AP Calculus AB study guide
Work through each official AP unit: original study notes, the key terms you must know, and topic-question drills written for this site. Then test yourself with the timed AP Calculus AB practice exam and see where you stand with the score calculator.
Units
Unit 1: Limits and Continuity8 drills
The foundation of calculus: how functions behave as inputs approach a value, and how to tell when a function is continuous.
What you need to know
What a limit is
A limit asks what a function gets close to as the input gets close to a value. We write lim(x→a) f(x) = L and say the outputs approach L, whether or not f is actually defined at a. Limits can be read from a graph or a table of values.
One-sided limits check behavior from the left (x→a⁻) and right (x→a⁺). The two-sided limit exists only when both one-sided limits exist and agree. That is why a function like f(x) = |x|/x has no two-sided limit at 0 even though it is defined everywhere nearby.
Evaluating limits algebraically
The fastest path is direct substitution: if plugging in a works and gives a number, that is the limit. When substitution gives 0/0 (an indeterminate form), simplify first: factor and cancel, rationalize radicals, or combine fractions.
Common limits worth memorizing: the limit of sin x / x as x→0 is 1, and the limit of (1 − cos x)/x as x→0 is 0. Horizontal asymptotes are found with limits at infinity: if the degree of the numerator equals the denominator, the limit is the ratio of leading coefficients.
Continuity
A function f is continuous at a point a when three things hold: f(a) is defined, the limit exists at a, and the limit equals f(a). A function is continuous on an interval when it is continuous at every point there.
The Intermediate Value Theorem says a continuous function on [a, b] takes every value between f(a) and f(b). That makes it a powerful tool for proving roots exist. Discontinuities are removable (hole), jump, or infinite (vertical asymptote).
Key terms
- Limit — The value a function approaches as its input approaches a given value.
- One-sided limit — A limit approached from only the left or only the right.
- Indeterminate form — A limit expression like 0/0 that needs simplification before evaluation.
- Vertical asymptote — A line x = a approached by the function as it grows unboundedly near a.
- Horizontal asymptote — A line y = L approached by the function as x goes to ±∞.
- Continuous function — A function with no breaks, holes, or jumps on its domain.
- Intermediate Value Theorem — A continuous function takes every value between its outputs at two endpoints.
- Removable discontinuity — A hole where the limit exists but the function is undefined or off-value.
- Jump discontinuity — A break where the one-sided limits differ.
- Infinite discontinuity — A vertical asymptote where the function grows without bound.
Topic drills
Tap an answer to check it. Every question is original and written for this site.
Q1.
The limit of f(x) = (x² − 4)/(x − 2) as x → 2 is
Answer: B — Factor to x + 2; the limit is 4 even though f is undefined at 2.
Q2.
A two-sided limit exists at x = a only if
Answer: B — The two-sided limit equals both one-sided limits when they agree.
Q3.
The limit of sin x / x as x → 0 is
Answer: B — This is a fundamental limit equal to 1.
Q4.
For continuity at a point a, which is required?
Answer: A — All three conditions must hold for continuity.
Q5.
A function with a hole where the limit exists is said to have a
Answer: B — A hole with a defined limit is a removable discontinuity.
Q6.
The Intermediate Value Theorem guarantees that a continuous function
Answer: B — Continuity ensures every intermediate value occurs on the interval.
Q7.
As x → ∞, f(x) = (3x² + 2)/(x² − 1) approaches
Answer: B — Same-degree rational function: limit is the ratio of leading coefficients, 3/1 = 3.
Q8.
Which type of discontinuity occurs at a vertical asymptote?
Answer: C — A vertical asymptote is an infinite discontinuity.
Free response practice
FRQ1.
Let f(x) = (x − 3)/(x² − 9). (a) Simplify f. (b) Find lim(x→3) f(x). (c) Describe the discontinuity at x = 3.
6 points · rubric: Simplify 2 pts, limit 2 pts, classification 2 pts.
Model answer
(a) f = 1/(x + 3). (b) The limit as x→3 is 1/6. (c) The function is undefined at 3 but the limit exists, so x = 3 is a removable discontinuity.
FRQ2.
Use the Intermediate Value Theorem to show that g(x) = x³ + x − 1 has a root between 0 and 1.
5 points · rubric: Check endpoints 2 pts, continuity 1 pt, IVT conclusion 2 pts.
Model answer
g is a polynomial, hence continuous. g(0) = −1 and g(1) = 1. Since 0 lies between −1 and 1, the IVT guarantees some c in (0,1) with g(c) = 0.
Unit 2: Differentiation: Definition and Fundamental Properties8 drills
The derivative as a rate of change, its rules, and the shapes of derivative graphs.
What you need to know
The definition of the derivative
The derivative f'(x) measures instantaneous rate of change. It is the limit of a difference quotient: f'(a) = lim(h→0) [f(a+h) − f(a)]/h. Geometrically, the derivative at a point is the slope of the tangent line there.
If the derivative exists at every point on an interval, the function is differentiable there. Differentiability implies continuity, but continuity does not imply differentiability: a sharp corner is continuous yet has no derivative.
Differentiation rules
The power rule says d/dx(xⁿ) = nxⁿ⁻¹. Constant multiples and sums follow linearly. The product rule is d/dx(fg) = f'g + fg', and the quotient rule handles ratios.
The chain rule differentiates compositions: d/dx f(g(x)) = f'(g(x)) · g'(x). It is the workhorse of calculus and appears in nearly every later topic, including implicit differentiation and related rates.
Reading derivative graphs
Where the original function increases, its derivative is positive; where it decreases, the derivative is negative; at a flat spot (local max or min) the derivative is zero.
The sign of f' tells whether f is rising or falling. The sign of f'' (the second derivative, derivative of the derivative) tells concavity: f'' > 0 means concave up, f'' < 0 means concave down. Inflection points occur where concavity changes.
Key terms
- Derivative — The instantaneous rate of change of a function, the slope of the tangent line.
- Difference quotient — The expression [f(a+h) − f(a)]/h whose limit defines the derivative.
- Tangent line — A line that touches a curve at one point with the same slope as the curve there.
- Power rule — d/dx(xⁿ) = nxⁿ⁻¹.
- Product rule — d/dx(fg) = f'g + fg'.
- Quotient rule — d/dx(f/g) = (f'g − fg')/g².
- Chain rule — d/dx f(g(x)) = f'(g(x))·g'(x).
- Differentiability — The property of having a derivative at each point of an interval.
- Second derivative — The derivative of the derivative, f'', which signals concavity.
- Concavity — Whether a curve bends upward (f''>0) or downward (f''<0).
- Inflection point — A point where concavity changes.
Topic drills
Tap an answer to check it. Every question is original and written for this site.
Q1.
The derivative of f(x) = x⁵ is
Answer: A — Power rule: bring down 5, reduce exponent to 4.
Q2.
If f is differentiable at a, then f is
Answer: A — Differentiability implies continuity.
Q3.
The derivative of f(x) = x²·eˣ (product rule) is
Answer: A — Product rule: f'g + fg' = 2x·eˣ + x²·eˣ.
Q4.
The derivative of sin(3x) is
Answer: A — Chain rule: cos(3x)·3.
Q5.
Where a function has a local maximum, its derivative
Answer: A — Flat spots have derivative zero (or undefined at corners).
Q6.
If f''(x) > 0 on an interval, then f is
Answer: A — Positive second derivative means concave up.
Q7.
A continuous function with a sharp corner at x = a is
Answer: A — Corners are continuous yet have no derivative there.
Q8.
The derivative of f(x) = x³ − 4x evaluated at x = 2 is
Answer: A — f'(x) = 3x² − 4; at x = 2, f'(2) = 12 − 4 = 8.
Free response practice
FRQ1.
Use the limit definition of the derivative to find f'(x) for f(x) = x².
6 points · rubric: Correct difference quotient 3 pts, correct algebra 2 pts, correct limit 1 pt.
Model answer
f'(x) = lim(h→0)[(x+h)² − x²]/h = lim(h→0)(2xh + h²)/h = lim(h→0)(2x + h) = 2x.
FRQ2.
State the product rule and use it to differentiate f(x) = (x² + 1)(x³ − 2).
5 points · rubric: Rule statement 2 pts, correct application 3 pts.
Model answer
Product rule: (x²+1)'·(x³−2) + (x²+1)·(x³−2)' = 2x(x³−2) + (x²+1)(3x²).
Unit 3: Differentiation: Composite, Implicit, and Inverse Functions8 drills
The chain rule deep-dive: composite functions, implicit differentiation, inverse functions, and their derivatives.
What you need to know
The chain rule and its power
The chain rule lets us differentiate any composition: d/dx f(g(x)) = f'(g(x))·g'(x). Practically, differentiate the outer function, keep the inner function intact, then multiply by the inner derivative.
This one rule powers the rest of the course: exponential, logarithmic, and trigonometric compositions are all chain-rule problems. For example, d/dx e^(3x) = 3e^(3x), and d/dx ln(x²) = 2x/x² = 2/x.
Implicit differentiation
When y is not isolated, treat y as a function of x and differentiate both sides, adding dy/dx wherever you differentiate a y. Then solve for dy/dx algebraically.
Implicit differentiation handles curves like x² + y² = 25, which cannot be written as a single function. It also appears in related rates problems, where both x and y depend on time.
Derivatives of inverse functions
The derivative of an inverse function obeys the simple reciprocal rule: (f⁻¹)'(x) = 1 / f'(f⁻¹(x)). For example, the derivative of ln x (inverse of eˣ) is 1/x, and d/dx arcsin(x) = 1/√(1−x²).
Knowing f'(a) tells us (f⁻¹)' at the point f(a) without solving for the inverse explicitly.
Key terms
- Chain rule — Differentiates compositions: d/dx f(g(x)) = f'(g(x))·g'(x).
- Implicit differentiation — Differentiating an equation treating y as a function of x, then solving for dy/dx.
- Inverse function — A function that reverses another; f(f⁻¹(x)) = x.
- Inverse derivative rule — (f⁻¹)'(x) = 1/f'(f⁻¹(x)).
- Composite function — A function made by applying one function inside another.
- dy/dx — Leibniz notation for the derivative of y with respect to x.
- Logarithmic differentiation — Taking logs before differentiating to simplify products and powers.
Topic drills
Tap an answer to check it. Every question is original and written for this site.
Q1.
The derivative of e^(2x) is
Answer: A — Chain rule: e^(2x)·2.
Q2.
The derivative of ln(5x) is
Answer: A — Chain rule: (1/(5x))·5 = 1/x.
Q3.
Using implicit differentiation on x² + y² = 25 gives dy/dx =
Answer: A — 2x + 2y·dy/dx = 0, so dy/dx = −x/y.
Q4.
The derivative of the inverse of f at the point f(a) is
Answer: A — (f⁻¹)'(f(a)) = 1/f'(a).
Q5.
The derivative of arcsin(x) is
Answer: A — The standard inverse trig derivative is 1/√(1−x²).
Q6.
d/dx cos(x²) equals
Answer: A — Chain rule: −sin(x²)·2x.
Q7.
The derivative of ln(x² + 1) is
Answer: A — Chain rule: (1/(x²+1))·2x.
Q8.
If f(x) = √x, then f⁻¹(x) = x² and (f⁻¹)'(x) is
Answer: A — The derivative of x² is 2x; equivalently 1/f'(f⁻¹(x)) = 1/(1/(2x)) = 2x.
Free response practice
FRQ1.
Find dy/dx for x²y + y³ = 10 using implicit differentiation.
6 points · rubric: Correct derivative of each term 4 pts, correct algebra solving for dy/dx 2 pts.
Model answer
Differentiating: 2xy + x²·dy/dx + 3y²·dy/dx = 0. Factor: dy/dx(x² + 3y²) = −2xy, so dy/dx = −2xy/(x² + 3y²).
FRQ2.
If f(2) = 5 and f'(2) = 7, find the derivative of the inverse function at x = 5.
5 points · rubric: Identifies the point 3 pts, applies inverse rule 2 pts.
Model answer
(f⁻¹)'(5) = 1/f'(f⁻¹(5)) = 1/f'(2) = 1/7.
Unit 4: Contextual Applications of Differentiation8 drills
Using derivatives in the real world: rates of change, related rates, motion, and linearization.
What you need to know
Motion and rates
If s(t) is position, then velocity is v(t) = s'(t) and acceleration is a(t) = v'(t). Speed is |v|. A particle speeds up when velocity and acceleration have the same sign, and slows when they differ.
Rates of change in context: population growth, water flow, and temperature change are all derivative problems. The units of a derivative are the output units per input unit, which helps confirm answers.
Related rates
In related-rates problems, several quantities change together. Steps: draw and label, write a relationship between the quantities (Pythagorean theorem, volume formulas), differentiate with respect to time, substitute known values, and solve.
The classic example is a ladder sliding down a wall: relate the ladder length to the base and height, then differentiate to find how fast the top falls.
Linearization and tangent lines
The tangent line at a point is the best linear approximation to the curve nearby. The linearization is L(x) = f(a) + f'(a)(x − a), used to estimate f near a.
Euler's method builds on this idea, stepping along tangent lines to approximate solutions of differential equations. It appears again in Unit 7.
Key terms
- Velocity — The rate of change of position; the first derivative of position.
- Acceleration — The rate of change of velocity; the second derivative of position.
- Speed — The absolute value of velocity.
- Related rates — Problems where multiple quantities change together and are connected by an equation.
- Linearization — The tangent-line approximation L(x) = f(a) + f'(a)(x − a).
- Euler's method — A stepwise tangent-line approximation for solving differential equations.
- Instantaneous rate of change — The derivative evaluated at a single point.
Topic drills
Tap an answer to check it. Every question is original and written for this site.
Q1.
If s(t) = t³ − 6t², the velocity at t = 2 is
Answer: A — v(t) = 3t² − 12t; at t = 2, v = 12 − 24 = −12.
Q2.
A particle speeds up when velocity and acceleration
Answer: A — Same sign means speeding up; opposite signs means slowing down.
Q3.
A ladder sliding down a wall is a classic problem in
Answer: A — It connects moving quantities through geometry, a related-rates problem.
Q4.
The linearization of f at a is
Answer: A — L(x) = f(a) + f'(a)(x − a).
Q5.
Acceleration is the derivative of
Answer: B — Acceleration is the derivative of velocity (second derivative of position).
Q6.
The units of a derivative f'(x) are
Answer: A — A rate has output units per input unit.
Q7.
Euler's method approximates a solution curve by
Answer: A — Euler's method uses small tangent-line steps.
Q8.
If v(t) = 4 − t, the particle changes direction when
Answer: A — v = 0 at t = 4, where the particle reverses.
Free response practice
FRQ1.
A 10-ft ladder rests against a wall with its base sliding out at 2 ft/s. When the base is 6 ft from the wall, how fast is the top falling?
6 points · rubric: Set up Pythagorean relationship 2 pts, differentiate with respect to time 2 pts, substitute and solve 2 pts.
Model answer
x² + y² = 100 with x = 6 gives y = 8. Differentiating: 2x·dx/dt + 2y·dy/dt = 0. So 2(6)(2) + 2(8)(dy/dt) = 0, giving dy/dt = −1.5 ft/s.
FRQ2.
A particle moves along a line with position s(t) = t³ − 6t² + 9t. Find (a) the velocity function, (b) when the particle is at rest.
5 points · rubric: Velocity 2 pts, at-rest times 3 pts.
Model answer
(a) v(t) = 3t² − 12t + 9. (b) Set v = 0: 3(t² − 4t + 3) = 3(t−1)(t−3), so the particle is at rest at t = 1 and t = 3.
Unit 5: Analytical Applications of Differentiation8 drills
Using the first and second derivatives to analyze functions: extrema, concavity, and curve sketching.
What you need to know
The Mean Value Theorem
The Mean Value Theorem (MVT) says a differentiable function on [a, b] has a point where the tangent slope equals the average (secant) slope. Its special case, Rolle's Theorem, applies when f(a) = f(b).
The MVT justifies the connection between a function's behavior and its derivative: a function with derivative zero everywhere is constant.
Finding extrema
A function's absolute maximum and minimum on a closed interval occur either at critical points (where f' = 0 or is undefined) or at the endpoints. To find them, evaluate f at all candidates.
For local extrema, use the first-derivative test (where f' changes sign) or the second-derivative test (f'' > 0 means local min, f'' < 0 means local max).
Curve sketching and optimization
The sign of f' gives increase/decrease; the sign of f'' gives concavity; points where f'' changes sign are inflection points. Together these describe the full shape of a graph.
Optimization applies all of this to word problems: identify what to maximize or minimize, build a function of one variable, find critical points, and justify the answer on the interval.
Key terms
- Mean Value Theorem — At some point on [a,b], f'(c) equals the average rate of change (f(b)−f(a))/(b−a).
- Rolle's Theorem — MVT with f(a) = f(b): a horizontal secant guarantees a horizontal tangent.
- Critical point — A point in the domain where f' = 0 or f' is undefined.
- Absolute maximum — The largest output of a function on an interval.
- First-derivative test — Locating extrema by where f' changes sign.
- Second-derivative test — Using f'' at a critical point to classify a local max or min.
- Optimization — Using derivatives to find maximum or minimum values in applied problems.
- Closed interval method — Checking critical points and endpoints to find absolute extrema.
Topic drills
Tap an answer to check it. Every question is original and written for this site.
Q1.
The Mean Value Theorem requires that f is
Answer: A — Continuity on the closed interval and differentiability on the open interval are required.
Q2.
A function's absolute maximum on [a, b] must occur at
Answer: A — Candidates are critical points and endpoints.
Q3.
If f' changes from positive to negative at c, then c is a
Answer: A — Positive to negative means a local maximum.
Q4.
The second-derivative test says f''(c) < 0 at a critical point means
Answer: A — Negative second derivative at a critical point indicates a local max.
Q5.
An inflection point occurs where
Answer: A — Concavity changes at inflection points.
Q6.
Rolle's Theorem is a special case of the Mean Value Theorem where
Answer: A — Rolle's Theorem applies when the endpoints have equal outputs.
Q7.
If f'(x) > 0 on an interval, then f is
Answer: A — Positive derivative means the function increases.
Q8.
To maximize a quantity in a word problem, you should
Answer: A — Optimization: find critical points and justify the extremum.
Free response practice
FRQ1.
Find and classify all local extrema of f(x) = 2x³ − 3x² − 12x using the first-derivative test.
6 points · rubric: Critical points 2 pts, sign analysis 2 pts, classification 2 pts.
Model answer
f'(x) = 6x² − 6x − 12 = 6(x−2)(x+1). Critical at x = −1, 2. Sign of f': + to − at −1 (local max), − to + at 2 (local min).
FRQ2.
State the Mean Value Theorem and verify it for f(x) = x² on [1, 3] by finding the guaranteed c.
6 points · rubric: Statement 2 pts, setup 2 pts, correct c 2 pts.
Model answer
MVT: for a continuous-on-[a,b] and differentiable-on-(a,b) function, some c satisfies f'(c) = (f(b)−f(a))/(b−a). Here (9−1)/(3−1) = 4, and f'(c) = 2c = 4 gives c = 2, which lies in (1,3).
Unit 6: Integration and Accumulation of Change8 drills
Antiderivatives, definite integrals, the Fundamental Theorem of Calculus, and area accumulation.
What you need to know
Antiderivatives and indefinite integrals
An antiderivative F of f satisfies F' = f. The indefinite integral ∫f(x) dx = F(x) + C records the whole family, with C the constant of integration.
Basic rules mirror differentiation: power rule (add 1 to the exponent, divide by the new exponent), and known integrals such as ∫cos x dx = sin x + C, ∫eˣ dx = eˣ + C, ∫1/x dx = ln|x| + C.
The definite integral and the FTC
The definite integral ∫ₐᵇ f(x) dx measures the signed area between the curve and the x-axis. It is defined as a limit of Riemann sums (left, right, midpoint, trapezoidal).
The Fundamental Theorem of Calculus has two parts: (1) d/dx ∫ₐˣ f(t) dt = f(x), connecting antidifferentiation to integration; and (2) ∫ₐᵇ f(x) dx = F(b) − F(a) for any antiderivative F.
Accumulation and area
Definite integrals accumulate change: total distance = ∫|v| dt, net displacement = ∫v dt, and area between curves = ∫(top − bottom) dx.
When functions change sign, split the integral at the zeros so area stays positive. Substitution (u-substitution) is the integration counterpart of the chain rule for more complicated integrands.
Key terms
- Antiderivative — A function F whose derivative is the given function f.
- Indefinite integral — The family of antiderivatives, written ∫f(x) dx = F(x) + C.
- Definite integral — The signed area under a curve from a to b, written ∫ₐᵇ f(x) dx.
- Riemann sum — A rectangular approximation of a definite integral.
- Fundamental Theorem of Calculus — The theorem linking differentiation and integration.
- Constant of integration — The arbitrary constant C in an indefinite integral.
- u-substitution — The integration technique that reverses the chain rule.
- Signed area — Area counted as negative below the x-axis.
Topic drills
Tap an answer to check it. Every question is original and written for this site.
Q1.
The antiderivative of x³ is
Answer: A — Power rule for integration: add 1 to the exponent, divide by the new exponent.
Q2.
The Fundamental Theorem of Calculus (part 2) states that ∫ₐᵇ f(x) dx =
Answer: A — The definite integral equals the antiderivative difference F(b) − F(a).
Q3.
∫₀¹ 3x² dx =
Answer: A — Antiderivative is x³; x³ from 0 to 1 is 1.
Q4.
Total distance traveled equals
Answer: A — Integrating absolute velocity gives total distance.
Q5.
u-substitution is the reverse of
Answer: A — u-substitution reverses the chain rule.
Q6.
∫cos x dx =
Answer: A — The antiderivative of cos x is sin x.
Q7.
A Riemann sum with rectangles underestimates an increasing function when using
Answer: A — Left sums underestimate increasing functions.
Q8.
The area between y = x and y = x² on [0,1] is
Answer: A — ∫₀¹(x − x²) dx = x²/2 − x³/3 evaluated from 0 to 1 = 1/2 − 1/3 = 1/6.
Free response practice
FRQ1.
Evaluate ∫₀² (3x² + 2x) dx and explain the meaning of the definite integral.
6 points · rubric: Correct antiderivative 2 pts, correct evaluation 2 pts, meaning 2 pts.
Model answer
Antiderivative: x³ + x². Evaluated from 0 to 2: (8 + 4) − 0 = 12. The integral gives the signed area between y = 3x² + 2x and the x-axis from 0 to 2.
FRQ2.
A particle moves with velocity v(t) = 2t − 4 for 0 ≤ t ≤ 6. Find the net displacement and the total distance traveled.
6 points · rubric: Net displacement 2 pts, split at v=0 2 pts, total distance 2 pts.
Model answer
Net displacement = ∫₀⁶(2t−4) dt = [t²−4t]₀⁶ = 36−24 = 12. v = 0 at t = 2. Total distance = ∫₀²(4−2t) dt + ∫₂⁶(2t−4) dt = 4 + 16 = 20.
Unit 7: Differential Equations8 drills
Equations involving derivatives: slope fields, solving separable equations, and modeling growth and decay.
What you need to know
Slope fields
A slope field draws tiny tangent segments at grid points, each with slope given by dy/dx. Following the segments sketches solution curves without solving the equation.
Slope fields reveal the qualitative behavior of solutions: where they increase, decrease, level off, or approach asymptotes. They are especially useful for equations that are hard to solve exactly.
Separable differential equations
A separable equation can be written dy/dx = g(x)·h(y). Separate variables so all y's and dy are on one side and all x's and dx on the other, integrate both sides, then solve for y and use the initial condition to find C.
The result is a particular solution satisfying the initial value y(x₀) = y₀.
Exponential models
The equation dy/dt = ky models exponential growth (k > 0) or decay (k < 0), with solution y = y₀e^(kt). It describes populations, radioactive decay, and Newton's law of cooling.
The logistic equation dy/dt = ky(1 − y/L) models growth that slows as a population approaches its carrying capacity L. Its solutions level off at the horizontal asymptote y = L.
Key terms
- Differential equation — An equation involving a function and its derivatives.
- Slope field — A grid of tangent segments that visualize solutions of a differential equation.
- Separable equation — A differential equation writable as dy/dx = g(x)·h(y).
- Initial condition — A value y(x₀) = y₀ used to determine the constant C.
- General solution — The family of solutions including the constant C.
- Particular solution — A solution fixed by an initial condition.
- Exponential growth/decay — Models dy/dt = ky with solutions y = y₀e^(kt).
- Logistic growth — Model dy/dt = ky(1 − y/L) that levels off at carrying capacity L.
- Euler's method — Numerical tangent-line stepping to approximate solutions.
Topic drills
Tap an answer to check it. Every question is original and written for this site.
Q1.
A differential equation is called separable if it can be written as
Answer: A — Separable equations factor into a function of x times a function of y.
Q2.
The solution of dy/dt = 3y is
Answer: A — Separate and integrate to get y = Ce^(3t).
Q3.
In the model dy/dt = ky, k < 0 models
Answer: A — Negative k gives decay.
Q4.
The logistic model dy/dt = ky(1 − y/L) levels off at
Answer: A — The carrying capacity L is the horizontal asymptote.
Q5.
An initial condition is used to
Answer: A — Initial conditions determine the particular constant.
Q6.
A slope field shows
Answer: A — Slope fields plot tangent segments from dy/dx.
Q7.
Newton's law of cooling is modeled by
Answer: A — Cooling rates depend on the temperature difference from the environment.
Q8.
Euler's method produces
Answer: A — Euler's method approximates solutions numerically.
Free response practice
FRQ1.
Solve the separable differential equation dy/dx = 2xy with initial condition y(0) = 3.
6 points · rubric: Separate variables 2 pts, integrate both sides 2 pts, apply initial condition 2 pts.
Model answer
dy/y = 2x dx. Integrating: ln|y| = x² + C. So y = Ce^(x²). Using y(0) = 3 gives C = 3, so y = 3e^(x²).
FRQ2.
A population grows with dP/dt = 0.02P. If P(0) = 1000, (a) find P(t) and (b) predict the population after 10 years.
6 points · rubric: General solution 2 pts, particular solution 2 pts, evaluation 2 pts.
Model answer
(a) P = 1000e^(0.02t). (b) P(10) = 1000e^(0.2) ≈ 1221.
Unit 8: Applications of Integration8 drills
Using integrals in geometry and physics: average value, area between curves, volumes, and more.
What you need to know
Average value and area
The average value of a continuous function on [a, b] is (1/(b−a)) ∫ₐᵇ f(x) dx. It is the height of a rectangle with the same area as the region under the curve.
The area between two curves y = f(x) and y = g(x) is ∫(top − bottom) dx over the region, found by subtracting the lower curve from the upper curve and integrating.
Volumes of solids
Volumes by disk/washer: rotate a region and slice perpendicular to the axis; each slice is a disk (πr²) or washer (π(R²−r²)), integrated along the axis.
Volumes by cross sections: if the cross sections perpendicular to an axis are squares or other shapes of area A(x), the volume is ∫A(x) dx.
Motion and accumulation
Integrals finish motion problems: net change = ∫a dt gives velocity from acceleration, and ∫v dt gives position from velocity.
More applications include work (∫ force dx), displacement vs distance, and the fundamental connection that derivatives and integrals undo each other. These are favorite AP FRQ scenarios.
Key terms
- Average value — The value (1/(b−a))∫ₐᵇ f(x) dx.
- Disk method — Volume by rotating slices into circular disks: V = π∫r² dx.
- Washer method — Volume using rings: V = π∫(R² − r²) dx.
- Cross section — A slice of a solid, whose area is integrated to find volume.
- Volume of revolution — A solid formed by rotating a region around an axis.
- Net change — The total change given by a definite integral.
- Accumulated rate — Integrating a rate function gives total accumulation.
Topic drills
Tap an answer to check it. Every question is original and written for this site.
Q1.
The average value of f(x) = x on [0, 4] is
Answer: A — (1/4)∫₀⁴ x dx = (1/4)(8) = 2.
Q2.
The disk method computes volume as
Answer: A — Disks have area πr²; integrate along the axis.
Q3.
The area between y = x and y = 0 on [0, 3] is
Answer: A — ∫₀³ x dx = x²/2 = 9/2.
Q4.
To find the volume of a solid with square cross sections of side s(x), integrate
Answer: A — Square cross-section area is s².
Q5.
Integrating acceleration with respect to time gives
Answer: A — The integral of acceleration is velocity.
Q6.
The net change in F from a to b is
Answer: A — Net change equals the integral of the derivative.
Q7.
The washer method is used when rotating produces
Answer: A — Washers handle holes by subtracting the inner radius squared.
Q8.
If v(t) = 3t², the displacement from t = 0 to t = 2 is
Answer: A — ∫₀² 3t² dt = t³ = 8.
Free response practice
FRQ1.
Find the area of the region bounded by y = x² and y = x between their intersection points.
6 points · rubric: Intersection points 2 pts, correct setup 2 pts, correct evaluation 2 pts.
Model answer
Intersections: x² = x → x = 0, 1. Area = ∫₀¹(x − x²) dx = x²/2 − x³/3 = 1/2 − 1/3 = 1/6.
FRQ2.
The region in the first quadrant under y = √x from x = 0 to x = 4 is rotated about the x-axis. Find the volume using disks.
6 points · rubric: Correct radius 2 pts, correct integral 2 pts, correct evaluation 2 pts.
Model answer
V = π∫₀⁴ (√x)² dx = π∫₀⁴ x dx = π·x²/2 = π·16/2 = 8π.
Printable cheat sheetone pager
Limits
lim(x→a) exists iff both one-sided limits exist and agree.
0/0 indeterminate → factor/cancel, rationalize, or combine.
lim(x→0) sin x/x = 1; lim(x→0) (1−cos x)/x = 0.
Continuity: f(a) defined, limit exists, limit = f(a).
IVT: continuous on [a,b] takes every value between f(a) and f(b).
Differentiation rules
Power: d(xⁿ) = nxⁿ⁻¹.
Product: (fg)' = f'g + fg'.
Quotient: (f/g)' = (f'g − fg')/g².
Chain: f(g(x))' = f'(g(x))·g'(x).
Inverse: (f⁻¹)'(x) = 1/f'(f⁻¹(x)).
Curve behavior
f' > 0 increasing; f' < 0 decreasing; f' = 0 flat.
f'' > 0 concave up (min); f'' < 0 concave down (max).
Inflection: f'' changes sign.
Absolute extrema: check critical points + endpoints.
MVT: some c where f'(c) = (f(b)−f(a))/(b−a).
Integration
∫xⁿ dx = xⁿ⁺¹/(n+1) + C.
∫eˣ dx = eˣ + C; ∫1/x dx = ln|x| + C; ∫cos x dx = sin x + C.
FTC: ∫ₐᵇ f(x) dx = F(b) − F(a).
u-substitution reverses the chain rule.
Total distance = ∫|v| dt; displacement = ∫v dt.
Average value = (1/(b−a))∫ₐᵇ f(x) dx.
Differential equations
Separable: dy/dx = g(x)·h(y) → separate, integrate, use initial condition.
Exponential: dy/dt = ky → y = y₀e^(kt).
Logistic: dy/dt = ky(1 − y/L) → levels at L.
Slope fields: follow tangent segments for qualitative solutions.
Volumes & motion
Disk: V = π∫r² dx. Washer: V = π∫(R² − r²) dx.
Cross sections: V = ∫A(x) dx.
a → v → s: integrate; s → v → a: differentiate.
Area between curves: ∫(top − bottom) dx.