midpoint riemann sum

3 min read 24-10-2024

Midpoint Riemann Sums: A Precise Way to Approximate Area

Imagine you want to find the area under a curve. It's not always easy to find an exact solution, especially for complex functions. This is where Riemann sums come in handy, offering a way to approximate the area using rectangles. While there are different types of Riemann sums, the midpoint Riemann sum stands out for its often higher accuracy and ease of implementation.

What is a Midpoint Riemann Sum?

The midpoint Riemann sum is a method for approximating the definite integral of a function. It involves dividing the area under the curve into a series of equally-sized rectangles, where the height of each rectangle is determined by the function's value at the midpoint of its corresponding subinterval.

Let's break it down:

Divide the interval: Divide the interval over which you want to find the area (let's say it's from a to b) into n equal subintervals. Each subinterval will have a width of Δx = (b - a) / n.
Find the midpoints: For each subinterval, find the midpoint. You can calculate this by averaging the endpoints of the subinterval.
Evaluate the function: Evaluate the function at each midpoint you calculated. This will give you the height of the rectangle for each subinterval.
Calculate the areas: Multiply the height of each rectangle by the width (Δx) to find the area of each rectangle.
Sum the areas: Add up the areas of all the rectangles to obtain the approximate area under the curve.

Why Use the Midpoint Riemann Sum?

The midpoint Riemann sum offers several advantages over other methods, like the left or right Riemann sum:

Often More Accurate: The midpoint Riemann sum tends to be more accurate because the rectangle's height is determined at the midpoint of the subinterval, making it more likely to capture the average value of the function within that interval.
Easier to Implement: Finding the midpoint of each subinterval is relatively straightforward compared to other methods that might require more complex calculations.

Example: Calculating the Midpoint Riemann Sum

Let's take an example. Say we want to approximate the area under the curve of the function f(x) = x² from x = 0 to x = 2 using 4 subintervals.

Divide the interval: We divide the interval [0, 2] into 4 subintervals, each with a width of Δx = (2 - 0) / 4 = 0.5.
Find the midpoints: The midpoints of these subintervals are:
- Subinterval 1: (0 + 0.5) / 2 = 0.25
- Subinterval 2: (0.5 + 1) / 2 = 0.75
- Subinterval 3: (1 + 1.5) / 2 = 1.25
- Subinterval 4: (1.5 + 2) / 2 = 1.75
Evaluate the function: We evaluate f(x) at each midpoint:
- f(0.25) = 0.0625
- f(0.75) = 0.5625
- f(1.25) = 1.5625
- f(1.75) = 3.0625
Calculate the areas: We multiply the height of each rectangle by Δx:
- Area 1: 0.0625 * 0.5 = 0.03125
- Area 2: 0.5625 * 0.5 = 0.28125
- Area 3: 1.5625 * 0.5 = 0.78125
- Area 4: 3.0625 * 0.5 = 1.53125
Sum the areas: We add up the areas of all the rectangles:
- Approximate Area = 0.03125 + 0.28125 + 0.78125 + 1.53125 = 2.625

The midpoint Riemann sum provides an approximation of the area under the curve. As you increase the number of subintervals, the approximation becomes more accurate, approaching the exact value of the definite integral.