Unit 6 · Topic 6.3

6.3 Riemann Sums, Summation Notation, and Definite Integral Notation

A Riemann sum can be written in sigma (Σ) notation, and as the rectangles get thinner and thinner, the sum approaches an exact value called the definite integral. You need to translate in both directions: from a limit of a sum to an integral, and from an integral to a limit of a sum.

Key terms

summation notation (Σ)
definite integral
limit of a Riemann sum
Δx
limits of integration

Sigma notation

Σ (the Greek capital letter sigma) means “add up.” The expression Σ (k = 1 to n) f(xₖ)Δx means: plug in k = 1, 2, 3, … up to n, compute f(xₖ)Δx each time, and add the results. The letter k is the index. It just counts the rectangles.

A Riemann sum is a sum of products: a function value at a point in each subinterval times that subinterval's width. In sigma notation, a right Riemann sum on [a, b] with n equal pieces is Σ (k = 1 to n) f(a + kΔx)·Δx, where Δx = (b − a)/n. A left sum uses a + (k − 1)Δx instead of a + kΔx.

From the sum to the definite integral

If f is continuous on [a, b], Riemann sums get closer and closer to one number as the widths of all the subintervals shrink to 0. That number is the definite integral, written ∫ₐᵇ f(x) dx. With equal widths, this means:

∫ₐᵇ f(x) dx = lim (n → ∞) Σ (k = 1 to n) f(xₖ)·Δx, where Δx = (b − a)/n and xₖ is a point in the kth subinterval.

It doesn't matter whether you use left endpoints, right endpoints or any other point in each piece: in the limit, they all give the same value. The parts of the notation match up: ∫ replaces lim Σ, f(x) replaces f(xₖ), and dx replaces Δx. The numbers a and b are the limits of integration, and f(x) is the integrand.

Translating a limit of a sum into an integral

Exam questions often show something like lim (n → ∞) Σ (k = 1 to n) (3/n)·√(1 + 3k/n) and ask which integral it equals. Work in this order:

Find Δx: it's the factor that looks like (constant)/n. Here Δx = 3/n, so b − a = 3.
Find xₖ: it's what's inside the function, usually of the form a + kΔx. Here xₖ = 1 + 3k/n, so a = 1.
Then b = a + (b − a) = 1 + 3 = 4, and f(x) = √x.
Result: ∫₁⁴ √x dx. Other answers are also correct if they match the same sum, like ∫₀³ √(1 + x) dx, where you let xₖ = 3k/n and f(x) = √(1 + x).

Why it matters

This definition is the reason integrals measure accumulation. Each term f(xₖ)Δx is a small piece, rate × short time or height × thin width, and the integral adds infinitely many of them. When you set up area, volume or distance problems in Unit 8, you're really building a Riemann sum and letting the slices get thin.

You'll rarely have to evaluate a limit of a sum directly. Once you've translated it to an integral, you compute the integral with the Fundamental Theorem of Calculus (6.7).

Worked examples

Try each one yourself first, then open the solution.

Example 1
Limit of a sum → definite integral
Write lim (n → ∞) Σ (k = 1 to n) (3/n)·√(1 + 3k/n) as a definite integral, and evaluate it.
Show the solutionHide the solution
1. Step 1: Δx = 3/n, so the interval has length 3.
2. Step 2: xₖ = 1 + 3k/n = 1 + kΔx, so the interval starts at a = 1 and ends at b = 1 + 3 = 4.
3. Step 3: The function applied to xₖ is the square root, so f(x) = √x.
4. Step 4: The sum is a right Riemann sum for ∫₁⁴ √x dx.
5. Step 5: Evaluate (using 6.7): an antiderivative of x^(1/2) is (2/3)x^(3/2). (2/3)(4^(3/2) − 1^(3/2)) = (2/3)(8 − 1) = 14/3.
Answer: ∫₁⁴ √x dx = 14/3
Example 2
Definite integral → limit of a right Riemann sum
Write ∫₂⁵ x³ dx as the limit of a right Riemann sum.
Show the solutionHide the solution
1. Step 1: Δx = (5 − 2)/n = 3/n.
2. Step 2: Right endpoints: xₖ = 2 + kΔx = 2 + 3k/n, for k = 1 to n.
3. Step 3: Each term is f(xₖ)Δx = (2 + 3k/n)³ · (3/n).
Answer: ∫₂⁵ x³ dx = lim (n → ∞) Σ (k = 1 to n) (2 + 3k/n)³ · (3/n)
Example 3
Trap: read the interval from Δx, not from a guess
Which definite integral equals lim (n → ∞) Σ (k = 1 to n) (2/n)(2k/n)²?
Show the solutionHide the solution
1. Step 1: Tempting wrong answer: ∫₀¹ x² dx, because k/n runs from 0 to 1.
2. Step 2: But Δx = 2/n, so the interval has length 2, and xₖ = 2k/n = 0 + kΔx starts at 0.
3. Step 3: So the interval is [0, 2], f(x) = x², and the sum approaches ∫₀² x² dx = 8/3.
4. Step 4: Check: ∫₀¹ x² dx = 1/3, which doesn't match, because the factor 2/n isn't 1/n.
Answer: ∫₀² x² dx (which equals 8/3)

Common mistakes

Mixing up the upper limit with the interval's length. If xₖ = 1 + 3k/n, then b − a = 3 and b = 4, not 3.
Leaving out the Δx factor when you write a sum from an integral.
Using k = 0 to n − 1 and calling it a right sum. Right sums run k = 1 to n with xₖ = a + kΔx; left sums run k = 0 to n − 1 (or use a + (k − 1)Δx with k = 1 to n).

On the exam

Multiple-choice questions often show a limit of a sum and four integrals. Match Δx to the interval length and the inside of the function to xₖ, then check one answer quickly.
Several integrals can match the same sum (shifted versions). If your answer isn't listed, try shifting the variable.

Connected topics

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