Twelve lessons covering the ten BC units: limits, derivative applications, integration techniques, improper integrals, differential equations, parametric, polar and vector motion, and the full series toolkit.
A reasoning guide to AP Calculus BC, not a full course. Unit titles and weightings follow the College Board course page; all numbers in examples are invented for practice.
AP Calculus AB level algebra and trigonometry. Calculus AB is helpful but BC repeats the core.
Course outline
Limits and continuity: what the function approaches, not where it sits
Evaluate limits algebraically and graphically, and justify continuity and the Intermediate Value Theorem.
Derivative rules and the chain rule, implicit and inverse derivatives
Differentiate products, quotients, composites, implicit relations and inverse functions.
Applications of derivatives: rates, shape, optimization and L'Hopital
Use derivatives for related rates, extrema, the Mean Value Theorem and indeterminate limits.
Integration and accumulation: the Fundamental Theorem
Connect definite integrals, Riemann sums and antiderivatives, and apply both parts of the Fundamental Theorem.
Integration techniques: substitution, parts and partial fractions
Choose among u-substitution, integration by parts and partial fractions and justify each choice.
Improper integrals: infinite intervals and unbounded integrands
Evaluate improper integrals as limits and decide convergence for p-integrals.
Differential equations: slope fields, separation, Euler and logistic growth
Read every lesson below. The interactive reader above contains the same explanations, with visual tools and quizzes.
1. Limits and continuity: what the function approaches, not where it sits
Learning goal: Evaluate limits algebraically and graphically, and justify continuity and the Intermediate Value Theorem.
A limit describes the value a function approaches as x gets close to a point, not the value at the point. The limit exists only when the left-hand and right-hand limits agree. A hole, a jump and a vertical asymptote are three different failures, and each one shows up differently in a table or graph.
A function is continuous at x = c when three things hold: f(c) is defined, the limit as x approaches c exists, and the two are equal. On the free-response section you must state all three, not just say the graph has no break. A removable discontinuity can be repaired by redefining the single value f(c).
The Intermediate Value Theorem needs continuity on a closed interval [a, b]. If f is continuous there and k lies between f(a) and f(b), some c in [a, b] has f(c) = k. Without continuity the conclusion can fail. The Squeeze Theorem handles oscillating limits such as x sin(1/x): trap the function between -|x| and |x| and both bounds go to 0.
Limits at infinity compare growth rates. For a rational function, compare the highest powers of the numerator and denominator. Equal degrees give the ratio of leading coefficients, a smaller numerator degree gives 0, and a larger one means the limit does not exist as a finite number.
Worked example
Find the limit as x approaches 0 of sin(3x)/x (invented practice).
Rewrite as 3 times sin(3x)/(3x).
As x goes to 0, sin(3x)/(3x) goes to 1.
The limit is 3 times 1.
Answer: 3.
Practice problem and solution
Find the limit as x approaches 2 of (x^2 - 4)/(x - 2).
Factor: (x-2)(x+2)/(x-2) = x + 2, which is 4 at x = 2.
Mental model: Check three conditions. Name the discontinuity by its limits. The IVT needs continuity on a closed interval.
Common trap: Plugging in the point and calling 0/0 'undefined' as if that meant no limit.
2. Derivative rules and the chain rule, implicit and inverse derivatives
The derivative is the instantaneous rate of change, defined as a limit of difference quotients. Rules save time but they are all consequences of that definition. Product rule: (fg)' = f'g + fg'. Quotient rule: (f/g)' = (f'g - fg')/g^2. Learn the structure, not just the letters.
The chain rule differentiates a composite from the outside in: (f(g(x)))' = f'(g(x)) times g'(x). Most lost points are a missing inner derivative. Trigonometric, exponential and logarithmic derivatives all need it: d/dx sin(x^2) = 2x cos(x^2).
Implicit differentiation treats y as a function of x, so every y-term picks up a dy/dx factor. For x^2 + y^2 = 25, differentiate to get 2x + 2y dy/dx = 0, so dy/dx = -x/y. At the point (3, 4) the slope is -3/4.
For an invertible differentiable function, (f^-1)'(b) = 1 / f'(a) where f(a) = b. The slopes of a function and its inverse are reciprocals at corresponding points. This gives d/dx arcsin x = 1/sqrt(1 - x^2) and d/dx arctan x = 1/(1 + x^2).
Worked example
Find dy/dx for x^2 + y^2 = 25 at (3, 4) (invented practice).
Differentiate: 2x + 2y dy/dx = 0.
Solve: dy/dx = -x/y.
Substitute x = 3, y = 4.
Slope is -3/4.
Practice problem and solution
Let f(x) = x^3 + x, so f(1) = 2. Find (f^-1)'(2) as a decimal.
f'(x) = 3x^2 + 1, so f'(1) = 4 and the inverse derivative is 1/4.
Mental model: Differentiate outside in. Put dy/dx on every y. Inverse slopes are reciprocals at matching points.
Common trap: Evaluating the inverse derivative at 2 instead of at f's matching input.
3. Applications of derivatives: rates, shape, optimization and L'Hopital
Learning goal: Use derivatives for related rates, extrema, the Mean Value Theorem and indeterminate limits.
In a related rates problem two quantities are linked by an equation and both change with time. Write the equation, differentiate with respect to t, and substitute the instant values only after differentiating. A ladder of length 10 against a wall obeys x^2 + y^2 = 100, so x dx/dt + y dy/dt = 0.
The Mean Value Theorem says that for f continuous on [a, b] and differentiable on (a, b), some c has f'(c) = (f(b) - f(a))/(b - a). Rolle's Theorem is the case where f(a) = f(b). The Extreme Value Theorem guarantees a maximum and minimum on a closed interval when f is continuous there.
To find extrema, check critical points where f' = 0 or is undefined, and the endpoints. The First Derivative Test reads the sign change of f'; the Second Derivative Test uses the sign of f'' at a critical point. For a justification, write the sign chart and the conclusion in words.
L'Hopital's Rule applies to limits of the form 0/0 or infinity/infinity. If the limit of f'/g' exists, it equals the limit of f/g. For (e^x - 1 - x)/x^2 as x goes to 0, the first pass gives (e^x - 1)/(2x), still 0/0, and a second pass gives e^x/2, which is 1/2. Always confirm the form before using the rule.
Worked example
A 10 ft ladder slides. The base is 6 ft from the wall and moving away at 2 ft/s (invented). How fast is the top moving?
y = 8 when x = 6.
x dx/dt + y dy/dt = 0.
6(2) + 8 dy/dt = 0.
dy/dt = -1.5 ft/s, so the top moves down.
Practice problem and solution
Find the limit as x approaches 0 of (e^x - 1 - x)/x^2 as a decimal.
Apply L'Hopital twice to get e^x/2, which equals 1/2 at 0.
Mental model: Differentiate before substituting. Check endpoints. Verify the form before L'Hopital.
Common trap: Plugging instant values into the equation before differentiating.
4. Integration and accumulation: the Fundamental Theorem
Learning goal: Connect definite integrals, Riemann sums and antiderivatives, and apply both parts of the Fundamental Theorem.
A definite integral is the limit of Riemann sums and measures net accumulated change. Left, right and midpoint sums are approximations, and a trapezoidal sum averages left and right. Areas above the axis count positive and areas below count negative, so the integral is net signed area.
The Fundamental Theorem of Calculus has two parts. Part 1: if F(x) is the integral from a to x of f(t) dt, then F'(x) = f(x). With a variable upper limit g(x), use the chain rule: d/dx of the integral from 0 to x^2 of sin t dt equals 2x sin(x^2).
Part 2: the integral from a to b of f(x) dx equals F(b) - F(a), where F is any antiderivative. For 3x^2, the antiderivative is x^3, so the integral from 0 to 2 is 8 - 0 = 8. Do not forget constants in indefinite integrals.
In context, the integral of a rate gives total change: the integral of r(t) from a to b equals the change in the quantity over [a, b]. To find a final amount, add the initial amount. Average value of f on [a, b] is the integral divided by (b - a). Units of the integral are rate units times time units.
Worked example
Find d/dx of the integral from 0 to x^2 of sin t dt at x = 1 (invented).
Apply FTC part 1 with the chain rule.
Derivative is sin(x^2) times 2x.
At x = 1: 2 sin(1).
About 1.683.
Practice problem and solution
Evaluate the integral from 0 to 2 of 3x^2 dx.
The antiderivative is x^3, so 8 - 0 = 8.
Mental model: Part 1 differentiates accumulation; part 2 evaluates. Add the initial amount for totals.
Common trap: Reporting net change as if it were the final quantity.
5. Integration techniques: substitution, parts and partial fractions
Learning goal: Choose among u-substitution, integration by parts and partial fractions and justify each choice.
Substitution reverses the chain rule. If the integrand contains a function and (a constant multiple of) its derivative, let u be the inner function. For the integral of 2x cos(x^2) dx, set u = x^2, so du = 2x dx, giving sin(x^2) + C. For definite integrals, change the limits to u-values or switch back.
Integration by parts reverses the product rule: the integral of u dv equals uv minus the integral of v du. Pick u to simplify when differentiated, and dv to be easy to integrate. For x e^x, take u = x and dv = e^x dx to get x e^x - e^x + C.
Partial fractions split a rational function into simpler terms. For 1/(x(x+1)), write A/x + B/(x+1); solving gives A = 1 and B = -1, so the integral is ln|x| - ln|x+1| + C. The BC exam uses linear factors in the denominator, so the method is limited to that case.
Choose a technique by looking at the integrand. A composite with its derivative suggests substitution, a product of unrelated types suggests parts, and a rational function with factorable denominator suggests partial fractions. Check every antiderivative by differentiating it.
Worked example
Evaluate the integral from 0 to 1 of x e^x dx (invented practice).
Take u = x, dv = e^x dx.
Parts gives x e^x - e^x.
Evaluate from 0 to 1: (e - e) - (0 - 1).
Answer: 1.
Practice problem and solution
Evaluate the integral from 0 to 1 of x e^x dx as an integer.
By parts, x e^x - e^x is 0 - (-1) = 1 over [0, 1].
Mental model: Match the method to the structure. Update limits with u. Check by differentiating.
Common trap: Forgetting to change the limits after a substitution.
6. Improper integrals: infinite intervals and unbounded integrands
Learning goal: Evaluate improper integrals as limits and decide convergence for p-integrals.
An improper integral has an infinite limit of integration or an integrand that blows up at an endpoint. It is defined as a limit. The integral from 1 to infinity of f(x) dx means the limit as b goes to infinity of the integral from 1 to b. If the limit is finite, it converges; otherwise it diverges.
For 1/x^2 from 1 to infinity, the antiderivative is -1/x. Evaluate at b to get 1 - 1/b, which goes to 1. The area is finite. For 1/x, the antiderivative ln x grows without bound, so the integral diverges even though the integrand goes to 0.
The p-integral rule: the integral from 1 to infinity of 1/x^p converges exactly when p > 1. The integral from 0 to 1 of 1/x^p converges exactly when p < 1, and 1/sqrt(x) has p = 1/2, so the integral from 0 to 1 of x^(-1/2) dx equals 2.
The comparison test uses a known integral: if 0 <= f <= g and the integral of g converges, then the integral of f converges; if f >= g and g diverges, f diverges. Always write the limit explicitly on the exam; naming the improper integral without a limit loses the argument.
Worked example
Evaluate the integral from 1 to infinity of 1/x^2 dx (invented practice).
Integrate on [1, b]: 1 - 1/b.
Take b to infinity.
The limit is 1.
The integral converges to 1.
Practice problem and solution
Evaluate the integral from 1 to infinity of x^(-3) dx as a decimal.
The antiderivative is -1/(2x^2), so the value is 0 - (-1/2) = 1/2.
Mental model: Write the limit. Memorize the p-integral rule both ways. Comparison needs the right inequality.
Common trap: Concluding convergence just because the integrand goes to zero.
7. Differential equations: slope fields, separation, Euler and logistic growth
A differential equation relates a function to its derivative. A slope field draws the slope dy/dx at sample points. A solution curve follows the segments. Check for horizontal segments where dy/dx = 0, and for fields that depend only on x, only on y, or both.
To solve a separable equation, move all y terms to one side and all x terms to the other, integrate both sides and add one constant. For dy/dx = ky with y(0) = 5, you get ln|y| = kx + C, so y = 5e^(kx). Use the initial condition to find the constant, and state the domain of the solution.
Euler's method estimates a solution with tangent steps: y_new = y_old + h times f(x_old, y_old). For dy/dx = x + y with y(0) = 1 and h = 0.5, the first step gives 1.5 at x = 0.5, and the second gives 1.5 + 0.5(0.5 + 1.5) = 2.5 at x = 1. Euler estimates are underestimates when the solution is concave up, because the tangent lies below the curve.
The logistic model dP/dt = kP(1 - P/K) has equilibria at 0 and K. Growth is fastest at P = K/2, where dP/dt is largest, and the population approaches K when it starts above 0. The BC exam also asks you to read these facts from the equation and from a graph.
Worked example
Use Euler with h = 0.5 for dy/dx = x + y, y(0) = 1, to estimate y(1) (invented).
Step 1: slope is 1, y = 1 + 0.5(1) = 1.5.
x = 0.5.
Step 2: slope is 2, y = 1.5 + 0.5(2) = 2.5.
Estimate: 2.5.
Practice problem and solution
For dy/dx = y with y(0) = 2 and h = 0.5, Euler's method after two steps gives y(1) = ?
y1 = 2 + 0.5(2) = 3, y2 = 3 + 0.5(3) = 4.5.
Mental model: Separate, integrate, use the initial condition. Euler is tangent steps. Logistic peaks at K/2.
Common trap: Adding a constant on both sides or forgetting to apply the initial condition.
8. Applications of integration: area, volume, average value and arc length
Learning goal: Set up integrals for area between curves, solids of revolution, average value and arc length.
Area between curves is the integral of top minus bottom over the interval. For y = x and y = x^2 on [0, 1], the line is on top, so the area is the integral of x - x^2, which is 1/2 - 1/3 = 1/6. If the curves cross, split the integral at the crossing.
For a solid made by revolving a region around the x-axis, disks use pi times radius squared, and washers subtract the inner radius squared: pi(R^2 - r^2) dx. Revolving y = sqrt(x) on [0, 4] gives pi times the integral of x dx, which is 8 pi. Axes other than the coordinate axes shift the radius.
Volumes can also come from known cross sections: the area formula for each slice (square, semicircle, triangle) is integrated over the interval. Say clearly which variable the slices are perpendicular to.
The average value of f on [a, b] is (1/(b - a)) times the integral. Arc length for a smooth curve y = f(x) from a to b is the integral of sqrt(1 + (f'(x))^2) dx. BC students also meet arc length for parametric and polar curves, covered in the next lessons.
Worked example
Find the area between y = x and y = x^2 from 0 to 1 (invented practice).
Line is on top for 0 < x < 1.
Integrate x - x^2.
1/2 - 1/3.
Area is 1/6.
Practice problem and solution
Revolve y = sqrt(x) on [0, 4] about the x-axis. The volume is k pi. Find k.
pi times the integral of x dx on [0, 4] is pi times 8.
Mental model: Sketch first. Disk radius is distance to the axis. Average value divides by the interval length.
Common trap: Using the top minus bottom in the wrong order and getting negative area.
9. Parametric equations and polar coordinates
Learning goal: Differentiate parametric curves, find arc length, and compute area and slope in polar form.
A parametric curve gives x(t) and y(t). The slope is dy/dx = (dy/dt)/(dx/dt), as long as dx/dt is not 0. For x = t^2 and y = t^3 at t = 2, dy/dt = 12 and dx/dt = 4, so the slope is 3. The second derivative is d/dt(dy/dx) divided by dx/dt, not the derivative of dy/dt alone.
Arc length of a parametric curve from t = a to t = b is the integral of sqrt((dx/dt)^2 + (dy/dt)^2) dt. This is the same integrand as speed of a particle. Distance traveled is always the integral of speed, not the integral of velocity.
Polar coordinates use r and theta with x = r cos(theta) and y = r sin(theta). The area enclosed by r = f(theta) from alpha to beta is one half the integral of r^2 dtheta. For r = 2, a circle of radius 2, the area is (1/2)(4)(2 pi) = 4 pi. Find limits from where r is 0 or where the region starts and stops.
For polar curves, convert to parametric form to get slope: x = r cos(theta), y = r sin(theta), then apply dy/dx = (dy/dtheta)/(dx/dtheta). Watch the sign of r: a negative r plots on the opposite side. Pick limits that trace the curve once.
Worked example
For x = t^2 and y = t^3, find dy/dx at t = 2 (invented practice).
dy/dt = 3t^2 = 12.
dx/dt = 2t = 4.
Divide: 12/4.
Slope is 3.
Practice problem and solution
For x = 2t + 1 and y = t^2, find dy/dx at t = 4.
dy/dt = 2t = 8, dx/dt = 2, ratio 4.
Mental model: Slope is a ratio of rates. Speed integrates to distance. Polar area has a one half.
Common trap: Using the integral of r instead of one half r squared for area.
10. Vector-valued functions and motion in the plane
Learning goal: Differentiate and integrate vector-valued functions and distinguish velocity, speed, acceleration and distance.
A vector-valued function r(t) = gives position. Velocity is r'(t) = , acceleration is r''(t), and each component is differentiated or integrated separately. The velocity vector is tangent to the path.
Speed is the magnitude of the velocity vector, sqrt((x')^2 + (y')^2). It is a nonnegative scalar and different from velocity. For r(t) = , velocity is <2t, 3t^2>, so at t = 1 the speed is sqrt(13).
Total distance traveled from t = a to b is the integral of speed. The displacement is the vector r(b) - r(a). A particle moving in a circle returns to its start and has zero displacement but positive distance.
A particle is speeding up when velocity and acceleration point the same direction. In components, check whether the dot product of velocity and acceleration is positive. Given initial position and velocity, integrate components and use the initial values to find constants.
Worked example
A particle has r(t) = <3t, 4t>. Find the distance from t = 0 to t = 2 (invented practice).
Velocity is <3, 4>.
Speed is 5.
Integrate 5 from 0 to 2.
Distance is 10.
Practice problem and solution
A particle has r(t) = <6t, 8t>. Find the distance traveled from t = 0 to t = 3.
Speed is sqrt(36 + 64) = 10, times 3 units of time.
Mental model: Velocity is a vector, speed is its magnitude. Distance integrates speed.
Common trap: Reporting displacement when the question asks for distance.
11. Series convergence tests
Learning goal: Choose and justify convergence tests, from the nth-term test to the ratio test and alternating series.
An infinite series is the limit of its partial sums. If the terms do not go to 0, the series diverges. If they do go to 0, you still need another test; the harmonic series, with terms 1/n going to 0, diverges. This nth-term test is a divergence test only.
A geometric series with first term a and ratio r converges to a/(1 - r) when |r| < 1. A p-series, the sum of 1/n^p, converges exactly when p > 1. The integral test links a series with positive decreasing terms to an improper integral and gives the same p rule.
Comparison tests compare terms with a known series; the limit comparison test uses the limit of a_n/b_n. The ratio test takes the limit of |a_(n+1)/a_n|. If it is less than 1 the series converges absolutely, greater than 1 it diverges, and equal to 1 the test is silent. For the sum of n/2^n the ratio is (n + 1)/(2n), which goes to 1/2, so it converges.
An alternating series converges if the terms decrease in absolute value to 0. The error after n terms is bounded by the first omitted term. A series converges absolutely if the sum of absolute values converges, and conditionally if it converges but not absolutely, as with the alternating harmonic series.
Worked example
Does the sum of n/2^n converge (invented practice)?
Ratio is ((n+1)/2^(n+1)) / (n/2^n).
This simplifies to (n+1)/(2n).
The limit is 1/2.
1/2 < 1, so it converges.
Practice problem and solution
Find the sum from n = 0 to infinity of 3(1/4)^n.
a = 3, r = 1/4, so the sum is 3/(1 - 1/4) = 4.
Mental model: Terms to 0 is necessary, not sufficient. Ratio test is silent at 1. Alternating error is the next term.
Common trap: Claiming the series converges because the terms go to zero.
12. Power series, Taylor series and the Lagrange error bound
Learning goal: Find radius and interval of convergence, build Taylor polynomials and bound their error.
A power series in (x - a) converges on an interval centered at a with radius R. Use the ratio test to find R, then test each endpoint separately because convergence at the ends can differ. For the sum of x^n/n, the radius is 1; at x = -1 it converges by the alternating series test, at x = 1 it diverges as the harmonic series, so the interval is [-1, 1).
A Taylor polynomial of degree n about a has coefficients f^(k)(a)/k!. Maclaurin series are centered at 0. Memorize e^x = 1 + x + x^2/2! + ..., sin x = x - x^3/3! + x^5/5! - ..., cos x = 1 - x^2/2! + ..., and 1/(1 - x) = 1 + x + x^2 + ... for |x| < 1.
You can build new series by substitution, differentiation and integration. Replace x by -x^2 in 1/(1 - x) to get a series for 1/(1 + x^2), then integrate term by term to get arctan x. Differentiation and integration keep the same radius but may change endpoints.
The Lagrange error bound says that if |f^(n+1)(t)| <= M on the interval between a and x, then the error of the degree n polynomial is at most M |x - a|^(n+1)/(n + 1)!. For e^x at x = 0.5 with degree 2, the polynomial gives 1.625 and the bound is at most 3(0.125)/6 = 0.0625 using M = 3.
Worked example
Approximate e^0.5 with a degree 2 Maclaurin polynomial and bound the error (invented practice).
Polynomial: 1 + 0.5 + 0.25/2 = 1.625.
Third derivative of e^x is e^x, at most e^0.5 < 3 on [0, 0.5].
Bound: 3(0.5^3)/3! = 0.0625.
Error is at most 0.0625.
Practice problem and solution
Find the radius of convergence of the sum of x^n/3^n.
The ratio is |x|/3, which is less than 1 for |x| < 3.
Mental model: Test endpoints separately. Know the core series. State M for error bounds.
Common trap: Skipping the endpoint tests and reporting an open interval by default.