Is the limit of a Riemann sum always equal to the definite integral?

Yes, for a continuous function on a closed interval [a,b], the limit of any Riemann sum (left, right, midpoint, or any choice of sample point) as the mesh of the partition goes to zero equals the definite integral. This is the definition from Stewart Calculus section 5.2.

What if the sum uses a non-uniform partition?

The same definition holds: the definite integral is the limit of Riemann sums as the mesh (the widest subinterval width) goes to zero. Non-uniform partitions appear rarely on AP exams but are common in college-level proofs. The AP Calculus CED Topic 6.3 uses uniform partitions for simplicity.

How do I know whether a sum uses left or right endpoints?

Examine the sample point xᵢ*. If it equals a + (i-1)Δx, it is left; if it equals a + iΔx, it is right. For the limit, the distinction disappears because both converge to the same integral. In conversion problems, the problem will specify the endpoint choice.

Can the limit of a Riemann sum give a negative answer?

Yes, because the definite integral is signed area. If the function is negative on the interval, the terms f(xᵢ*) are negative, and the limit is negative. Forgetting the sign convention is a common failure mode: a student computes a positive sum for a function below the x-axis.

What is the most common error when converting a sum to an integral?

The most common error is misidentifying the interval [a,b] by reading the coefficient of i/n incorrectly. If the sum has a term like (2 + 5i/n), the interval is [2,7] only if Δx = 5/n. If Δx = 3/n, the interval is [2,5], not [2,7]. Always compare the step size to Δx.

Do I need to know the summation formulas for Σi² and Σi³?

Yes, for evaluating the limit algebraically. The AP Calculus BC exam and first-year college calculus problems expect you to use Σi = n(n+1)/2 and Σi² = n(n+1)(2n+1)/6. The formula for Σi³ is [n(n+1)/2]², which appears in problems with cubic integrands.

Why does the limit definition seem harder than using antiderivatives?

The limit definition is the theoretical foundation, not a practical method for most integrals. Antiderivatives via the Fundamental Theorem of Calculus are faster. The limit definition is used to prove the theorem and to evaluate integrals that have no simple antiderivative, like ∫₀¹ e^{x²} dx, where the limit is the only exact expression.

Limit of a Riemann Sum as a Definite Integral

Converting a Limit of a Riemann Sum to an Integral

You see a limit of a Riemann sum on an exam. Write it as a definite integral. The direct conversion rule is: ∫ₐᵇ f(x) dx = lim_{n→∞} Σᵢ₌₁ⁿ f(xᵢ*) Δx. This formula is the bridge between a finite approximation and the exact signed area under a curve. The limit of a Riemann sum becomes the definite integral when the mesh of the partition goes to zero. Stewart Calculus section 5.2 and AP Calculus CED Topic 6.3 anchor this definition. Apply it by identifying three pieces: the width Δx, the sample point xᵢ*, and the function f.

Reading Δx, a, and b Off a Given Sum

Every uniform partition Riemann sum in standard form uses Δx = (b - a)/n. The lower limit a is the starting x-value of the interval. b is the ending x-value. See a sum like lim_{n→∞} Σᵢ₌₁ⁿ (2 + 3i/n)² * (3/n). The term (3/n) is Δx. Since Δx = (b - a)/n, and the coefficient of i/n inside the function is the step size from a, solve for a and b. Δx = 3/n implies b - a = 3. The term (2 + 3i/n) shows xᵢ = a + iΔx = a + 3i/n, so a = 2 and b = 5. This is the reverse of building a Riemann sum: read the structure from the summand back to the integral limits.

Worked Examples: Sum to Integral

Example 1: Linear Function

Convert lim_{n→∞} Σᵢ₌₁ⁿ (1 + 2i/n) * (2/n) to an integral. Δx = 2/n, so b - a = 2. The term xᵢ = 1 + 2i/n gives a = 1 and b = 3. The function f(x) = x. The integral is ∫₁³ x dx.

Example 2: Quadratic with Right Endpoints

Convert lim_{n→∞} Σᵢ₌₁ⁿ ( (i/n)² + 1 ) * (1/n). Δx = 1/n gives b - a = 1. The right endpoint xᵢ = i/n implies a = 0, b = 1. Function f(x) = x² + 1. The integral is ∫₀¹ (x² + 1) dx. A common trap: the term inside the sum is the function value, not a sample point. Here the sample point is clearly xᵢ = i/n.

Example 3: Non-Standard Index

Convert lim_{n→∞} Σᵢ₌₁ⁿ (3 + 4(i-1)/n)³ * (4/n). The width Δx = 4/n, so b - a = 4. The sample point is xᵢ = 3 + 4(i-1)/n, the left endpoint of each subinterval. As n→∞, the difference between left and right endpoints vanishes in the limit. The integral is ∫₃⁷ x³ dx.

Example 4: Trigonometric Function

Convert lim_{n→∞} Σᵢ₌₁ⁿ sin(π/2 + iπ/(2n)) * (π/(2n)). Δx = π/(2n) gives b - a = π/2. The sample point xᵢ = π/2 + iπ/(2n) gives a = π/2, b = π. The integral is ∫_{π/2}^{π} sin(x) dx.

Worked Example: Integral to Sum

Convert ∫₁⁴ (3x - 2) dx into a limit of a Riemann sum using right endpoints. For a uniform partition on [1,4], Δx = (4 - 1)/n = 3/n. Right endpoints are xᵢ = 1 + 3i/n. The Riemann sum is Σᵢ₌₁ⁿ f(xᵢ) Δx = Σᵢ₌₁ⁿ (3(1 + 3i/n) - 2) * (3/n). Simplify to Σᵢ₌₁ⁿ (1 + 9i/n) * (3/n). The limit as n→∞ of this sum equals the integral. The failure mode: using the wrong sample point. Left endpoints change the sum, though the limit is the same for any choice of sample point in a continuous function.

Evaluating With the Limit Definition Using Σi and Σi² Formulas

Evaluate ∫₀¹ x² dx from the Limit

Use closed-form summation formulas. With right endpoints on [0,1], Δx = 1/n, xᵢ = i/n. The sum is Σᵢ₌₁ⁿ (i/n)² * (1/n) = (1/n³) Σᵢ₌₁ⁿ i². The formula Σᵢ₌₁ⁿ i² = n(n+1)(2n+1)/6 gives (1/n³) * n(n+1)(2n+1)/6 = (2n³ + 3n² + n) / (6n³) = 1/3 + 1/(2n) + 1/(6n²). Take the limit as n→∞ to get 1/3. This matches ∫₀¹ x² dx = 1/3.

Essential Summation Formulas

Σi = n(n+1)/2 and Σi² are necessary for polynomial integrands. For Σi³, use [n(n+1)/2]². The table below summarizes these formulas.

Summation Formulas for Riemann Sum Limits

Sum	Closed Form	Example Use
Σᵢ₌₁ⁿ i	n(n+1)/2	∫₀¹ x dx
Σᵢ₌₁ⁿ i²	n(n+1)(2n+1)/6	∫₀¹ x² dx
Σᵢ₌₁ⁿ i³	[n(n+1)/2]²	∫₀¹ x³ dx
Σᵢ₌₁ⁿ c	c ⋅ n	Constant function on any interval

Traps: Several Integrals Can Match One Sum

A single limit of a Riemann sum can correspond to multiple integrals. Consider lim_{n→∞} Σᵢ₌₁ⁿ (i/n) * (1/n). You might read this as ∫₀¹ x dx. But the same limit also equals ∫₀¹ t dt, ∫₀¹ u du, or any integral with the same area on [0,1] for the identity function. The variable of integration is a dummy variable. A more subtle trap appears when the sum uses a different partition or sample point. The sum lim_{n→∞} Σᵢ₌₁ⁿ f(2 + 3i/n) * (3/n) maps to ∫₂⁵ f(x) dx only if the sample points are the right endpoints. Using left endpoints, the sum still converges to the same integral for a continuous function, but the intermediate algebraic form looks different. The most frequent mistake on AP exams: misidentifying the function. A student sees a sum with (i/n)² and writes the integral of x², but misses that the coefficient of i/n defines the interval endpoint.