Riemann Sums

Riemann Sum is a certain kind of approximation of an integral by a finite sum. A Riemann sum is the sum of rectangles or trapezoids that approximate vertical slices of the area in question. German mathematician Bernhard Riemann developed the concept of Riemann Sums.

In this article, we will look into the Riemann sums, their approximation, sum notation, and solved examples in detail.

What is Riemann Sums?

Riemann sum is a method used for approximating the definite integral of a function over a given interval by dividing the interval into subintervals and then evaluating the function at specific points within each subinterval. It is named after the German mathematician Bernhard Riemann.

Riemann Approximation

Riemann's sums are a method for approximating the area under the curve. The intuition behind it is, that if we divide the area into very small rectangles, we can calculate the area of each rectangle and then add them to find the area of the total region. These sums can also be used to approximate and define the definite integrals.

Now let's start with dividing the given area into several rectangles, assuming the area is divided into 'n' rectangles of equal width. Notice that these rectangles do not cover the full area, so it is an approximation of the area. But as the number of rectangles increases, the approximation comes closer and closer to the actual area. 

In the definite integral notation, this area will be represented as, 

\int^{b}_{a}f(x)dx

This area can be approximated by dividing the area under the curve into n equally sized rectangles. So, the interval [a, b] is divided into n-subintervals defined by the points. 

a = x₀ < x₁ < x₂ < .... x_n-2< x_n-1 < x_n = b

Then, the n intervals are, 

[x₀, x₁], [x₁, x₂], .... [x_n-1, x_n]

So, for the i^th rectangle, the width will be, [x_i-1, x_i]. 

The area for i^th rectangle A_i = f(x_i)(x_{i —} x_i-1)

So, the total area will be \sum^{n}_{i = 1}A_{i}

This sum is called the Riemann sum. Since the height of the rectangle is determined by the right limit of the interval, this is called the right-Riemann sum. The figure below shows the left-Riemann sum.

Summation Notation of Riemann Sum

Steps given below should be followed to find the summation notation of the Riemann Integral.

• Step 1: Find out the width of each interval. Let's denote the width of interval with \Delta x

• Step 2: Let x_i denote the right-endpoint of the rectangle x_i = a + \Delta x.i

• Step 3: Define Area of Each Rectangle

• Step 4: Sum the Areas 

Let's say the goal is to calculate the area under the graph of the function f(x) = x³, the area will be calculated between the limits x = 0 to x = 4.

Divide the interval into four equal parts, the intervals will be [0, 1], [1, 2], [2, 3], and [3, 4]. 

Riemann sum then, can be written as follows, 

A(1) + A(2) + A(3) + A(4) = \sum^{4}_{i=1}A_{i}

Let's calculate the right sum Riemann sum. Assume x_i denotes the right endpoint of the i^th rectangle. 

So, the formula for x_i = i. Now, the value of the function at these points becomes, 

f(x_i) = (i)³

So, A(i) = (height)(width) = (i)³

Riemann sum becomes, 

 A(1) + A(2) + A(3) + A(4) = \sum^{4}_{i=1}A_{i}

⇒ A(1) + A(2) + A(3) + A(4) = \sum^{4}_{i=1}i^3

So, this way almost all the Riemann sums can be represented in a sigma notation. 

Let's work out some problems with these concepts. 

Articles Related to Riemann Sum:

• Trapezoidal Rule
• Riemann Sums in Summation Notation
• Definite Integral as the Limit of a Riemann Sum

Examples Using Riemann Sum Formula

Example 1: Choose which type of the Riemann integral is shown below in the figure. 

1. Left-Riemann Sum 
2. Right-Riemann Sum
3. Mid-point Riemann Sum 

Solution: 

Since the values of the intervals are decided according to the left-end point of the interval. This is a left-Riemann Sum 

Answer-(1).

Example 2: Calculate the Left-Riemann Sum for the function given in the figure above. 

Solution: 

Dividing the interval into four equal parts that is n = 4. The width of each interval will be, 

\Delta x = \frac{8 - 0}{n} = \frac{8 - 0}{4} = 2

x₀ = 0, x₁ = 1, x₂ = 2, x₃ = 0 and x₄ = 0

The value of the function in each interval will be the value of the function at the right-end of the interval. 

A = \sum^{n}_{i = 1}f(x_i)\Delta x

⇒A =  f(x_1)\Delta x + f(x_2)\Delta x + f(x_3)\Delta x + f(x_4)\Delta x

⇒A =  f(1)\Delta x + f(2)\Delta x + f(3)\Delta x + f(4)\Delta x

⇒A = f(1)(2) + f(2)(2)+ f(3)(2) + f(4)(2)

⇒A = (f(1) + f(2) + f(3)+ f(4))(2)

⇒A = (1 + 2 + 3+ 4)(2)

⇒A = (10)(2)

⇒A = 20

Example 3: Consider a function f(x) = 5 - 2x, its area is calculated from riemann sum from x = 0 to x = 4, the whole area is divided into 4 rectangles. Find the riemann sum in sigma notation

Solution: 

Step (i): Calculate the width

Whole length is divided into 4 equal parts, 

x_i = 0 and x_l = 4, 

Width of an interval is given by = \frac{x_{l} -x_{i}}{n}

Where x_i = initial point, and x_l - last point and n= number of parts 

n = 4 

\Delta x = \frac{x_l - x_i}{n} \\ = \frac{4 - 0}{4} \\ = 1

Step(ii): 

a = 0, 

x_i = 0 +  \Delta x i 

⇒ x_i = i

Step (iii): 

A_i = Height x Width 

    = f(x_i)  \Delta x 

    = (5 - 2i)(1)

    = 5 - 2i

Total Area = A(1) + A(2) + A(3) + A(4) + A(5)

= \sum^{5}_{i=1}5 - 2i

Example 4: Consider a function f(x) = √x, its area is calculated from riemann sum from x = 0 to x = 4, the whole area is divided into 4 rectangles. Find the riemann sum in sigma notation

Solution: 

Step (i): Calculate the width

Whole length is divided into 4 equal parts, 

x_i = 0 and x_l = 4, 

Width of an interval is given by = \frac{x_{l} -x_{i}}{n}

Where x_i = initial point, and x_l - last point and n= number of parts 

n = 4 

\Delta x = \frac{x_l - x_i}{n} \\ = \frac{4 - 0}{4} \\ = 1

Step(ii): 

a = 0, 

x_i = 0 +  \Delta x.i 

⇒ x_i = i

Step (iii) 

A_i = Height x Width 

    = f(x_i) \Delta x

    = (√i)(1)

    = √i

Total Area = A(1) + A(2) + A(3) + A(4) 

= \sum^{4}_{i=1}\sqrt{i}

Example 5: Consider a function f(x) = 3(x + 3), its area is calculated from Riemann sum from x = 0 to x = 6, and the whole area is divided into 6 rectangles. Find the Riemann sum in sigma notation

Solution: 

Step (i): Calculate the width

Whole length is divided into 4 equal parts, 

x_i = 0 and x_l = 6, 

Width of an interval is given by = \frac{x_{l} -x_{i}}{n}

Where x_i = initial point, and x_l - last point and n= number of parts 

n = 6 

\Delta x = \frac{x_l - x_i}{n} \\ = \frac{6 - 0}{6} \\ = 1

Step(ii): 

a = 0, 

x_i = 0 +  \Delta x   i 

⇒ x_i = i

Step (iii):

A_i = Height x Width 

    = f(x_i)  \Delta x 

    = (3(i + 3))(1)

    = 3(i + 3)

Total Area = A(1) + A(2) + A(3) + A(4) 

= \sum^{4}_{i=1}3(i + 3)