Riemann Sums in Integrals
Hello friends, how are you today? Hopefully well. In this article we will discuss Riemann Sums in Integrals which are directly related to the area of a region and definite integral forms. As the name suggests, Riemann was a German scientist born in Breselenz, a village near Dannenberg in the Kingdom of Hanover, Germany, with the full name George Friedrich Bernhard Riemann. One of his contributions that is still famous today is that he introduced the modern definition of definite integrals. To honor him, it is called Riemann Integral. In this writing, we will study a little of the science explained by Riemann. To make it easier to study this material, friends must first master sigma notation.
The Basic Idea of Riemann Sums
If we are asked to calculate the area of the shaded region above, how do we do it?
This is where the genius idea of Riemann comes out. Riemann's method is to approximate by dividing the shaded region into several rectangles, then summing all the areas of these rectangles as shown in the following image.
Definition of Riemann Sum
Given a function \( f(x) \) on the interval \([a, b]\), divide the interval into \( n \) subintervals of width \( \Delta x_i \). Choose a sample point \( x_i^* \) in each subinterval. The Riemann sum is:
Where:
- \( \Delta x_i \) = width of the i-th subinterval
- \( x_i^* \) = sample point in the i-th subinterval
- \( n \) = number of subintervals
Use the left endpoint of each subinterval as the sample point:
Use the right endpoint of each subinterval as the sample point:
Use the midpoint of each subinterval as the sample point:
Example Problems
Determine the Riemann sum of the function shown in the following graph.
Given: Function appears to be \( f(x) = x^2 - 4x + 3 \) on [0, 4] with 4 rectangles
Solution:
From the graph, we have 4 rectangles with different widths:
| Rectangle | Width (Δx) | Sample Point (xᵢ) | Height f(xᵢ) | Area |
|---|---|---|---|---|
| 1 | 0.7 | 0.5 | f(0.5) = (0.5)² - 4(0.5) + 3 = 1.25 | 0.7 × 1.25 = 0.875 |
| 2 | 1.0 | 1.5 | f(1.5) = (1.5)² - 4(1.5) + 3 = 0.75 | 1.0 × 0.75 = 0.75 |
| 3 | 1.0 | 2.0 | f(2) = (2)² - 4(2) + 3 = 1.0 | 1.0 × 1.0 = 1.0 |
| 4 | 1.3 | 3.5 | f(3.5) = (3.5)² - 4(3.5) + 3 = 1.25 | 1.3 × 1.25 = 1.625 |
Total Riemann sum = 0.875 + 0.75 + 1.0 + 1.625 = 4.25
Result: Riemann sum = 4.25
For \( f(x) = x \) on [0, 3], find Riemann sums using 6 equal subintervals with:
a) Right endpoints
b) Midpoints
c) Left endpoints
Solution: First, calculate \( \Delta x = \frac{3-0}{6} = 0.5 \)
a) Right endpoints:
Sample points: x₁ = 0.5, x₂ = 1.0, x₃ = 1.5, x₄ = 2.0, x₅ = 2.5, x₆ = 3.0
Heights: f(0.5)=0.5, f(1.0)=1.0, f(1.5)=1.5, f(2.0)=2.0, f(2.5)=2.5, f(3.0)=3.0
Sum of heights = 0.5 + 1.0 + 1.5 + 2.0 + 2.5 + 3.0 = 10.5
Riemann sum = 10.5 × 0.5 = 5.25
b) Midpoints:
Sample points: x₁ = 0.25, x₂ = 0.75, x₃ = 1.25, x₄ = 1.75, x₅ = 2.25, x₆ = 2.75
Heights: f(0.25)=0.25, f(0.75)=0.75, f(1.25)=1.25, f(1.75)=1.75, f(2.25)=2.25, f(2.75)=2.75
Sum of heights = 0.25 + 0.75 + 1.25 + 1.75 + 2.25 + 2.75 = 9.0
Riemann sum = 9.0 × 0.5 = 4.5
c) Left endpoints:
Sample points: x₁ = 0, x₂ = 0.5, x₃ = 1.0, x₄ = 1.5, x₅ = 2.0, x₆ = 2.5
Heights: f(0)=0, f(0.5)=0.5, f(1.0)=1.0, f(1.5)=1.5, f(2.0)=2.0, f(2.5)=2.5
Sum of heights = 0 + 0.5 + 1.0 + 1.5 + 2.0 + 2.5 = 7.5
Riemann sum = 7.5 × 0.5 = 3.75
Different sample point methods give different Riemann sums: Right endpoint (5.25) > Midpoint (4.5) > Left endpoint (3.75)
For \( f(x) = x^2 \) on [0, 3], find the Riemann sum using 6 equal subintervals with right endpoints.
Solution: \( \Delta x = \frac{3-0}{6} = 0.5 \)
| i | Subinterval | xᵢ (Right endpoint) | f(xᵢ) = xᵢ² | Area = f(xᵢ) × Δx |
|---|---|---|---|---|
| 1 | [0, 0.5] | 0.5 | 0.25 | 0.125 |
| 2 | [0.5, 1.0] | 1.0 | 1.0 | 0.5 |
| 3 | [1.0, 1.5] | 1.5 | 2.25 | 1.125 |
| 4 | [1.5, 2.0] | 2.0 | 4.0 | 2.0 |
| 5 | [2.0, 2.5] | 2.5 | 6.25 | 3.125 |
| 6 | [2.5, 3.0] | 3.0 | 9.0 | 4.5 |
Sum of areas = 0.125 + 0.5 + 1.125 + 2.0 + 3.125 + 4.5 = 11.375
Result: Riemann sum = 11.375
Definite Integral as a Limit of Riemann Sums
As we take more and more subintervals (n → ∞), the Riemann sum approaches the definite integral:
Where \( \Delta x = \frac{b-a}{n} \)
Express the limit as a definite integral: \[ \lim_{n \to \infty} \sum_{i=1}^n \frac{4i}{n} \cdot \frac{4}{n} \]
Solution:
Compare with the Riemann sum formula: \( \sum_{i=1}^n f(x_i) \Delta x \)
Here, \( \Delta x = \frac{4}{n} \), so \( b - a = 4 \)
Also, \( x_i = \frac{4i}{n} \)
So \( f(x_i) = \sqrt{x_i} \) (since \( \frac{4i}{n} = \sqrt{\frac{4i}{n}} \) is not correct - let's check: actually \( f(x_i) = \frac{4i}{n} \) and \( x_i = \frac{4i}{n} \), so \( f(x) = x \))
Wait, let's re-examine: The given sum is \( \sum \frac{4i}{n} \cdot \frac{4}{n} \)
So \( f(x_i) = \frac{4i}{n} \) and \( \Delta x = \frac{4}{n} \)
If \( x_i = \frac{4i}{n} \), then \( f(x_i) = x_i \), so \( f(x) = x \)
Also \( b - a = 4 \), and typically we start at a = 0, so b = 4
Thus: \[ \lim_{n \to \infty} \sum_{i=1}^n \frac{4i}{n} \cdot \frac{4}{n} = \int_0^4 x \, dx \]
Result: \( \int_0^4 x \, dx \)
Useful Sigma Notation Formulas
Calculate \( \int_0^3 x \, dx \) using the limit of Riemann sums.
Solution:
Divide [0, 3] into n equal subintervals: \( \Delta x = \frac{3}{n} \)
Using right endpoints: \( x_i = 0 + i\Delta x = \frac{3i}{n} \)
For \( f(x) = x \): \( f(x_i) = \frac{3i}{n} \)
Riemann sum: \[ \sum_{i=1}^n f(x_i) \Delta x = \sum_{i=1}^n \frac{3i}{n} \cdot \frac{3}{n} = \frac{9}{n^2} \sum_{i=1}^n i \]
Using the sigma formula: \( \sum_{i=1}^n i = \frac{n(n+1)}{2} \)
So: \[ \frac{9}{n^2} \cdot \frac{n(n+1)}{2} = \frac{9(n+1)}{2n} \]
Take the limit as n → ∞: \[ \lim_{n \to \infty} \frac{9(n+1)}{2n} = \lim_{n \to \infty} \frac{9}{2} \left(1 + \frac{1}{n}\right) = \frac{9}{2} \]
Result: \( \int_0^3 x \, dx = \frac{9}{2} = 4.5 \)
Notice this matches the midpoint Riemann sum with 6 subintervals from Example 2!
Key Concepts
• Left endpoint: Usually underestimates if f is increasing
• Right endpoint: Usually overestimates if f is increasing
• Midpoint: Often gives the best approximation
The definite integral \( \int_a^b f(x) dx \) is defined as the limit of Riemann sums:
This definition connects the geometric concept of area with the analytic concept of integration.
Georg Friedrich Bernhard Riemann (1826-1866) was a German mathematician who made profound contributions to analysis, number theory, and differential geometry. His work on integrals laid the foundation for modern integration theory. While we use the Fundamental Theorem of Calculus for most calculations today, understanding Riemann sums gives us insight into what an integral really represents: an infinite sum of infinitesimally small products.
For calculating definite integrals, we don't need to use Riemann sums like the examples above. We use the Fundamental Theorem of Calculus II, which makes it easier to compute all definite integral forms. However, understanding Riemann sums helps us understand what integrals really mean!