AP Calculus AB, AP exams, 17–20%

Unit 6: Integration and Accumulation of Change

Unit 6 carries the largest published weighting in the course. Riemann sums lead to the definite integral, the Fundamental Theorem of Calculus links integration to differentiation, and accumulation functions formalize the idea of total change. Antiderivative techniques at the AB level include substitution.

Position: 6 of 8 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 AB. 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 that defines the definite integral.
Accumulation functions: Interpreting an integral with a variable upper limit, and reading its behavior from the integrand's graph.
Fundamental Theorem of Calculus: Both parts of the theorem: evaluating definite integrals and differentiating an accumulation function.
Properties of definite integrals: Additivity over intervals, reversing limits, and using symmetry to simplify evaluation.
Antiderivatives and indefinite integrals: Reversing the basic derivative rules, including the transcendental functions.
Integration by substitution: Recognizing a composite integrand and changing variables, including adjusting the limits.

Study tips

How to practice Integration.

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

Practice both parts of the Fundamental Theorem separately until the second part, differentiating an accumulation function, is instant.
Do Riemann sum estimates from tables of values and state whether the estimate is an overestimate or underestimate, with a reason.
Give this unit the most practice time; its published weighting is the highest of the eight.

Where this fits

Unit 6: Integration and Accumulation of Change is 6 of 8 units in the AP Calculus AB 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 AB 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 0 to 2 of 3x^2 dx?
1. A4
2. B6
3. C8
4. D12
Show answer
Answer: C. 8
An antiderivative of 3x^2 is x^3, so the definite integral is 2^3 - 0^3 = 8. Answering 12 evaluates the integrand 3x^2 at x = 2 instead of evaluating the antiderivative.
Key insight: Antidifferentiate first, then substitute the limits; never evaluate the integrand itself.
Question 2Foundational
For x > 0, what is ∫ (1/x) dx?
1. Aln(x) + C
2. B1/(2x^2) + C
3. C-1/x^2 + C
4. Dx ln(x) + C
Show answer
Answer: A. ln(x) + C
The power rule for antiderivatives fails when the exponent is -1, and the antiderivative of 1/x is the natural logarithm. For x > 0 that is ln(x) + C. The choice -1/x^2 + C is the derivative of 1/x rather than an antiderivative.
Key insight: 1/x is the one power whose antiderivative is a logarithm rather than another power.
Question 3Intermediate
If g(x) = ∫ from 2 to x of √(t^3 + 1) dt, what is g'(x)?
1. A√(x^3 + 1)
2. B√(x^3 + 1) - 3
3. C3x^2 √(x^3 + 1)
4. D(3x^2)/(2√(x^3 + 1))
Show answer
Answer: A. √(x^3 + 1)
By the second part of the Fundamental Theorem of Calculus, differentiating an accumulation function with a variable upper limit of x simply returns the integrand evaluated at x. The constant lower limit contributes nothing, so no subtraction appears, and no chain rule factor is needed because the upper limit is x itself.
Key insight: Differentiating an accumulation function undoes the integral: g'(x) is just the integrand at x.
Question 4Intermediate
What is the value of ∫ from 0 to 1 of 2x(x^2 + 1)^3 dx?
1. A15/4
2. B15/8
3. C17/4
4. D4
Show answer
Answer: A. 15/4
Substituting u = x^2 + 1 gives du = 2x dx and changes the limits to u = 1 and u = 2, so the integral becomes ∫ from 1 to 2 of u^3 du = (2^4 - 1^4)/4 = 15/4. Dividing by an extra factor of 2 gives the tempting 15/8.
Key insight: When you substitute in a definite integral, change the limits too, or convert back before evaluating.
Question 5Advanced
For f(x) = x^2 on the interval [0, 8], what is the right Riemann sum approximation of ∫ from 0 to 8 of f(x) dx using four subintervals of equal width?
1. A112
2. B170
3. C240
4. D256
Show answer
Answer: C. 240
The width is Δx = 2 and the right endpoints are 2, 4, 6, and 8, where f takes the values 4, 16, 36, and 64. The sum is 2(4 + 16 + 36 + 64) = 240. The value 112 is the left Riemann sum, and 170 is close to the exact value 512/3, so both are overestimate or underestimate traps rather than the requested sum.
Key insight: For an increasing function a right Riemann sum overestimates the integral, and a left sum underestimates it.

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Unit 6: Integration and Accumulation of Change