Difficulty: Intermediate | Prerequisites: Function evaluation, sigma notation basics, understanding of area as a geometric concept.
Before the course introduces the definite integral, it needs a way to approximate the area under a curve. Riemann sums do this by slicing the region into thin rectangles and adding up their areas. As the number of rectangles increases, the approximation improves. This section is the conceptual bridge between geometry (area) and calculus (the definite integral). Everything about integration that follows in Chapters 4 and 5 rests on this idea.
A Riemann sum estimates the area under a curve by dividing the interval into n equal subintervals, evaluating the function at a chosen sample point in each, and summing the resulting rectangle areas. Left endpoints give one estimate, right endpoints another. For an increasing function, left endpoints underestimate and right endpoints overestimate. As n grows, both estimates converge to the true area.
Riemann sum
An approximation of the area under a curve, computed as the sum of areas of rectangles. Each rectangle has width Δx and height f(xᵢ), where xᵢ is the sample point.
Partition
The division of an interval [a, b] into n subintervals of equal width Δx = (b − a)/n.
Left endpoint approximation (Lₙ)
A Riemann sum where the height of each rectangle is the function value at the left edge of the subinterval. Think of it as: each rectangle's top-left corner touches the curve.
Right endpoint approximation (Rₙ)
A Riemann sum where the height of each rectangle is the function value at the right edge of the subinterval.
Underestimate
An approximation that falls below the true area. For an increasing function on [a, b], the left endpoint sum is an underestimate because each rectangle sits below the curve.
Overestimate
An approximation that exceeds the true area. For an increasing function on [a, b], the right endpoint sum is an overestimate because each rectangle pokes above the curve.
Δx (delta x)
The width of each subinterval: Δx = (b − a)/n. As n increases, Δx shrinks and the approximation improves.
Choose the interval [a, b] and the number of rectangles n.
Compute Δx = (b − a)/n.
Identify the sample points: for left endpoints, use xᵢ = a + iΔx for i = 0, 1, ..., n−1. For right endpoints, use xᵢ = a + iΔx for i = 1, 2, ..., n.
Sum: S = Σ f(xᵢ) · Δx.
For a function that is increasing on [a, b]:
Left endpoint sum (Lₙ) is an underestimate, because each rectangle's height is the smallest value of f on that subinterval.
Right endpoint sum (Rₙ) is an overestimate, because each rectangle's height is the largest value of f on that subinterval.
For a decreasing function, the roles reverse: left endpoints overestimate, right endpoints underestimate.
This function is increasing on [2, 8] (both x² and 3x grow as x increases from 2 to 8).
Setup: Δx = (8 − 2)/n = 6/n.
With n = 10: Δx = 0.6. The left endpoints are x = 2, 2.6, 3.2, ..., 7.4. The right endpoints are x = 2.6, 3.2, ..., 8. Evaluate f at each point, multiply by 0.6, and sum.
With n = 50 and n = 100: The same process with Δx = 0.12 and Δx = 0.06 respectively. As n increases, both the underestimate and overestimate tighten around the true area.
The gap between Lₙ and Rₙ shrinks as n increases.
Both converge to the same value, which is the definite integral ∫ from a to b of f(x) dx.
For this example, the true area is ∫₂⁸ (x² + 3x) dx. Once you learn the Fundamental Theorem, you can compute it exactly. The Riemann sum approach is the motivation for that theorem.
The recitation references a Desmos activity where a parameter c controls the fraction of each subinterval at which the sample point is taken.
c = 0 corresponds to left endpoints.
c = 1 corresponds to right endpoints.
c = 1/2 corresponds to midpoints (the midpoint rule, which often gives a better estimate than either endpoint alone).
The function s(x) in the activity computes the Riemann sum for a given n and sample-point rule.
\Delta x = \frac{b - a}{n}L_n = \sum_{i=0}^{n-1} f(a + i \Delta x) \cdot \Delta xR_n = \sum_{i=1}^{n} f(a + i \Delta x) \cdot \Delta xRiemann sums are the mathematical foundation for numerical integration, which is how computers approximate integrals that have no closed-form solution. Engineers use numerical integration (trapezoidal rule, Simpson's rule, both refinements of this idea) to compute areas, volumes and accumulated quantities from measured data. If you have a table of velocities at discrete time points from a sensor, summing v(tᵢ) · Δt is a Riemann sum that estimates total distance.
Students often mix up which endpoint gives the overestimate and which gives the underestimate. The rule depends on whether the function is increasing or decreasing. Draw a quick sketch to check.
Students sometimes compute Δx as (b − a) without dividing by n. Always divide by the number of rectangles.
Forgetting that the left sum uses i = 0 to n−1 and the right sum uses i = 1 to n is a frequent indexing error.
Students assume more rectangles always means the answer is "closer to the right answer" without understanding that both Lₙ and Rₙ converge to the same limit. The over/underestimate property tells you which side of the true value each sum sits on.
⚠️ Expect problems asking you to set up and compute a Riemann sum for a given function and n, using left or right endpoints.
⚠️ Determining whether a sum is an overestimate or underestimate (and explaining why) is a common short-answer or true/false question.
⚠️ Problems may ask you to guess the true area from your Lₙ and Rₙ values and state a bound on the error: the true area lies between the underestimate and overestimate.
⚠️ Understanding the limit concept (as n → ∞ the Riemann sum becomes the definite integral) is tested conceptually, even before you learn to compute definite integrals.
Fill in the blank: For f(x) = x² + 3x on [2, 8] with n = 10, Δx = ______ . (0.6)
True or false: For an increasing function, the right endpoint sum underestimates the true area. (False, it overestimates.)
True or false: As n → ∞, both Lₙ and Rₙ converge to the definite integral. (True.)
Fill in the blank: The left endpoint sum uses sample points xᵢ = a + iΔx for i = 0, 1, ..., ______ . (n − 1)
True or false: If f is decreasing on [a, b], the left endpoint sum is an underestimate. (False, for a decreasing function, left endpoints overestimate.)
Q: Set up (but do not evaluate) the left endpoint Riemann sum for f(x) = x³ on [0, 4] with n = 4.
A: Δx = (4 − 0)/4 = 1. Left endpoints: x = 0, 1, 2, 3. L₄ = [f(0) + f(1) + f(2) + f(3)] · 1 = [0 + 1 + 8 + 27] · 1 = 36.
Q: Is L₄ from the previous question an overestimate or underestimate? Why?
A: f(x) = x³ is increasing on [0, 4], so the left endpoint sum is an underestimate. Each rectangle sits below the curve.
Q: Compute R₄ for f(x) = x³ on [0, 4].
A: Right endpoints: x = 1, 2, 3, 4. R₄ = [f(1) + f(2) + f(3) + f(4)] · 1 = [1 + 8 + 27 + 64] · 1 = 100.
Q: The true value of ∫₀⁴ x³ dx is 64. How close are L₄ and R₄?
A: L₄ = 36, R₄ = 100, true value = 64. L₄ undershoots by 28, R₄ overshoots by 36. The true value lies between the two, as expected.
Q: For f(x) = x² + 3x on [2, 8], if L₁₀₀ = 245.73 and R₁₀₀ = 246.27, estimate the true area and bound the error.
A: The true area lies between 245.73 and 246.27. One reasonable estimate is the average: about 246.00. The error is at most (246.27 − 245.73)/2 = 0.27.
Riemann sums lead directly to the definition of the definite integral (§4.2), which is the limit of the Riemann sum as n → ∞. The Fundamental Theorem of Calculus (§5.3) then connects the definite integral back to antiderivatives, closing the loop between this section and §3.9. The midpoint and trapezoidal rules (refinements of Riemann sums) appear in numerical methods courses and in applied engineering.
Riemann sum, left endpoint sum, right endpoint sum, midpoint rule, overestimate, underestimate, area under a curve, definite integral, partition, sigma notation, Δx, rectangles, numerical integration, MTH 132, Calculus I, Section 4.1