Riemann Sum Calculator Formula
Understand the math behind the riemann sum calculator. Each variable explained with a worked example.
Formulas Used
Exact Val
exact_val = exp_val != -1 ? (1 / (exp_val + 1)) * (pow(b_val, exp_val + 1) - pow(a_val, exp_val + 1)) : 0Rect Width
rect_width = dxError Est
error_est = abs(b_val - a_val) * abs(pow(b_val, exp_val) - pow(a_val, exp_val)) / (2 * num_rect)Variables
| Variable | Description | Default |
|---|---|---|
exp_val | Exponent of x | 2 |
a_val | Left Bound | 0 |
b_val | Right Bound | 1 |
num_rect | Number of Rectangles | 10 |
dx | Derived value= (b_val - a_val) / num_rect | calculated |
How It Works
Riemann Sums
Concept
A Riemann sum approximates the area under a curve by dividing the interval into rectangles:
1. Divide [a, b] into n equal subintervals, each of width Δx = (b-a)/n 2. For each subinterval, evaluate the function at the left endpoint 3. Sum the areas: S = sum of f(xᵢ) × Δx
As n → infinity
The Riemann sum converges to the exact definite integral. More rectangles = better approximation.
Worked Example
Approximate the integral of x² from 0 to 1 with 10 rectangles.
- 01Δx = (1-0)/10 = 0.1
- 02Left endpoints: 0, 0.1, 0.2, ..., 0.9
- 03Sum = (0² + 0.1² + 0.2² + ... + 0.9²) × 0.1
- 04Exact value = 1³/3 = 0.333333...
Frequently Asked Questions
What is a Riemann sum?
A Riemann sum is a method for approximating the area under a curve by dividing it into rectangles and summing their areas. It is the foundation for the definite integral.
Do more rectangles give a better answer?
Yes, as the number of rectangles increases, the approximation gets closer to the exact value. In the limit as n → infinity, it equals the integral.
What are left, right, and midpoint sums?
Left sums evaluate at the left endpoint of each subinterval, right sums at the right, and midpoint sums at the center. Midpoint sums are generally the most accurate.
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