate the area under a curve by summing rectangle areas (using left, right, or midpoint sample points). As the number of rectangles grows and their width shrinks toward zero, the sum approaches the exact area — the definite integral.

Definite Integral Notation and Meaning

∫[a to b] f(x) dx is the exact signed area between f and the x-axis from x=a to x=b — area above the axis counts positive, area below counts negative.

The Fundamental Theorem of Calculus, Part 1

d/dx[∫[a to x] f(t) dt] = f(x). Differentiating an accumulation function just returns the original integrand, evaluated at the variable bound (with a chain-rule adjustment if that bound is more complex than plain x).

The Fundamental Theorem of Calculus, Part 2

∫[a to b] f(x) dx = F(b) − F(a), where F is any antiderivative of f. This is the tool that actually lets you evaluate definite integrals without falling back on Riemann sums.

Properties of Definite Integrals

Reversing the bounds negates the integral; a zero-width interval gives zero; and integrals over adjacent intervals add: ∫[a to b] + ∫[b to c] = ∫[a to c].

Basic Antiderivative Rules

The reverse power rule (∫x^n dx = x^(n+1)/(n+1) + C for n≠−1), plus the antiderivatives of the basic trig and exponential functions. Never drop the “+C” on an indefinite integral.

Integration by u-Substitution

Spot a composite function's derivative pattern inside the integral, let u equal the “inside” function, rewrite the whole integral in terms of u and du, integrate, then substitute back — the antiderivative counterpart to the chain rule.

What a Differential Equation Describes

An equation relating a function to its own derivative(s). Solving one means finding the family of functions — or one specific function — that satisfies that relationship.

Slope Fields

A slope field sketches short line segments at many points, each showing the slope dy/dx the differential equation predicts there. Sketching a solution curve means following a path tangent to the segments at every point it passes through.

Separable Differential Equations

When a differential equation can be rearranged so all y-terms (with dy) sit on one side and all x-terms (with dx) sit on the other, integrate both sides separately, then solve for y if possible.

General vs. Particular Solutions

Integrating a differential equation produces a general solution with an arbitrary constant C. An initial condition — a specific point the solution must pass through — lets you solve for C and pin down the one particular solution that fits.

Exponential Growth and Decay Models

The differential equation dy/dt = ky (rate of change proportional to the current amount) always solves to y = Ce^(kt). This single model underlies population growth, radioactive decay, and continuously compounded interest problems.

Average Value of a Function

The average value of f on [a,b] is (1/(b−a))·∫[a to b] f(x) dx — conceptually, the height of the rectangle over [a,b] that has the same area as the region under the curve.

Area Between Two Curves

∫[a to b] (top function − bottom function) dx, where “top” and “bottom” are whichever function has the greater y-value across that interval — if that relationship flips, you have to split the interval.

Volumes by Disks

When a region is revolved around an axis and the cross-sections perpendicular to that axis are solid circles, integrate π·(radius)² across the interval.

Volumes by Washers

When the revolved region leaves a hole — an inner and outer radius — integrate π·[(outer radius)² − (inner radius)²] across the interval.

Volumes with Known Cross-Sections

When cross-sections perpendicular to an axis are a specified shape other than a circle (squares, equilateral triangles, semicircles), integrate that shape's area formula, expressed in terms of the boundary functions, across the interval.

Connecting Position and Distance via Integration

The definite integral of velocity gives displacement (net position change); integrating the absolute value of velocity gives total distance traveled. Mixing these two up is one of the highest-frequency scoring errors on motion FRQs.