AP Calculus BC, AP exams, 17–20%

Unit 6: Integration and Accumulation of Change

Unit 6 carries the largest published weighting among the shared units. Riemann sums lead to the definite integral, the Fundamental Theorem of Calculus links integration and differentiation, and BC adds integration by parts, partial fractions, and improper integrals to the technique set.

Position: 6 of 10 units
Published weighting: 17–20% in the official framework
Topics covered: 6 topics in this guide

What is covered

Topics inside this unit.

These topics describe the scope students most often review for this part of AP Calculus BC. Treat them as a study checklist rather than an exhaustive list of what a single exam form can sample.

Riemann sums and definite integrals: Left, right, midpoint, and trapezoidal approximations and the limit defining the integral.
Fundamental Theorem of Calculus: Evaluating definite integrals and differentiating accumulation functions.
Substitution: Changing variables for composite integrands, including adjusting limits of integration.
Integration by parts: Choosing the parts sensibly, and handling repeated applications and cyclic cases.
Partial fractions: Decomposing rational integrands with distinct linear factors before integrating.
Improper integrals: Rewriting an infinite limit or an unbounded integrand as a limit and testing for convergence.

Study tips

How to practice Integration.

Section-specific guidance that follows from how this material is usually tested.

Build a decision routine for choosing a technique: try substitution, then parts, then partial fractions.
Always convert an improper integral to a limit explicitly; the notation itself earns credit.
Give this unit heavy practice; it carries the largest weighting of the eight shared units.

Where this fits

Unit 6: Integration and Accumulation of Change is 6 of 10 units in the AP Calculus BC guide, and it carries a published weighting of 17–20% in the official framework. Use the full guide to balance it against everything else you have to review.

Back to the AP Calculus BC guide Official exam information

Sample Integration questions

Practice questions, not yet expert-reviewed

Original practice questions written to the published framework for this section. No expert has reviewed them, and they are not taken from any official exam. Try each one before you open the answer.

Question 1Foundational
What is the value of ∫ from 1 to e of (1/x) dx?
1. A1
2. Be
3. Ce - 1
4. D0
Show answer
Answer: A. 1
An antiderivative of 1/x is ln(x), so the integral equals ln(e) - ln(1) = 1 - 0 = 1. The choice e - 1 is the value of ∫ from 0 to 1 of e^x dx, a different integral entirely.
Key insight: ln(e) = 1 and ln(1) = 0, which makes this the cleanest definite integral of 1/x.
Question 2Foundational
What is ∫ e^(2x) dx?
1. Ae^(2x)/2 + C
2. B2e^(2x) + C
3. Ce^(2x) + C
4. De^(x^2)/2 + C
Show answer
Answer: A. e^(2x)/2 + C
Substituting u = 2x gives du = 2 dx, so the integral is (1/2)e^u + C = e^(2x)/2 + C. Differentiating the answer confirms it returns e^(2x). Multiplying by 2 instead of dividing reverses the chain rule adjustment.
Key insight: Antidifferentiating e^(kx) divides by k, while differentiating it multiplies by k.
Question 3Intermediate
What is ∫ x e^x dx?
1. Ax e^x - e^x + C
2. Bx e^x + e^x + C
3. C(x^2/2) e^x + C
4. Dx e^x + C
Show answer
Answer: A. x e^x - e^x + C
Integration by parts with u = x and dv = e^x dx gives x e^x - ∫ e^x dx = x e^x - e^x + C. Differentiating the result returns e^x + x e^x - e^x = x e^x, confirming the sign. Adding rather than subtracting the second term ignores the minus sign in the parts formula.
Key insight: The parts formula subtracts the remaining integral: ∫ u dv = uv - ∫ v du.
Question 4Intermediate
What is the value of ∫ from 2 to ∞ of (1/x^2) dx?
1. A1/2
2. B0
3. C2
4. DThe integral diverges
Show answer
Answer: A. 1/2
Evaluate the improper integral as a limit: ∫ from 2 to b of x^(-2) dx = 1/2 - 1/b, which approaches 1/2 as b → ∞. The comparable integral of 1/x diverges, which is why students often mark this one as divergent too.
Key insight: ∫ from a to ∞ of x^(-p) dx converges exactly when p > 1, and 1/x^2 clears that bar.
Question 5Advanced
What is the value of ∫ from 1 to 2 of 1/(x(x + 1)) dx?
1. Aln(4/3)
2. Bln(2/3)
3. Cln(3/2)
4. Dln(2)
Show answer
Answer: A. ln(4/3)
Partial fractions give 1/(x(x + 1)) = 1/x - 1/(x + 1), whose antiderivative is ln|x| - ln|x + 1| = ln(x/(x + 1)). Evaluating from 1 to 2 gives ln(2/3) - ln(1/2) = ln(4/3). Stopping at the upper-limit value alone produces the negative ln(2/3).
Key insight: Decompose the rational integrand first; each simple factor integrates to a logarithm.

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Unit 6: Integration and Accumulation of Change