Difficulty: Intermediate | Prerequisites: Read the companion notes on Riemann Sums and the Definite Integral first. You need to be comfortable identifying a, b, Δx, and f(x) from both integral and summation notation.
These worked examples walk through the two core exercise types you will see on Riemann sum problems: spotting errors in a given Riemann sum expression, and converting a Riemann sum into the definite integral it represents. Both appear regularly on the AP exam. Working through these with full solutions is the fastest way to build fluency with the notation.
This document provides step-by-step worked solutions for converting Riemann sums to definite integrals, and for identifying errors in Riemann sum notation. Each solution shows exactly how to extract a, b, Δx, and f(x) from the summation.
In these problems, a definite integral and a Riemann sum are paired, but exactly one number in the Riemann sum is wrong and should be replaced with a 2. The task is to find the error.
Given integral: ∫ from 4 to 6 of (3x² + 5) dx
Given Riemann sum: lim (n→∞) Σ 3(4 + 2i/n)² + 5
Solution: Here a = 4, b = 6, so Δx = (6 − 4)/n = 2/n. The sample point 4 + 2i/n is correct. The integrand 3x² + 5 evaluated at x = 4 + 2i/n is correct. The error is the trailing factor: it reads 6/n, but Δx = 2/n. Replace 6 with 2.
Given integral: ∫ from 2 to 7 of (4x − 1) dx
Given Riemann sum: lim (n→∞) Σ 4(4 + 5i/n) − 1
Solution: a = 2, b = 7, Δx = 5/n. The trailing factor 5/n is correct. The coefficient of i/n inside is 5, matching Δx. But the constant inside the function argument is 4 when it should be a = 2. Replace 4 with 2.
Given integral: ∫ from 3 to 4 of (6x² + 3x) dx
Given Riemann sum: lim (n→∞) Σ 6(3 + 1i/n)³ + 3(3 + 1i/n)
Solution: a = 3, b = 4, Δx = 1/n. The trailing factor and the sample point (3 + i/n) are both correct. The integrand is 6x² + 3x, which has x². But the sum shows (3 + i/n)³, a cube instead of a square. Replace 3 (the exponent) with 2.
Given integral: ∫ from 5 to 9 of (8 − 2x) dx
Given Riemann sum: lim (n→∞) Σ 8 − 4(5 + 4i/n)
Solution: a = 5, b = 9, Δx = 4/n. The trailing factor 4/n is correct. The sample point 5 + 4i/n is correct. The integrand is 8 − 2x. When you substitute x = 5 + 4i/n, the coefficient of x should be 2, but the sum shows 4. Replace 4 (the coefficient in front of the parenthetical) with 2.
Given integral: ∫ from −1 to 1 of 3x³ dx
Given Riemann sum: lim (n→∞) Σ 3(−1 + 1i/n)³
Solution: a = −1, b = 1, Δx = (1 − (−1))/n = 2/n. The trailing factor 2/n is correct. The integrand 3x³ is correct. But the sample point reads −1 + 1i/n, which means Δx = 1/n. Since Δx should be 2/n, replace 1 (the coefficient of i/n) with 2.
For each problem, the method is the same: identify Δx (the trailing factor), extract a (the constant in the sample point), compute b = a + numerator of Δx, and read f(x) by replacing the sample point expression with x.
Riemann sum: lim (n→∞) Σ √(2(6i/n)) + 1
Solution: Δx = 6/n, so b − a = 6. The sample point is 6i/n = 0 + 6i/n, so a = 0 and b = 6. Replace 6i/n with x: f(x) = √(2x) + 1.
Answer: ∫ from 0 to 6 of (√(2x) + 1) dx
Riemann sum: lim (n→∞) Σ (−2 + 5i/n)² − 3
Solution: Δx = 5/n, so b − a = 5. The sample point is −2 + 5i/n, so a = −2 and b = 3. Replace (−2 + 5i/n) with x: f(x) = x² − 3.
Answer: ∫ from −2 to 3 of (x² − 3) dx
Riemann sum: lim (n→∞) Σ 3(1 + 5i/n) − 4
Solution: Δx = 5/n, so b − a = 5. The sample point is 1 + 5i/n, so a = 1 and b = 6. Replace (1 + 5i/n) with x: f(x) = 3x − 4.
Answer: ∫ from 1 to 6 of (3x − 4) dx
Riemann sum: lim (n→∞) Σ (−2 + 6i/n)³
Solution: Δx = 6/n, so b − a = 6. The sample point is −2 + 6i/n, so a = −2 and b = 4. Replace (−2 + 6i/n) with x: f(x) = x³.
Answer: ∫ from −2 to 4 of x³ dx
Riemann sum: lim (n→∞) Σ √((−2 + 2i/n)² + 1)
Solution: Δx = 2/n, so b − a = 2. The sample point is −2 + 2i/n, so a = −2 and b = 0. Replace (−2 + 2i/n) with x: f(x) = √(x² + 1).
Answer: ∫ from −2 to 0 of √(x² + 1) dx
Riemann sum: lim (n→∞) Σ 5(2 + 4i/n) + 7
Solution: Δx = 4/n, so b − a = 4. The sample point is 2 + 4i/n, so a = 2 and b = 6. Replace (2 + 4i/n) with x: f(x) = 5x + 7.
Answer: ∫ from 2 to 6 of (5x + 7) dx
Riemann sum: lim (n→∞) Σ 6(4i/n)² − 2
Solution: Δx = 4/n, so b − a = 4. The sample point is 4i/n = 0 + 4i/n, so a = 0 and b = 4. Replace (4i/n) with x: f(x) = 6x² − 2.
Answer: ∫ from 0 to 4 of (6x² − 2) dx
Riemann sum: lim (n→∞) Σ (1 + 2i/n)³ − 1
Solution: Δx = 2/n, so b − a = 2. The sample point is 1 + 2i/n, so a = 1 and b = 3. Replace (1 + 2i/n) with x: f(x) = x³ − 1.
Answer: ∫ from 1 to 3 of (x³ − 1) dx
When the sample point looks like 6i/n with no constant added, students sometimes panic. The constant is 0, so a = 0. There is nothing wrong with having a = 0.
Students sometimes match Δx to the wrong part of the expression. The trailing factor (the one multiplied outside the brackets at the very end) is always Δx. The coefficient of i/n inside the function argument should be the same value, but it is part of the sample point, not a separate quantity.
On error-spotting problems, students tend to check only one component and stop. Always verify all four pieces: a, b, Δx, and the integrand f(x), including exponents and coefficients.
⚠️ Converting Riemann sums to integrals is a recurring multiple-choice question on both AB and BC exams.
⚠️ Error-spotting exercises train the same skill tested by "which of the following Riemann sums represents this integral?" questions.
⚠️ On the free-response section, you may be asked to set up (not evaluate) a Riemann sum for a given context. Practising these conversions in both directions builds that setup skill.
Q: Write the definite integral for lim (n→∞) Σ (3 + 7i/n)² + 2(3 + 7i/n).
A: Δx = 7/n, a = 3, b = 10. f(x) = x² + 2x. The integral is ∫ from 3 to 10 of (x² + 2x) dx.
Q: The Riemann sum for ∫ from 0 to 5 of (x² + 1) dx is written as lim (n→∞) Σ (5i/n)² + 1. Find the error.
A: Δx should be 5/n (since b − a = 5 − 0 = 5), but the trailing factor reads 3/n. Replace 3 with 5.
Q: Write the Riemann sum for ∫ from −3 to 1 of (2x + 4) dx.
A: a = −3, b = 1, Δx = 4/n. Sample point: −3 + 4i/n. The sum is lim (n→∞) Σ 2(−3 + 4i/n) + 4.
Q: A Riemann sum reads lim (n→∞) Σ √(9 − (1 + 3i/n)²). What is the definite integral?
A: Δx = 3/n, a = 1, b = 4. f(x) = √(9 − x²). The integral is ∫ from 1 to 4 of √(9 − x²) dx. (This represents part of a semicircle of radius 3.)
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