A student computes both a left-endpoint and a right-endpoint Riemann sum for the same integral using n = 100 rectangles. She concludes that these two approximations will converge to different values as n → ∞. What is wrong with this conclusion?
AShe is correct — left and right sums always converge to different limits
BBoth sums converge to the same limit — the definite integral — regardless of which endpoint rule is used
COnly the midpoint sum converges to the definite integral; endpoint rules do not converge
DThe sums converge to different limits only for non-monotone functions
The key insight is that all valid Riemann sum choices — left endpoint, right endpoint, midpoint, or any sample point — converge to the same limit as n → ∞, which is the definite integral. The sums differ for finite n (and the direction of the error depends on monotonicity), but the limit is unique. This is what makes the definite integral well-defined.
Question 2 Multiple Choice
You approximate ∫₀² x² dx using n = 4 right-endpoint rectangles. The right endpoints are x = 0.5, 1.0, 1.5, 2.0 and Δx = 0.5. What is the Riemann sum, and is it an overestimate or underestimate?
A2.67 — an exact value since x² is a polynomial
B3.75 — an overestimate, because x² is increasing on [0, 2] and right endpoints sample the taller side of each rectangle
C2.25 — an underestimate, because right-endpoint sums always undershoot
D3.75 — but whether it over- or underestimates depends on n, not the function
f(0.5) + f(1.0) + f(1.5) + f(2.0) = 0.25 + 1.0 + 2.25 + 4.0 = 7.5; multiply by Δx = 0.5 to get 3.75. Since x² is strictly increasing on [0, 2], each rectangle's right endpoint is the maximum of f on that subinterval, so the rectangle overshoots the curve — giving an overestimate. The exact integral is 8/3 ≈ 2.67.
Question 3 True / False
For an increasing function on [a, b], the left-endpoint Riemann sum underestimates the definite integral for any finite n.
TTrue
FFalse
Answer: True
For an increasing function, f is smallest at the left endpoint of each subinterval. The left-endpoint sum uses this minimum as the rectangle height, so every rectangle falls below the curve — meaning the sum underestimates the area. Symmetrically, the right-endpoint sum overestimates. These inequalities hold for any finite n; both converge to the integral as n → ∞.
Question 4 True / False
A Riemann sum with a very large number of rectangles (say, n = 10,000) is equal to the definite integral.
TTrue
FFalse
Answer: False
A Riemann sum, no matter how large n is, is always an approximation — it equals the definite integral only in the limit as n → ∞. The definite integral is defined as this limit, not as any finite sum. For n = 10,000 the error is tiny but nonzero. This distinction is precisely the conceptual leap from finite approximation to the exact integral.
Question 5 Short Answer
What is the definite integral, precisely defined in terms of Riemann sums? Why must we take a limit rather than simply use a very large n?
Think about your answer, then reveal below.
Model answer: The definite integral ∫_a^b f(x) dx is defined as the limit of the Riemann sum Σᵢ f(xᵢ*) Δx as n → ∞ (and Δx → 0). No finite Riemann sum equals the integral exactly; the integral is the value that all valid Riemann sums approach. Taking the limit is necessary because rectangles can only approximate curved regions — no matter how thin, they have flat tops and the curve is (generally) curved. The limit captures the precise area by making the approximation error arbitrarily small.
This distinction matters because it makes the integral a precisely defined mathematical object, not just an approximation. In practice, numerical integration uses large-n Riemann sums as approximations. In theory, the integral is the limit. Understanding this duality — finite approximation vs. exact limit — is the foundational insight of integral calculus.