Riemann sum

October 4, 2024October 9, 2024by Kurious Fox

A Riemann sum is a method used in calculus to approximate the integral (or area under a curve) of a function. It is named after the German mathematician Bernhard Riemann. The basic idea behind a Riemann sum is to break up the region under a curve into small rectangles, compute the area of each rectangle, and then sum those areas to approximate the total area under the curve.

Steps to compute a Riemann sum:

Divide the interval:
Suppose you want to approximate the integral of a function $f(x)$ over the interval $[a, b]$ . You divide this interval into n subintervals (small segments) of equal width, $\Delta x = \frac{b - a}{n}$ .
Choose sample points:
For each subinterval, you select a sample point $x_i^*$ (usually in the middle, left, or right of the subinterval) to evaluate the function $f(x)$ at that point.
Compute the area of rectangles:
For each subinterval, the height of the rectangle is $f(x_i^*)$ and the width is $\Delta x$ . The area of the rectangle is then $f(x_i^*) \cdot \Delta x$ .
Sum the areas:
Add up the areas of all the rectangles to get an approximation of the total area under the curve:
$R_n = \sum_{i=1}^{n} f(x_i^*) \cdot \Delta x$
Where:

$n$ is the number of rectangles (subintervals).
$f(x_i^*)$ is the function evaluated at the sample point $x_i^*$ .
$\Delta x$ is the width of each subinterval.

Types of Riemann sums:

Left Riemann sum: The sample points $x_i^*$ are taken to be the left endpoints of each subinterval.
Right Riemann sum: The sample points $x_i^*$ are taken to be the right endpoints of each subinterval.
Midpoint Riemann sum: The sample points $x_i^*$ are taken to be the midpoints of each subinterval.

As the number of rectangles (n) increases:

The approximation becomes more accurate because the rectangles fit the curve more closely.
In the limit, as $n \to \infty$ , the Riemann sum approaches the exact value of the definite integral of $f(x)$ over the interval $[a, b]$ :
$\lim_{n \to \infty} R_n = \int_a^b f(x) \, dx$

Example:

Let’s approximate the integral $\int_0^1 x^2 \, dx$ using a Riemann sum with 4 subintervals ( $n = 4$ ):

Divide the interval $[0, 1]$ into 4 equal parts, so $\Delta x = \frac{1-0}{4} = 0.25$ .
For a left Riemann sum, the sample points are $x_0 = 0$ , $x_1 = 0.25$ , $x_2 = 0.5$ , $x_3 = 0.75$ .
Evaluate $f(x) = x^2$ at those points:

$f(0) = 0^2 = 0$
$f(0.25) = 0.25^2 = 0.0625$
$f(0.5) = 0.5^2 = 0.25$
$f(0.75) = 0.75^2 = 0.5625$

Multiply each value by $\Delta x = 0.25$ and sum them:
$R_4 = 0 \cdot 0.25 + 0.0625 \cdot 0.25 + 0.25 \cdot 0.25 + 0.5625 \cdot 0.25 = 0 + 0.015625 + 0.0625 + 0.140625 = 0.21875$

Thus, the Riemann sum approximation of the integral is $0.21875$ .

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