Find a Formula for the Riemann Sum: Step-by-Step Guide
To find a formula for the Riemann sum, split the interval into (n) equal subintervals, write the width of each as (\Delta x = \frac{b-a}{n}), pick a sample point (x_i) in each subinterval, and add up the products (f(x_i),\Delta x). That sum is the formula, and taking the limit as (n \to \infty) turns it into the exact definite integral. This guide walks through each step with a worked example, then shows how the choices of sample point affect the result.
What a Riemann Sum Formula Actually Represents
A Riemann sum approximates the area under a curve (f(x)) on an interval ([a, b]) by stacking up thin rectangles. The height of each rectangle is a function value, and the width is the same for every piece when the subintervals are equal: (\Delta x = \frac{b-a}{n}).
The general form is compact:
[ S_n = \sum_{i=1}^{n} f(x_i),\Delta x ]
Everything hinges on that sample point (x_i). Change where you sample inside each subinterval and you change the sum — but not, importantly, the limit it approaches. All three common conventions (left, right, midpoint) converge to the same integral value as (n) grows.
Left, Right, and Midpoint Endpoints
- Left endpoint: (x_i = a + (i-1)\Delta x)
- Right endpoint: (x_i = a + i,\Delta x)
- Midpoint: (x_i = a + \left(i - \tfrac{1}{2}\right)\Delta x)
The right-endpoint version is the most common in textbooks and is the one we’ll build out below. Paul’s Online Math Notes uses this convention when it defines the definite integral as a limit of sums.
Step-by-Step: Building the Formula
Step 1 — Fix the interval and the number of pieces
Suppose we want the area under (f(x) = x^2) on ([0, 2]). Here (a = 0) and (b = 2), so:
[ \Delta x = \frac{2 - 0}{n} = \frac{2}{n} ]
Step 2 — Write a general sample point
Using right endpoints, the (i)-th point is:
[ x_i = a + i,\Delta x = 0 + i\cdot\frac{2}{n} = \frac{2i}{n} ]
Step 3 — Form each rectangle’s area
Evaluate the function at that point and multiply by the width:
[ f(x_i),\Delta x = \left(\frac{2i}{n}\right)^2 \cdot \frac{2}{n} = \frac{8i^2}{n^3} ]
Step 4 — Sum over all rectangles
[ S_n = \sum_{i=1}^{n} \frac{8i^2}{n^3} = \frac{8}{n^3}\sum_{i=1}^{n} i^2 ]
Step 5 — Use a known summation identity
This is where the algebra simplifies. The standard identity is:
[ \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6} ]
Substituting:
[ S_n = \frac{8}{n^3}\cdot\frac{n(n+1)(2n+1)}{6} = \frac{8(n+1)(2n+1)}{6n^2} ]
That single expression is the formula for the Riemann sum — a closed form in terms of (n) alone, with no summation sign left.
Step 6 — Take the limit to get the exact value
[ \lim_{n\to\infty} \frac{8(n+1)(2n+1)}{6n^2} = \frac{8 \cdot 2n^2}{6n^2} = \frac{16}{6} = \frac{8}{3} ]
Check it against the integral: (\int_0^2 x^2,dx = \left[\frac{x^3}{3}\right]_0^2 = \frac{8}{3}). The limit of the sums matches the integral exactly, which is the whole point of the construction.
The Summation Formulas You’ll Reuse
Almost every Riemann sum reduces to one of a handful of polynomial sum identities. Memorize these and the rest is arithmetic:
| Sum | Closed form |
|---|---|
| (\sum_{i=1}^{n} 1) | (n) |
| (\sum_{i=1}^{n} i) | (\frac{n(n+1)}{2}) |
| (\sum_{i=1}^{n} i^2) | (\frac{n(n+1)(2n+1)}{6}) |
| (\sum_{i=1}^{n} i^3) | (\left[\frac{n(n+1)}{2}\right]^2) |
A practical habit: once (\Delta x) is factored out of the sum, you’re left with a sum of powers of (i). Match it to the table, substitute, then simplify. MIT OpenCourseWare covers these identities in its single-variable calculus sequence precisely because they do the heavy lifting here.
Common Pitfalls to Avoid
A few mistakes trip up almost everyone the first time:
- Forgetting (\Delta x) multiplies the whole sum. The width applies to every rectangle, so factor it out but never drop it.
- Mixing up the index. Right endpoints use (i), left endpoints use (i-1). Off-by-one here changes your answer.
- Stopping at the sum. The formula (\frac{8(n+1)(2n+1)}{6n^2}) is the Riemann sum. The integral only appears once you take the limit.
- Over-expanding too early. Keep the sum in (\Sigma) form until you’ve matched an identity; expanding term by term invites arithmetic errors.
If you keep (\Delta x) separate and apply one identity at a time, the algebra stays manageable even for cubic functions.
Key Takeaways
- A Riemann sum formula has the form (\sum_{i=1}^{n} f(x_i),\Delta x), with (\Delta x = \frac{b-a}{n}).
- The sample point (x_i) differs by convention: left uses (a+(i-1)\Delta x), right uses (a+i,\Delta x), midpoint uses (a+(i-\tfrac{1}{2})\Delta x).
- Evaluating (f) at (x_i) and summing leaves you with a closed-form expression in (n) after applying a standard power-sum identity.
- Taking (\lim_{n\to\infty}) of that expression yields the exact definite integral (\int_a^b f(x),dx).
- All three endpoint conventions produce the same limit, so pick whichever makes the algebra easiest.
For the example (f(x)=x^2) on ([0,2]), the right-endpoint sum simplifies to (\frac{8(n+1)(2n+1)}{6n^2}), which approaches (\frac{8}{3}) — matching the integral exactly.
